Days between dates

From Rosetta Code
Days between dates is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
Task

Calculate the number of days between two dates.

Date input should be of the form   YYYY-MM-DD.


Motivation

To demonstrate one of the numerous ways this can be done.

11l

F parse_date(date)
   R time:strptime(date, ‘%Y-%m-%d’)

V date1 = parse_date(‘2019-01-01’)
V date2 = parse_date(‘2019-09-30’)
print((date2 - date1).days())
Output:
272

Action!

INCLUDE "D2:REAL.ACT" ;from the Action! Tool Kit

TYPE Date=[CARD year BYTE month,day]

BYTE FUNC IsLeapYear(CARD y)
  IF y MOD 100=0 THEN
    IF y MOD 400=0 THEN RETURN (1)
    ELSE RETURN (0) FI
  FI
  IF y MOD 4=0 THEN RETURN (1) FI
RETURN (0)

PROC DecodeDate(CHAR ARRAY s Date POINTER d)
  BYTE ARRAY maxD=[0 31 28 31 30 31 30 31 31 30 31 30 31]
  CHAR ARRAY tmp(5)
  BYTE m

  IF s(0)#10 THEN
    Break()
  FI
  SCopyS(tmp,s,1,4) d.year=ValC(tmp)
  SCopyS(tmp,s,6,7) d.month=ValB(tmp)
  SCopyS(tmp,s,9,10) d.day=ValB(tmp)

  IF d.year=0 OR d.month=0 OR d.month>12 OR d.day=0 THEN
    Break()
  FI
  IF IsLeapYear(d.year) THEN
    maxD(2)=29
  ELSE
    maxD(2)=28
  FI
  m=d.month
  IF d.day>maxD(m) THEN
    Break()
  FI
RETURN

PROC DateToDays(Date POINTER d REAL POINTER res)
  BYTE m
  INT y,tmp1
  REAL tmp2

  m=(d.month+9) MOD 12
  y=d.year-m/10
  tmp1=y/4-y/100+y/400+(m*306+5)/10+(d.day-1)

  IntToReal(365,res)
  IntToReal(y,tmp2)
  RealMult(res,tmp2,res)
  IntToReal(tmp1,tmp2)
  RealAdd(res,tmp2,res)
RETURN

PROC DiffDays(Date POINTER d1,d2 REAL POINTER diff)
  REAL days1,days2

  DateToDays(d1,days1)
  DateToDays(d2,days2)
  RealSub(days2,days1,diff)
RETURN

PROC Test(CHAR ARRAY s1,s2)
  Date d1,d2
  REAL diff

  DecodeDate(s1,d1)
  DecodeDate(s2,d2)
  DiffDays(d1,d2,diff)
  PrintF("%S .. %S -> ",s1,s2)
  PrintRE(diff)
RETURN

PROC Main()
  Put(125) PutE() ;clear the screen
  Test("1995-11-21","1995-11-21")
  Test("2019-01-01","2019-01-02")
  Test("2019-01-02","2019-01-01")
  Test("2019-01-01","2019-03-01")
  Test("2020-01-01","2020-03-01")
  Test("1902-01-01","1968-12-25")
  Test("2090-01-01","2098-12-25")
  Test("1902-01-01","2098-12-25")
RETURN
Output:

Screenshot from Atari 8-bit computer

1995-11-21 .. 1995-11-21 -> 0
2019-01-01 .. 2019-01-02 -> 1
2019-01-02 .. 2019-01-01 -> -1
2019-01-01 .. 2019-03-01 -> 59
2020-01-01 .. 2020-03-01 -> 60
1902-01-01 .. 1968-12-25 -> 24465
2090-01-01 .. 2098-12-25 -> 3280
1902-01-01 .. 2098-12-25 -> 71947

Ada

with Ada.Calendar;
with Ada.Text_IO;
with Ada.Integer_Text_IO;

with GNAT.Calendar.Time_IO;

procedure Days_Between_Dates is

   function Days_Between (Lower_Date  : in Ada.Calendar.Time;
                          Higher_Date : in Ada.Calendar.Time) return Integer
   is
      use Ada.Calendar;
      Diff : constant Duration := Higher_Date - Lower_Date;
   begin
      return Integer (Diff / Day_Duration'Last);
   end Days_Between;

   procedure Put_Days_Between (Lower_Date  : in String;
                               Higher_Date : in String;
                               Comment     : in String)
   is
      use Ada.Text_IO;
      use Ada.Integer_Text_IO;
      use GNAT.Calendar.Time_IO;

      Diff : constant Integer := Days_Between (Lower_Date  => Value (Lower_Date),
                                               Higher_Date => Value (Higher_Date));
   begin
      Put ("Days between " & Lower_Date & " and " & Higher_Date & " is ");
      Put (Diff, Width => 5);
      Put (" days  --  ");
      Put (Comment);
      New_Line;
   end Put_Days_Between;

begin
   Put_Days_Between ("1995-11-21", "1995-11-21", "Identical dates");
   Put_Days_Between ("2019-01-01", "2019-01-02", "Positive span");
   Put_Days_Between ("2019-01-02", "2019-01-01", "Negative span");
   Put_Days_Between ("2019-01-01", "2019-03-01", "Non-leap year");
   Put_Days_Between ("2020-01-01", "2020-03-01", "Leap year");
   Put_Days_Between ("1902-01-01", "1968-12-25", "Past");
   Put_Days_Between ("2090-01-01", "2098-12-25", "Future");
   Put_Days_Between ("1902-01-01", "2098-12-25", "Long span");
end Days_Between_Dates;
Output:
Days between 1995-11-21 and 1995-11-21 is     0 days  --  Identical dates
Days between 2019-01-01 and 2019-01-02 is     1 days  --  Positive span
Days between 2019-01-02 and 2019-01-01 is    -1 days  --  Negative span
Days between 2019-01-01 and 2019-03-01 is    59 days  --  Non-leap year
Days between 2020-01-01 and 2020-03-01 is    60 days  --  Leap year
Days between 1902-01-01 and 1968-12-25 is 24465 days  --  Past
Days between 2090-01-01 and 2098-12-25 is  3280 days  --  Future
Days between 1902-01-01 and 2098-12-25 is 71947 days  --  Long span

ALGOL 68

Translation of: FreeBASIC
BEGIN # calculate the number of days between a pair of dates                 #
      # based on a translation of the FreeBASIC sample                       #

    [,]STRING test cases
        = ( ( "1902-01-01", "1968-12-25" )
          , ( "2019-01-01", "2019-01-02" ), ( "2019-01-02", "2019-01-01" )
          , ( "2019-01-01", "2019-03-01" ), ( "2020-01-01", "2020-03-01" )
          , ( "1995-11-21", "1995-11-21" ), ( "2090-01-01", "2098-12-25" )
          );

    PROC gregorian = ( INT y, m, d )INT:
         BEGIN
             INT n = ( m + 9 ) - ( ( ( m + 9 ) OVER 12 ) * 12 );
             INT w = y - ( n OVER 10 );
             ( 365 * w ) + ( w OVER 4 ) - ( w OVER 100 ) + ( w OVER 400 )
                         + ( ( ( n * 306 ) + 5 ) OVER 10 ) + ( d - 1 )
         END # gregorian # ;

    OP TOINT = ( STRING s )INT:
       BEGIN
           INT v := 0;
           FOR s pos FROM LWB s TO UPB s DO
               v *:= 10 +:= ( ABS s[ s pos ] - ABS "0" )
           OD;
           v
       END # TOINT #;

    FOR n FROM LWB test cases TO UPB test cases DO
        STRING from date = test cases[ n, 1 ];
        STRING to date   = test cases[ n, 2 ];
        INT from g = gregorian( TOINT from date[ 1 :  4 ]
                              , TOINT from date[ 6 :  7 ]
                              , TOINT from date[ 9 : 10 ]
                              );
        INT to g   = gregorian( TOINT   to date[ 1 :  4 ]
                              , TOINT   to date[ 6 :  7 ]
                              , TOINT   to date[ 9 : 10 ]
                              );
        print( ( "Days between ", from date, " and ", to date, " is " ) );
        print( ( whole( to g - from g, -5 ), " days", newline ) )
    OD
END
Output:
Days between 1902-01-01 and 1968-12-25 is 24465 days
Days between 2019-01-01 and 2019-01-02 is     1 days
Days between 2019-01-02 and 2019-01-01 is    -1 days
Days between 2019-01-01 and 2019-03-01 is    59 days
Days between 2020-01-01 and 2020-03-01 is    60 days
Days between 1995-11-21 and 1995-11-21 is     0 days
Days between 2090-01-01 and 2098-12-25 is  3280 days

ALGOL W

Translation of: FreeBASIC
begin % calculate the number of days between a pair of dates                 %
      % based on a translation of the FreeBASIC sample                       %

    integer procedure gregorian ( integer value y, m, d ) ;
    begin
        integer n, w;
        n := ( m + 9 ) - ( ( ( m + 9 ) div 12 ) * 12 );
        w := y - ( n div 10 );
        ( 365 * w ) + ( w div 4 ) - ( w div 100 ) + ( w div 400 )
                    + ( ( ( n * 306 ) + 5 ) div 10 ) + ( d - 1 )
    end gregorian ;

    integer procedure toInt( string(4) value s ) ;
    begin
        integer v;
        v := 0;
        for sPos := 0 until 3 do begin
            string(1) c;
            c := s( sPos // 1 );
            if c not = " " then v := ( v * 10 ) + ( decode( c ) - decode( "0" ) )
        end;
        v
    end toInt ;

    procedure testGregorian ( string(10) value fromDate, toDate ) ;
    begin
        integer fromG, toG;
        fromG := gregorian( toInt( fromDate( 0 // 4 ) )
                          , toInt( fromDate( 5 // 2 ) )
                          , toInt( fromDate( 8 // 2 ) )
                          );
        toG   := gregorian( toInt(   toDate( 0 // 4 ) )
                          , toInt(   toDate( 5 // 2 ) )
                          , toInt(   toDate( 8 // 2 ) )
                          );
        writeon( s_w := 0, "Days between ", fromDate, " and ", toDate, " is " );
        writeon( i_w := 5, s_w := 0, toG - fromG, " days" );
        write();
    end testGregorian ;

    testGregorian( "1902-01-01", "1968-12-25" );testGregorian( "2019-01-01", "2019-01-02" );
    testGregorian( "2019-01-02", "2019-01-01" );testGregorian( "2019-01-01", "2019-03-01" );
    testGregorian( "2020-01-01", "2020-03-01" );testGregorian( "1995-11-21", "1995-11-21" );
    testGregorian( "2090-01-01", "2098-12-25" )

end.
Output:
Days between 1902-01-01 and 1968-12-25 is 24465 days
Days between 2019-01-01 and 2019-01-02 is     1 days
Days between 2019-01-02 and 2019-01-01 is    -1 days
Days between 2019-01-01 and 2019-03-01 is    59 days
Days between 2020-01-01 and 2020-03-01 is    60 days
Days between 1995-11-21 and 1995-11-21 is     0 days
Days between 2090-01-01 and 2098-12-25 is  3280 days

AppleScript

on daysBetweenDates(date1, date2)
    considering numeric strings -- Allows for leading zeros having been omitted.
        if (date1 = date2) then return date1 & " and " & date2 & " are the same date"
    end considering
    
    -- Get the components of each date.
    set astid to AppleScript's text item delimiters
    set AppleScript's text item delimiters to "-"
    set {y1, m1, d1} to date1's text items
    set {y2, m2, d2} to date2's text items
    set AppleScript's text item delimiters to astid
    
    -- Derive AppleScript date objects.
    -- The best way to do this generally is to use the 'current date' function to obtain a date object
    -- and then to set the properties of the result (or of a copy) to the required values.
    -- The initial setting of the day values to 1 is to prevent overflow due to possibly
    -- incompatible day and month values during the sequential setting of the properties.
    -- The integer values set here are automatically coerced from the text values obtained above.
    tell (current date) to set {its day, its year, its month, its day, its time, firstDate} to {1, y1, m1, d1, 0, it}
    copy firstDate to secondDate
    tell secondDate to set {its day, its year, its month, its day} to {1, y2, m2, d2}
    
    -- Do the math.
    set daysDifference to (firstDate - secondDate) div days
    
    -- Format some output.
    if (daysDifference > 0) then
        return date1 & " comes " & daysDifference & " day" & item (((daysDifference is 1) as integer) + 1) of {"s", ""} & ¬
            " after " & date2
    else
        return date1 & " comes " & -daysDifference & " day" & item (((daysDifference is -1) as integer) + 1) of {"s", ""} & ¬
            " before " & date2
    end if
end daysBetweenDates

return daysBetweenDates("2020-04-11", "2001-01-01") & linefeed & ¬
    daysBetweenDates("2020-04-11", "2020-04-12") & linefeed & ¬
    daysBetweenDates("2020-04-11", "2020-04-11")
Output:
"2020-04-11 comes 7040 days after 2001-01-01
2020-04-11 comes 1 day before 2020-04-12
2020-04-11 and 2020-04-11 are the same date"


Or, composing a function from reusable generics, and drawing on NSISO8601DateFormatter:

use AppleScript version "2.4"
use framework "Foundation"
use scripting additions


-- daysBetween :: String -> String -> Int
on daysBetween(iso8601From, iso8601To)
    set midnight to "T00:00:00+00:00"
    ((dateFromISO8601(iso8601To & midnight) - ¬
        dateFromISO8601(iso8601From & midnight)) / 86400) as integer
end daysBetween


-- dateFromISO8601 :: String -> Date
on dateFromISO8601(isoDateString)
    set ca to current application
    tell ca's NSISO8601DateFormatter's alloc's init()
        set its formatOptions to ¬
            (ca's NSISO8601DateFormatWithInternetDateTime as integer)
        (its dateFromString:(isoDateString)) as date
    end tell
end dateFromISO8601


---------------------------TEST----------------------------
on run
    script test
        on |λ|(ab)
            set {a, b} to ab
            set delta to daysBetween(a, b)
            (a & " -> " & b & " -> " & ¬
                justifyRight(5, space, delta as string)) & ¬
                pluralize(delta, " day")
        end |λ|
    end script
    
    unlines(map(test, {¬
        {"2020-04-11", "2001-01-01"}, ¬
        {"2020-04-11", "2020-04-12"}, ¬
        {"2020-04-11", "2020-04-11"}, ¬
        {"2019-01-01", "2019-09-30"}}))
end run



--------------------GENERIC FUNCTIONS ---------------------

-- Absolute value.
-- abs :: Num -> Num
on abs(x)
    if 0 > x then
        -x
    else
        x
    end if
end abs


-- justifyRight :: Int -> Char -> String -> String
on justifyRight(n, cFiller, strText)
    if n > length of strText then
        text -n thru -1 of ((replicate(n, cFiller) as text) & strText)
    else
        strText
    end if
end justifyRight


-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
    -- The list obtained by applying f
    -- to each element of xs.
    tell mReturn(f)
        set lng to length of xs
        set lst to {}
        repeat with i from 1 to lng
            set end of lst to |λ|(item i of xs, i, xs)
        end repeat
        return lst
    end tell
end map


-- mReturn :: First-class m => (a -> b) -> m (a -> b)
on mReturn(f)
    -- 2nd class handler function lifted into 1st class script wrapper. 
    if script is class of f then
        f
    else
        script
            property |λ| : f
        end script
    end if
end mReturn


-- pluralize :: Int -> String -> String
on pluralize(n, s)
    set m to abs(n)
    if 0 = m or 2  m then
        s & "s"
    else
        s
    end if
end pluralize


-- replicate :: Int -> String -> String
on replicate(n, s)
    set out to ""
    if n < 1 then return out
    set dbl to s
    
    repeat while (n > 1)
        if (n mod 2) > 0 then set out to out & dbl
        set n to (n div 2)
        set dbl to (dbl & dbl)
    end repeat
    return out & dbl
end replicate


-- unlines :: [String] -> String
on unlines(xs)
    -- A single string formed by the intercalation
    -- of a list of strings with the newline character.
    set {dlm, my text item delimiters} to ¬
        {my text item delimiters, linefeed}
    set str to xs as text
    set my text item delimiters to dlm
    str
end unlines
Output:
2020-04-11 -> 2001-01-01 -> -7040 days
2020-04-11 -> 2020-04-12 ->     1 day
2020-04-11 -> 2020-04-11 ->     0 days
2019-01-01 -> 2019-09-30 ->   272 days

Arturo

Translation of: Ruby
daysBetweenDates: function [startDate, endDate][
    a: to :date.format: "dd/MM/yyyy" startDate
    b: to :date.format: "dd/MM/yyyy" endDate

    return abs b\days - a\days
]

print [
    "days between the two dates:" 
    daysBetweenDates "01/01/2019" "19/10/2019"
]
Output:
days between the two dates: 291

AutoHotkey

db =
(
1995-11-21|1995-11-21
2019-01-01|2019-01-02
2019-01-02|2019-01-01
2019-01-01|2019-03-01
2020-01-01|2020-03-01
1902-01-01|1968-12-25
2090-01-01|2098-12-25
1902-01-01|2098-12-25
)

for i, line in StrSplit(db, "`n", "`r"){
    D := StrSplit(line, "|")
    result .= "Days between " D.1 " and " D.2 "  :  " Days_between(D.1, D.2) " Day(s)`n"
}
MsgBox, 262144, , % result
return

Days_between(D1, D2){
    D1 := RegExReplace(D1, "\D")
    D2 := RegExReplace(D2, "\D")
    EnvSub, D2, % D1, days
    return D2
}
Output:
Days between 1995-11-21 and 1995-11-21  :  0 Day(s)
Days between 2019-01-01 and 2019-01-02  :  1 Day(s)
Days between 2019-01-02 and 2019-01-01  :  -1 Day(s)
Days between 2019-01-01 and 2019-03-01  :  59 Day(s)
Days between 2020-01-01 and 2020-03-01  :  60 Day(s)
Days between 1902-01-01 and 1968-12-25  :  24465 Day(s)
Days between 2090-01-01 and 2098-12-25  :  3280 Day(s)
Days between 1902-01-01 and 2098-12-25  :  71947 Day(s)

AWK

# syntax: GAWK -f DAYS_BETWEEN_DATES.AWK
BEGIN {
    regexp = "^....-..-..$" # YYYY-MM-DD
    main("1969-12-31","1970-01-01","builtin has bad POSIX start date")
    main("1970-01-01","2038-01-19","builtin has bad POSIX stop date")
    main("1970-01-01","2019-10-02","format OK")
    main("1970-01-01","2019/10/02","format NG")
    main("1995-11-21","1995-11-21","identical dates")
    main("2019-01-01","2019-01-02","positive date")
    main("2019-01-02","2019-01-01","negative date")
    main("2019-01-01","2019-03-01","non-leap year")
    main("2020-01-01","2020-03-01","leap year")
    exit(0)
}
function main(date1,date2,comment,  d1,d2,diff) {
    printf("\t%s\n",comment)
    d1 = days_builtin(date1)
    d2 = days_builtin(date2)
    diff = (d1 == "" || d2 == "") ? "error" : d2-d1
    printf("builtin %10s to %10s = %s\n",date1,date2,diff)
    d1 = days_generic(date1)
    d2 = days_generic(date2)
    diff = (d1 == "" || d2 == "") ? "error" : d2-d1
    printf("generic %10s to %10s = %s\n",date1,date2,diff)
}
function days_builtin(ymd) { # use gawk builtin
    if (ymd !~ regexp) { return("") }
    if (ymd < "1970-01-01" || ymd > "2038-01-18") { return("") } # outside POSIX range
    gsub(/-/," ",ymd)
    return(int(mktime(sprintf("%s 0 0 0",ymd)) / (60*60*24)))
}
function days_generic(ymd,  d,m,y,result) { # use Python formula
    if (ymd !~ regexp) { return("") }
    y = substr(ymd,1,4)
    m = substr(ymd,6,2)
    d = substr(ymd,9,2)
    m = (m + 9) % 12
    y = int(y - int(m/10))
    result = 365*y + int(y/4) - int(y/100) + int(y/400) + int((m*306+5)/10) + (d-1)
    return(result)
}
Output:
    builtin has bad POSIX start date
builtin 1969-12-31 to 1970-01-01 = error
generic 1969-12-31 to 1970-01-01 = 1
    builtin has bad POSIX stop date
builtin 1970-01-01 to 2038-01-19 = error
generic 1970-01-01 to 2038-01-19 = 24855
    format OK
builtin 1970-01-01 to 2019-10-02 = 18171
generic 1970-01-01 to 2019-10-02 = 18171
    format NG
builtin 1970-01-01 to 2019/10/02 = error
generic 1970-01-01 to 2019/10/02 = error
    identical dates
builtin 1995-11-21 to 1995-11-21 = 0
generic 1995-11-21 to 1995-11-21 = 0
    positive date
builtin 2019-01-01 to 2019-01-02 = 1
generic 2019-01-01 to 2019-01-02 = 1
    negative date
builtin 2019-01-02 to 2019-01-01 = -1
generic 2019-01-02 to 2019-01-01 = -1
    non-leap year
builtin 2019-01-01 to 2019-03-01 = 59
generic 2019-01-01 to 2019-03-01 = 59
    leap year
builtin 2020-01-01 to 2020-03-01 = 60
generic 2020-01-01 to 2020-03-01 = 60

C

#include<stdbool.h>
#include<string.h>
#include<stdio.h>

typedef struct{
    int year, month, day;
}date;

date extractDate(char* str){
    return (date){.year = 1000 * (str[0]-'0') + 100 * (str[1]-'0') + 10 * (str[2]-'0') + (str[3]-'0'), .month = 10*(str[5]-'0') + (str[6]-'0'), .day = 10*(str[8]-'0') + (str[9]-'0')};
}

bool isValidDate(char* str){
    date newDate;
    
    if(strlen(str)!=10 && str[4]!='-' && str[7]!='-'){
        return false;
    }

    newDate = extractDate(str);

    if(newDate.year<=0 || newDate.month<=0 || newDate.day<=0 || newDate.month>12 || (newDate.month==2 && newDate.day>29) || 
      ((newDate.month==1 || newDate.month==3 || newDate.month==5 || newDate.month==7 || newDate.month==8 || newDate.month==10 || newDate.month==12) && newDate.day>31) || 
      newDate.day>30 || (newDate.year%4==0 && newDate.month==2 && newDate.month>28)){
          return false;
      }

    return true;
}

int diffDays(date date1,date date2){
    int days1, days2;
    
    date1.month = (date1.month + 9)%12;
    date1.year = date1.year - date1.month/10;

    date2.month = (date2.month + 9)%12;
    date2.year = date2.year - date2.month/10;

    days1 = 365*date1.year + date1.year/4 - date1.year/100 + date1.year/400 + (date1.month*306 + 5)/10 + ( date1.day - 1 );
    days2 = 365*date2.year + date2.year/4 - date2.year/100 + date2.year/400 + (date2.month*306 + 5)/10 + ( date2.day - 1 );

    return days2 - days1;
}

int main(int argc,char** argv)
{
    if(argc!=3){
        return printf("Usage : %s <yyyy-mm-dd> <yyyy-mm-dd>",argv[0]);
    }

    if(isValidDate(argv[1])&&isValidDate(argv[2]) == false){
        return printf("Dates are invalid.\n");
    }

    printf("Days Difference : %d\n", diffDays(extractDate(argv[1]),extractDate(argv[2])));

    return 0;
}

Output :

abhishek_ghosh@Azure:~/doodles$ ./a.out 2019-01-01 2019-12-02
Days Difference : 335

BQN

The date namespace is taken from bqn-libs. DBw is the final function which calculates the proper difference between two date strings.

DivMod  ÷˜  |
date  {
  o  719469
  y  365.25
  dur  (100×y)-0.75,    y, 30.6
  off  o-0.25,       0.75, 0.41

  From  {
    ymd  𝕩
    f  0 > m - 3
    (d-o) +´  off +(¯1) dur × (100 DivMod y-f)  m+12×f
  }
  To  {
    t𝕩
    cym  dur { dm  𝕨 DivMod 𝕩+t  tm  d }¨ off
    m - 12×10m
    (100×c)+y+m<0, 3+m, 1+t
  }
}
Split(⊢-˜+`׬)=⊔⊢
ToI10×+˜´⌽-'0'

S2DToI¨ '-'Split 
DBw  -(date.From S2D)
  "2019-09-30" DBw "2019-01-01"
272


C++

Translation of: c
#include <iomanip>
#include <iostream>

class Date {
private:
    int year, month, day;

public:
    Date(std::string str) {
        if (isValidDate(str)) {
            year = atoi(&str[0]);
            month = atoi(&str[5]);
            day = atoi(&str[8]);
        } else {
            throw std::exception("Invalid date");
        }
    }

    int getYear() {
        return year;
    }

    int getMonth() {
        return month;
    }

    int getDay() {
        return day;
    }

    // YYYY-MM-DD
    static bool isValidDate(std::string str) {
        if (str.length() != 10 || str[4] != '-' || str[7] != '-') {
            return false;
        }

        if (!isdigit(str[0]) || !isdigit(str[1]) || !isdigit(str[2]) || !isdigit(str[3])) {
            return false; // the year is not valid
        }
        if (!isdigit(str[5]) || !isdigit(str[6])) {
            return false; // the month is not valid
        }
        if (!isdigit(str[8]) || !isdigit(str[9])) {
            return false; // the day is not valid
        }

        int year = atoi(&str[0]);
        int month = atoi(&str[5]);
        int day = atoi(&str[8]);

        // quick checks
        if (year <= 0 || month <= 0 || day <= 0) {
            return false;
        }
        if (month > 12) {
            return false;
        }

        switch (month) {
        case 2:
            if (day > 29) {
                return false;
            }
            if (!isLeapYear(year) && day == 29) {
                return false;
            }
            break;
        case 1:
        case 3:
        case 5:
        case 7:
        case 8:
        case 10:
        case 12:
            if (day > 31) {
                return false;
            }
            break;
        default:
            if (day > 30) {
                return false;
            }
            break;
        }

        return true;
    }

    static bool isLeapYear(int year) {
        if (year > 1582) {
            return ((year % 4 == 0) && (year % 100 > 0))
                || (year % 400 == 0);
        }
        if (year > 10) {
            return year % 4 == 0;
        }
        // not bothering with earlier leap years
        return false;
    }

    friend std::ostream &operator<<(std::ostream &, Date &);
};

std::ostream &operator<<(std::ostream &os, Date &d) {
    os << std::setfill('0') << std::setw(4) << d.year << '-';
    os << std::setfill('0') << std::setw(2) << d.month << '-';
    os << std::setfill('0') << std::setw(2) << d.day;
    return os;
}

int diffDays(Date date1, Date date2) {
    int d1m = (date1.getMonth() + 9) % 12;
    int d1y = date1.getYear() - d1m / 10;

    int d2m = (date2.getMonth() + 9) % 12;
    int d2y = date2.getYear() - d2m / 10;

    int days1 = 365 * d1y + d1y / 4 - d1y / 100 + d1y / 400 + (d1m * 306 + 5) / 10 + (date1.getDay() - 1);
    int days2 = 365 * d2y + d2y / 4 - d2y / 100 + d2y / 400 + (d2m * 306 + 5) / 10 + (date2.getDay() - 1);

    return days2 - days1;
}

int main() {
    std::string ds1 = "2019-01-01";
    std::string ds2 = "2019-12-02";

    if (Date::isValidDate(ds1) && Date::isValidDate(ds2)) {
        Date d1(ds1);
        Date d2(ds2);
        std::cout << "Days difference : " << diffDays(d1, d2);
    } else {
        std::cout << "Dates are invalid.\n";
    }

    return 0;
}
Output:
Days difference : 335

Alternative using Boost

Library: Boost
#include <cstdlib>
#include <iostream>
#include <boost/date_time/gregorian/gregorian.hpp>

int main(int argc, char** argv) {
    using namespace boost::gregorian;
    if (argc != 3) {
        std::cerr << "usage: " << argv[0] << " start-date end-date\n";
        return EXIT_FAILURE;
    }
    try {
        date start_date(from_simple_string(argv[1]));
        date end_date(from_simple_string(argv[2]));
        std::cout << end_date - start_date << '\n';
    } catch (const std::exception& ex) {
        std::cerr << ex.what() << '\n';
        return EXIT_FAILURE;
    }
    return EXIT_SUCCESS;
}
Output:
./days_between_dates 2020-01-01 2020-09-06
249

C#

using System;
using System.Globalization;

public class Program
{
    public static void Main() => WriteLine(DateDiff("1970-01-01", "2019-10-18"));

    public static int DateDiff(string d1, string d2) {
        var a = DateTime.ParseExact(d1, "yyyy-MM-dd", CultureInfo.InvariantCulture);
        var b = DateTime.ParseExact(d2, "yyyy-MM-dd", CultureInfo.InvariantCulture);
        return (int)(b - a).TotalDays;
    }
}
Output:
18187

COBOL

Works with: GnuCOBOL
COBOL *> days-between
      *> Tectonics: cobc -xj days-between.cob

       identification division.
       program-id. days-between.

       procedure division.
       compute tally = 
         function integer-of-formatted-date('YYYY-MM-DD', '2019-11-24')
         -
         function integer-of-formatted-date('YYYY-MM-DD', '2000-01-01')
       display tally

       compute tally = 
         function integer-of-formatted-date('YYYYMMDD', '20191124')
         -
         function integer-of-formatted-date('YYYYMMDD', '20000101')
       display tally

       goback.
       end program days-between.
Output:
prompt$ cobc -xj days-between-dates.cob
07267
07267


Commodore BASIC

100 REM ===============================
110 REM DAYS BETWEEN 2 DATES
120 REM
130 REM CONVERT FROM TEXT TO PRG
140 REM USING C64LIST.EXE V3.5
150 REM COMMAND LINE UTILITY ON WINDOWS
160 REM
170 REM V1, 2021-10-09, ALVALONGO
180 REM ===============================
190 REM
200 REM INSPIRED BY THE PYTHON VERSION
210 REM OF THE ALGORITHM
220 REM AND THE DISCUSSION AT
230 REM HTTPS://STACKOVERFLOW.COM/QUESTIONS/12862226
240 REM /THE-IMPLEMENTATION-OF-CALCULATING-THE-NUMBER-OF-DAYS-BETWEEN-2-DATES
250 REM ===============================
260 :
1000 REM INIT ============================
1010 PRINT CHR$(147);:REM CLEAR SCREEN
1020 PRINT CHR$(5);:REM INK WHITE COLOR
1030 POKE 53280,3:REM BORDER COLOR CYAN
1040 POKE 53281,14:REM BACKGROUND COLOR BLUE
1050 :
1100 REM MAIN ==========================
1110 PRINT "          DAYS BETWEEN 2 DATES"
1120 PRINT
1130 INPUT "FIRST DATE, YEAR";Y1
1140 INPUT "MONTH=";M1
1150 INPUT "DAY  =";D1
1160 INPUT "SECOND DATE, YEAR";Y2
1170 INPUT "MONTH=";M2
1180 INPUT "DAY  =";D2
1190 Y=Y1:M=M1:D=D1:GOSUB 9000:G1=G
1200 Y=Y2:M=M2:D=D2:GOSUB 9000:G2=G
1210 DI=ABS(G2-G1)
1220 PRINT "DAYS=";DI
1230 GET K$:IF K$="" THEN 1230
1240 PRINT
1250 GOTO 1120
1260 END
1270 :
9000 REM GREGORIAN ======================
9010 REM TRANSFORM A DATE INTO A DAY NUMBER IN THE GREGORIAN CALENDAR
9020 REM INPUT PARAMETERS: Y IS YEAR
9030 REM M IS MONTH
9040 REM D IS DAY
9050 N=(M+9)-INT((M+9)/12)*12
9060 W=Y-INT(N/10)
9070 G=365*W+INT(W/4)-INT(W/100)+INT(W/400)
9080 G=G+INT((N*306+5)/10)+(D-1)
9090 RETURN



D

import std.datetime.date;
import std.stdio;

void main() {
    auto fromDate = Date.fromISOExtString("2019-01-01");
    auto toDate = Date.fromISOExtString("2019-10-07");
    auto diff = toDate - fromDate;
    writeln("Number of days between ", fromDate, " and ", toDate, ": ", diff.total!"days");
}
Output:
Number of days between 2019-Jan-01 and 2019-Oct-07: 279

Delphi

Translation of: C#
program Days_between_dates;

{$APPTYPE CONSOLE}

uses
  System.SysUtils;

function CreateFormat(fmt: string; Delimiter: Char): TFormatSettings;
begin
  Result := TFormatSettings.Create();
  with Result do
  begin
    DateSeparator := Delimiter;
    ShortDateFormat := fmt;
  end;
end;

function DaysBetween(Date1, Date2: string): Integer;
var
  dt1, dt2: TDateTime;
  fmt: TFormatSettings;
begin
  fmt := CreateFormat('yyyy-mm-dd', '-');
  dt1 := StrToDate(Date1, fmt);
  dt2 := StrToDate(Date2, fmt);
  Result := Trunc(dt2 - dt1);
end;

begin
  Writeln(DaysBetween('1970-01-01', '2019-10-18'));
  readln;
end.
Output:
18187

Erlang

-module(daysbetween).
-export([between/2,dateToInts/2]).

% Return Year or Month or Date from datestring
dateToInts(String, POS) ->
  list_to_integer( lists:nth( POS, string:tokens(String, "-") ) ).

% Alternative form of above
% dateToInts(String,POS) ->
%   list_to_integer( lists:nth( POS, re:split(String ,"-", [{return,list},trim]) ) ).

% Return the number of days between dates formatted "2019-09-30"
between(DateOne,DateTwo) ->
  L = [1,2,3],
  [Y1,M1,D1] =  [ dateToInts(DateOne,X) || X <- L],
  [Y2,M2,D2] =  [ dateToInts(DateTwo,X) || X <- L],
  GregOne = calendar:date_to_gregorian_days(Y1,M1,D1),
  GregTwo = calendar:date_to_gregorian_days(Y2,M2,D2),
  GregTwo - GregOne.
Output:

erlang shell:

30> c(daysbetween).
c(daysbetween).
{ok,daysbetween}
31> daysbetween:between("2019-01-01", "2019-09-30").
daysbetween:between("2019-01-01", "2019-09-30").
272

Excel

LAMBDA

Excel dates are numbers counting days since the start of January 1900, and arithmetic operations can be applied to them directly.

(Unless entered as strings, they differ from other numbers only in the display format applied to the containing cell).

The lambda expression below maps over an array of date strings, defining a new array of the differences, in units of one day, between today's date and those earlier dates.

(See LAMBDA: The ultimate Excel worksheet function)

Output:

The formula in cell B2 defines the array of day counts which populate the range B2:B5

fx =LAMBDA(dateString, TODAY() - DATEVALUE(dateString) )(C2#)
A B C D
1 Difference in days Earlier date Today
2 39020 1914-07-28 2021-05-27
3 37088 1919-11-11
4 29916 1939-07-01
5 27661 1945-09-02

Alternatively, we can express the same mapping over the whole C2# array by directly writing:

fx = TODAY() - DATEVALUE(C2#)
A B C D
1 Difference in days Earlier date Today
2 39020 1914-07-28 2021-05-27
3 37088 1919-11-11
4 29916 1939-07-01
5 27661 1945-09-02

F#

// Days between dates: Nigel Galloway. June 3rd., 2021
let n,g=System.DateTime.Parse("1792-9-22"),System.DateTime.Parse("1805-12-31")
printfn "There are %d days between %d-%d-%d and %d-%d-%d" (g-n).Days n.Year n.Month n.Day g.Year g.Month g.Day
Output:
There are 4847 days between 1792-9-22 and 1805-12-31

Factor

Factor supports the addition and subtraction of timestamps and durations with the time+ and time- words.

USING: calendar calendar.parser kernel math prettyprint ;

: days-between ( ymd-str ymd-str -- n )
    [ ymd>timestamp ] bi@ time- duration>days abs ;

"2019-01-01" "2019-09-30" days-between .
"2016-01-01" "2016-09-30" days-between .  ! leap year
Output:
272
273


FreeBASIC

Dim Shared As Integer M, Y, D
Dim As Integer Y1, M1, D1, Y2, M2, D2, G1, G2
Dim As String DaysBetween(7, 2) = {{"1902-01-01","1968-12-25"}, _
{"2019-01-01","2019-01-02"}, {"2019-01-02","2019-01-01"}, _
{"2019-01-01","2019-03-01"}, {"2020-01-01","2020-03-01"}, _
{"1995-11-21","1995-11-21"}, {"2090-01-01","2098-12-25"}}

Function Gregorian() As Integer
    Dim As Integer N = (M+9) - Int((M+9)/12) * 12
    Dim As Integer W = Y - Int(N/10)
    Dim As Integer G = 365 * W + Int(W/4) - Int(W/100) + Int(W/400)
    G += Int((N*306+5)/10)+(D-1)
    Return G
End Function

For n As Integer = 0 To Ubound(DaysBetween)-1
    Y1 = Val(Left(DaysBetween(n,0),4))
    M1 = Val(Mid(DaysBetween(n,0),6,2))
    D1 = Val(Right(DaysBetween(n,0),2))
    Y2 = Val(Mid(DaysBetween(n,1),1,4))
    M2 = Val(Mid(DaysBetween(n,1),6,2))
    D2 = Val(Mid(DaysBetween(n,1),9,2))
    Y = Y1 : M = M1 : D = D1 : G1 = Gregorian
    Y = Y2 : M = M2 : D = D2 : G2 = Gregorian
    Print "Days between "; DaysBetween(n,0); " and "; DaysBetween(n,1); " is "; 
    Print Using "##### days"; (G2-G1)
Next n
Sleep
Output:
Days between 1902-01-01 and 1968-12-25 is 24465 days
Days between 2019-01-01 and 2019-01-02 is     1 days
Days between 2019-01-02 and 2019-01-01 is    -1 days
Days between 2019-01-01 and 2019-03-01 is    59 days
Days between 2020-01-01 and 2020-03-01 is    60 days
Days between 1995-11-21 and 1995-11-21 is     0 days
Days between 2090-01-01 and 2098-12-25 is  3280 days


Frink

Frink handles dates and intervals between dates robustly, including time zones. The sample below could use minutes or seconds instead of days.

Frink can even track leap seconds as they are added to the calendar if desired. The code d1-d2 in the sample below could be replaced by subtractLeap[d1,d2] to obtain exact counts of seconds between dates.

See Frink's Date/Time Handling documentation to see how easy it is to work with dates, times, timezones, calendrical systems, and even leap seconds correctly and easily.

d1 = # 2020-12-25 #
d2 = # 2020-12-06 #
println[d1-d2 -> days]
Output:
19

Fōrmulæ

Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for storage and transfer purposes more than visualization and edition.

Programs in Fōrmulæ are created/edited online in its website.

In this page you can see and run the program(s) related to this task and their results. You can also change either the programs or the parameters they are called with, for experimentation, but remember that these programs were created with the main purpose of showing a clear solution of the task, and they generally lack any kind of validation.

Solution

Note. For this script, the time zone is intentionally set to America/Los_Angeles, because it observes daylight saving time, It is necesary to solve this exercise.

Provided that the ToNumber expression applied to a time expression reduces to the number of milliseconds of such that time from the epoch:

The solution seems easy, calculating the difference between two times (in milliseconds), and dividing by 86,4000,000 (number of milliseconds in a day):

However, it does not work if one time is in daylight saving time, and the other one is in standard time:

Solution 1

The first solution consists in simply rounding the result to the nearest integer:

Solution 2

The expression GetTimeZoneOffset reduces to the offset (in minutes) of the given time.

The solution consist in taking this difference in account.

So, the function that works correctly is:

Test cases

Notice that it works even for fractions of days:

Go

package main

import (
    "fmt"
    "log"
    "time"
)

const layout = "2006-01-02" // template for time.Parse

// Parameters assumed to be in YYYY-MM-DD format.
func daysBetween(date1, date2 string) int {
    t1, err := time.Parse(layout, date1)
    check(err)
    t2, err := time.Parse(layout, date2)
    check(err)
    days := int(t1.Sub(t2).Hours() / 24)
    if days < 0 {
        days = -days
    }
    return days
}
    
func check(err error) {
    if err != nil {
        log.Fatal(err)
    }
}

func main() {
    date1, date2 := "2019-01-01", "2019-09-30"
    days := daysBetween(date1, date2)
    fmt.Printf("There are %d days between %s and %s\n", days, date1, date2)

    date1, date2 = "2015-12-31", "2016-09-30"
    days = daysBetween(date1, date2)
    fmt.Printf("There are %d days between %s and %s\n", days, date1, date2)
}
Output:
There are 272 days between 2019-01-01 and 2019-09-30
There are 274 days between 2015-12-31 and 2016-09-30

Groovy

Translation of: Kotlin
Translation of: Java
import java.time.LocalDate

def fromDate = LocalDate.parse("2019-01-01")
def toDate = LocalDate.parse("2019-10-19")
def diff = fromDate - toDate
println "Number of days between ${fromDate} and ${toDate}: ${diff}"
Output:
Number of days between 2019-01-01 and 2019-10-19: 291

Haskell

import Data.Time (Day)
import Data.Time.Calendar (diffDays)
import Data.Time.Format (parseTimeM,defaultTimeLocale)

main = do
    putStrLn $ task "2019-01-01" "2019-09-30"
    putStrLn $ task "2015-12-31" "2016-09-30"

task :: String -> String -> String
task xs ys =  "There are " ++ (show $ betweenDays xs ys) ++ " days between " ++ xs ++ " and " ++ ys ++ "."

betweenDays :: String -> String -> Integer
betweenDays date1 date2 = go (stringToDay date1) (stringToDay date2) 
    where
    go (Just x) (Just y) = diffDays y x
    go Nothing _ = error "Exception: Bad format first date"
    go _ Nothing = error "Exception: Bad format second date"

stringToDay :: String -> Maybe Day
stringToDay date = parseTimeM True defaultTimeLocale "%Y-%-m-%-d" date
Output:
There are 272 days between 2019-01-01 and 2019-09-30.
There are 274 days between 2015-12-31 and 2016-09-30.

Or, composing rather than raising errors:

import Data.Time (Day)
import Data.Time.Calendar (diffDays)
import Data.Time.Format (defaultTimeLocale, parseTimeM)

-------------------- DAYS BETWEEN DATES ------------------

daysBetween :: String -> String -> Maybe Integer
daysBetween s1 s2 =
  dayFromString s2
    >>= \d2 -> diffDays d2 <$> dayFromString s1

dayFromString :: String -> Maybe Day
dayFromString =
  parseTimeM
    True
    defaultTimeLocale
    "%Y-%-m-%-d"

--------------------------- TEST -------------------------
main :: IO ()
main =
  mapM_
    (putStrLn . uncurry showDateDiff)
    [ ("2019-01-01", "2019-09-30"),
      ("2015-12-31", "2016-09-30")
    ]

showDateDiff :: String -> String -> String
showDateDiff s1 s2 =
  maybe
    (unlines ["Unparseable as date string pair:", s1, s2])
    ( \n ->
        concat
          [ "There are ",
            show n,
            " days between ",
            s1,
            " and ",
            s2,
            "."
          ]
    )
    $ daysBetween s1 s2
Output:
There are 272 days between 2019-01-01 and 2019-09-30.
There are 274 days between 2015-12-31 and 2016-09-30.

J

   require'~addons/types/datetime/datetime.ijs'
   2021 03 16 daysDiff 1960 05 04
22231

   vectorize=: [: evaluate hyphen_to_space
   hyphen_to_space=: (=&'-')`(,:&' ')}
   evaluate=: ".
   but_first=: &


   '2021-03-16' daysDiff but_first vectorize  '1960-05-04'
22231

Java

Translation of: Kotlin
import java.time.LocalDate;
import java.time.temporal.ChronoUnit;

public class DaysBetweenDates {
    public static void main(String[] args) {
        LocalDate fromDate = LocalDate.parse("2019-01-01");
        LocalDate toDate = LocalDate.parse("2019-10-19");
        long diff = ChronoUnit.DAYS.between(fromDate, toDate);
        System.out.printf("Number of days between %s and %s: %d\n", fromDate, toDate, diff);
    }
}
Output:
Number of days between 2019-01-01 and 2019-10-19: 291

JavaScript

const timeRatio = 1000 * 60 * 60 * 24;
var floor = Math.floor, abs = Math.abs;
var daysBetween = (d1, d2) => floor(abs(new Date(d1) - new Date(d2)) / timeRatio);

console.log('Days between 2021-10-27 and 2020-03-03: %s', daysBetween('2021-10-27', '2020-03-03'));
Output:
Days between 2021-10-27 and 2020-03-03: 603

jq

Works with: jq

Works with gojq, the Go implementation of jq

Two solutions are provided: the first uses jq's built-ins and is very brief. The second starts from first principles, and is of potential interest because of the various generic functions that are provided. The output is the same in both cases.

Using jq's built-ins

def days_between(yyyymmddBefore; yyyymmddAfter):
  (yyyymmddBefore | strptime("%Y-%m-%d") | mktime) as $before
  | (yyyymmddAfter | strptime("%Y-%m-%d") | mktime) as $after
  # leap seconds are always inserted
  | (($after - $before) / (24*60*60) | floor) ;

def task:
  def prolog: "In the following, if the dates are not the same,",
  "the \"from\" date is included in the count of days, but the \"to\" date is not.\n",
  "If the first date is later than the second date, the count of days is negated.";

  def dates:
    ["1995-11-21", "1995-11-21"],
    ["2019-01-01", "2019-01-02"],
    ["2019-01-02", "2019-01-01"],
    ["2019-01-01", "2019-03-01"],
    ["2020-01-01", "2020-03-01"],
    ["1902-01-01", "1968-12-25"],
    ["2090-01-01", "2098-12-25"],
    ["1902-01-01", "2098-12-25"],
    ["1970-01-01", "2019-10-18"],
    ["2019-03-29", "2029-03-29"],
    ["2020-02-29", "2020-03-01"] ;

  prolog,
   (dates | "The number of days from \(.[0]) until \(.[1]) is \(days_between(.[0]; .[1] ))")
  ;

task
Output:
In the following, if the dates are not the same,
the "from" date is included in the count of days, but the "to" date is not.

If the first date is later than the second date, the count of days is negated.
The number of days from 1995-11-21 until 1995-11-21 is 0
The number of days from 2019-01-01 until 2019-01-02 is 1
The number of days from 2019-01-02 until 2019-01-01 is -1
The number of days from 2019-01-01 until 2019-03-01 is 59
The number of days from 2020-01-01 until 2020-03-01 is 60
The number of days from 1902-01-01 until 1968-12-25 is 24465
The number of days from 2090-01-01 until 2098-12-25 is 3280
The number of days from 1902-01-01 until 2098-12-25 is 71947
The number of days from 1970-01-01 until 2019-10-18 is 18187
The number of days from 2019-03-29 until 2029-03-29 is 3653
The number of days from 2020-02-29 until 2020-03-01 is 1

From first principles

Adapted from Wren

# In general, dates should be valid Julian dates on or after Jan 1, 0001, but
# for the most part, this is not checked, in part because some
# computations based on ostensibly invalid dates do produce useful
# results, e.g. days(2000; 1; 1) computes the number of days from Jan 1, 0001
# up to and including Jan 1, 2000 whereas days(2000; 1; 0) excludes Jan 1, 2000.

# Output: the number of days from and including Jan 1, 0001,
# up to but excluding Jan 1 in the year $y for $y >= 1
def days_before:
  if . < 1
  then "The input to days_before should be a positive integer, not \(.)"|error
  else (. - 1 | floor) as $y
  | $y*365 + (($y/4)|floor) - (($y/100)|floor) + (($y/400)|floor)
  end;

def isLeapYear:
   .%4 == 0 and (.%100 != 0 or .%400 == 0);

# The day of the year (Jan 1 is 1)
def day_of_year($y; $m; $d):
  def diy: [0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334, 365];
  def diy2: [0, 31, 60, 91, 121, 152, 182, 213, 244, 274, 305, 335, 366];
  $d + if ($y|isLeapYear) then diy2[$m-1] else diy[$m-1] end;

# Output: the number of days from Jan 1, 0001 to the specified date, inclusive.
def days($y; $m; $d):
  ($y | days_before) + day_of_year($y; $m; $d);

# Output: the signed difference in the "days" values of the two dates.
# If the first specified date is later than the second specified date,
# then the result is the number of days from and including the earlier date,
# up to but excluding the later date.
def days_between(Year; Month; Day; laterYear; laterMonth; laterDay):
  days(laterYear; laterMonth; laterDay) -
   days(Year; Month; Day);

# Dates in yyyy-mm-dd format or as a numeric array [y,m,d]
def days_between(date; later):
  def toa: if type == "string" then split("-") | map(tonumber) else . end;
  (later  | toa) as $later
  | (date | toa) as $date
  | days_between($date[0]; $date[1]; $date[2]; $later[0]; $later[1]; $later[2]);

The Tasks

As above.

Output:

Same as above.


Julia

using Dates

@show Day(DateTime("2019-09-30") - DateTime("2019-01-01"))
 
@show Day(DateTime("2019-03-01") - DateTime("2019-02-01"))
 
@show Day(DateTime("2020-03-01") - DateTime("2020-02-01"))
 
@show Day(DateTime("2029-03-29") - DateTime("2019-03-29"))
Output:
Day(DateTime("2019-09-30") - DateTime("2019-01-01")) = 272 days
Day(DateTime("2019-03-01") - DateTime("2019-02-01")) = 28 days
Day(DateTime("2020-03-01") - DateTime("2020-02-01")) = 29 days
Day(DateTime("2029-03-29") - DateTime("2019-03-29")) = 3653 days

Kotlin

import java.time.LocalDate
import java.time.temporal.ChronoUnit

fun main() {
    val fromDate = LocalDate.parse("2019-01-01")
    val toDate = LocalDate.parse("2019-10-19")
    val diff = ChronoUnit.DAYS.between(fromDate, toDate)
    println("Number of days between $fromDate and $toDate: $diff")
}
Output:
Number of days between 2019-01-01 and 2019-10-19: 291

Lua

This uses os.difftime to compare two Epoch times. Not to be used with dates before 1970.

SECONDS_IN_A_DAY = 60 * 60 * 24

-- Convert date string as YYYY-MM-DD to Epoch time.
function parseDate (str)
  local y, m, d = string.match(str, "(%d+)-(%d+)-(%d+)")
  return os.time({year = y, month = m, day = d})
end

-- Main procedure
io.write("Enter date 1: ")
local d1 = parseDate(io.read())
io.write("Enter date 2: ")
local d2 = parseDate(io.read())
local diff = math.ceil(os.difftime(d2, d1) / SECONDS_IN_A_DAY)
print("There are " .. diff .. " days between these dates.")
Output:
Enter date 1: 1970-01-01
Enter date 2: 2019-10-02
There are 18171 days between these dates.

M2000 Interpreter

Version 12 has date type, so we can handle easy dates.

module Days_between_dates{
	date a="2019-01-01", b="2019-09-30"
	long z=b-a
	// Use the system default to display dates (DD/MM/YYYY)
	Print "Days from "+a+" to "+b+" = "+z
	// using locale 1033 to display dates (MM/DD/YYYY)
	Print "Days from "+date$(a, 1033)+" to "+date$(b, 1033)+" = "+z
}
Days_between_dates
Output:
Days from 1/1/2019 to 30/9/2019 = 272
Days from 1/1/2019 to 9/30/2019 = 272


Mathematica / Wolfram Language

DateDifference["2020-01-01", "2020-03-01"]
DateDifference["2021-01-01", "2021-03-01"]
Output:
Quantity[60, "Days"]
Quantity[59, "Days"]

Nim

import times

proc daysBetween(date1, date2: string): int64 =
  const Fmt = initTimeFormat("yyyy-MM-dd")
  (date2.parse(Fmt, utc()) - date1.parse(Fmt, utc())).inDays

const Dates = [("1995-11-21","1995-11-21"),
               ("2019-01-01","2019-01-02"),
               ("2019-01-02","2019-01-01"),
               ("2019-01-01","2019-03-01"),
               ("2020-01-01","2020-03-01"),
               ("1902-01-01","1968-12-25"),
               ("2090-01-01","2098-12-25")]

for (date1, date2) in Dates:
  echo "Days between ", date1, " and ", date2, ": ", daysBetween(date1, date2)
Output:
Days between 1995-11-21 and 1995-11-21: 0
Days between 2019-01-01 and 2019-01-02: 1
Days between 2019-01-02 and 2019-01-01: -1
Days between 2019-01-01 and 2019-03-01: 59
Days between 2020-01-01 and 2020-03-01: 60
Days between 1902-01-01 and 1968-12-25: 24465
Days between 2090-01-01 and 2098-12-25: 3280

Perl

Would not reinvent this wheel.

use feature 'say';
use Date::Calc qw(Delta_Days);

say Delta_Days(2018,7,13, 2019,9,13);   # triskaidekaphobia
say Delta_Days(1900,1,1,  2000,1,1);    # a century
say Delta_Days(2000,1,1,  2100,1,1);    # another, with one extra leap day
say Delta_Days(2020,1,1,  2019,10,1);   # backwards in time
say Delta_Days(2019,2,29, 2019,3,1);    # croaks
Output:
427
36524
36525
-92
Date::Calc::PP::Delta_Days(): Date::Calc::Delta_Days(): not a valid date at Days_between_dates line 10

Phix

Library: Phix/basics
include builtins\timedate.e 
-- specify as many or as few permitted formats as you like:
set_timedate_formats({"YYYY-MM-DD","DD/MM/YYYY","YYYY/MM/DD"})
 
constant SECONDS_TO_DAYS = 60*60*24
 
procedure test(string d1, d2, desc="")
    timedate td1 = parse_date_string(d1),
             td2 = parse_date_string(d2)
    atom s = timedate_diff(td1,td2,DT_DAY),
         d = s/SECONDS_TO_DAYS
    string e = elapsed(s)&desc
    printf(1,"Days between %s and %s: %d [%s]\n",{d1,d2,d,e})
end procedure
test("1969-12-31","1970-01-01")
test("1995-11-21","1995-11-21",", same date")
test("2019-01-02","2019-01-01",", negative date")
test("2019-01-01","2019-03-01",", non-leap year")
test("2020-01-01","2020-03-01",", leap year")
test("1970-01-01", "2019/10/18")
test("1970-01-01", "18/10/2019")

As shown, timedate_diff() can optionally round to the nearest whole number of days [else omit DT_DAY].
Note that elapsed() assumes all years are exactly 365 days, and in no way takes leap years into consideration (as opposed, of course, to timedate_diff() which handles them flawlessly), not that you sh/c/would ever use the string output of elapsed() in any further calculations anyway.

Output:
Days between 1969-12-31 and 1970-01-01: 1 [1 day]
Days between 1995-11-21 and 1995-11-21: 0 [0s, same date]
Days between 2019-01-02 and 2019-01-01: -1 [minus 1 day, negative date]
Days between 2019-01-01 and 2019-03-01: 59 [59 days, non-leap year]
Days between 2020-01-01 and 2020-03-01: 60 [60 days, leap year]
Days between 1970-01-01 and 2019/10/18: 18187 [49 years, 302 days]
Days between 1970-01-01 and 18/10/2019: 18187 [49 years, 302 days]

PicoLisp

(de diffDates (A B)
   (abs (- ($dat A "-") ($dat B "-"))) )
(println (diffDates "2019-1-1" "2019-9-30"))
(println (diffDates "2015-12-31" "2016-09-30"))
Output:
272
274

PL/M

Translation of: FreeBASIC
Works with: 8080 PL/M Compiler
... under CP/M (or an emulator)

Note that as the 8080 PL/M compiler only supports 8 and 16 bit unsigned integers, the dates must be at most 65535 days apart.

100H: /* CALCULATE THE NUMBER OF DAYS BETWEEN TWO DATES; BASED ON FREEBASIC */

   /* CP/M BDOS SYSTEM CALL AND I/O ROUTINES                                */
   BDOS: PROCEDURE( FN, ARG ); DECLARE FN BYTE, ARG ADDRESS; GOTO 5; END;
   PR$CHAR:   PROCEDURE( C ); DECLARE C BYTE;    CALL BDOS( 2, C );  END;
   PR$STRING: PROCEDURE( S ); DECLARE S ADDRESS; CALL BDOS( 9, S );  END;
   PR$NL:     PROCEDURE;   CALL PR$CHAR( 0DH ); CALL PR$CHAR( 0AH ); END;
   PR$NUMBER: PROCEDURE( N ); /* PRINTS A NUMBER IN THE MINIMUN FIELD WIDTH */
      DECLARE N ADDRESS;
      DECLARE V ADDRESS, N$STR ( 6 )BYTE, W BYTE;
      V = N;
      W = LAST( N$STR );
      N$STR( W ) = '$';
      N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
      DO WHILE( ( V := V / 10 ) > 0 );
         N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
      END;
      CALL PR$STRING( .N$STR( W ) );
   END PR$NUMBER;
   PR$SIGNED: PROCEDURE( N ); /* PRINTS N AS A SIGNED INTEGER               */
      DECLARE N ADDRESS;
      IF N <= 32767
      THEN CALL PR$NUMBER( N );
      ELSE DO;
         CALL PR$CHAR( '-' );
         CALL PR$NUMBER( - N );
      END;
   END PR$SIGNED ;
         

   /* TASK                                                                  */

   /* RETURNS THE GREGORIAN DAY CORRESPONDING TO YYYY/MM/DD                 */
   GREGORIAN: PROCEDURE( YYYY$MM$DD )ADDRESS;
      DECLARE YYYY$MM$DD ADDRESS;
      DECLARE DATE BASED YYYY$MM$DD ( 10 )BYTE;
      DECLARE ( YYYY, MM, DD, N, W ) ADDRESS;

      DIGIT: PROCEDURE( D )BYTE; DECLARE D BYTE; RETURN D - '0'; END;

      YYYY = ( DIGIT( DATE( 0 ) ) * 1000 ) + ( DIGIT( DATE( 1 ) ) * 100 )
           + ( DIGIT( DATE( 2 ) ) *   10 ) +   DIGIT( DATE( 3 ) );
      MM   = ( DIGIT( DATE( 5 ) ) *   10 ) +   DIGIT( DATE( 6 ) );
      DD   = ( DIGIT( DATE( 8 ) ) *   10 ) +   DIGIT( DATE( 9 ) );
      N = ( MM + 9 ) - ( ( ( MM + 9 ) / 12 ) * 12 );
      W = YYYY - ( N / 10 );
      RETURN ( 365 * W ) + ( W / 4 ) - ( W / 100 ) + ( W / 400 )
                         + ( ( ( N * 306 ) + 5 ) / 10 ) + ( DD - 1 );
   END GREGORIAN ;

   /* SHOWS TTHE DAYS DIFFERENCE BETWEEN FROM$G AND TO$$G                   */
   PR$DAYS$DIFFERENCE: PROCEDURE( FROM$DATE, TO$DATE );
      DECLARE ( FROM$DATE, TO$DATE )ADDRESS;
      CALL PR$STRING( .'DAYS BETWEEN $' );CALL PR$STRING( FROM$DATE );
      CALL PR$STRING( .' AND $'         );CALL PR$STRING( TO$DATE   );
      CALL PR$STRING( .' IS $'          );
      CALL PR$SIGNED( GREGORIAN( TO$DATE ) - GREGORIAN( FROM$DATE ) );
      CALL PR$NL;
   END PR$DAYS$DIFFERENCE ;

   CALL PR$DAYS$DIFFERENCE( .'1902-01-01$', .'1968-12-25$' );
   CALL PR$DAYS$DIFFERENCE( .'2019-01-01$', .'2019-01-02$' );
   CALL PR$DAYS$DIFFERENCE( .'2019-01-02$', .'2019-01-01$' );
   CALL PR$DAYS$DIFFERENCE( .'2019-01-01$', .'2019-03-01$' );
   CALL PR$DAYS$DIFFERENCE( .'2020-01-01$', .'2020-03-01$' );
   CALL PR$DAYS$DIFFERENCE( .'1995-11-21$', .'1995-11-21$' );
   CALL PR$DAYS$DIFFERENCE( .'2090-01-01$', .'2098-12-25$' );

EOF
Output:
DAYS BETWEEN 1902-01-01 AND 1968-12-25 IS 24465
DAYS BETWEEN 2019-01-01 AND 2019-01-02 IS 1
DAYS BETWEEN 2019-01-02 AND 2019-01-01 IS -1
DAYS BETWEEN 2019-01-01 AND 2019-03-01 IS 59
DAYS BETWEEN 2020-01-01 AND 2020-03-01 IS 60
DAYS BETWEEN 1995-11-21 AND 1995-11-21 IS 0
DAYS BETWEEN 2090-01-01 AND 2098-12-25 IS 3280

Python

#!/usr/bin/python
import sys

''' Difference between two dates = g(y2,m2,d2) - g(y1,m1,d1) 
    Where g() gives us the Gregorian Calendar Day
    Inspired  by discussion at:
    https://stackoverflow.com/questions/12862226/the-implementation-of-calculating-the-number-of-days-between-2-dates
'''

def days( y,m,d ):
  ''' input year and month are shifted to begin the year in march'''
  m = (m + 9) % 12 
  y = y - m/10

  ''' with (m*306 + 5)/10 the number of days from march 1 to the current 'm' month '''
  result = 365*y + y/4 - y/100 + y/400 + (m*306 + 5)/10 + ( d - 1 )
  return result

def diff(one,two):
  [y1,m1,d1] = one.split('-')
  [y2,m2,d2] = two.split('-')
  # strings to integers
  year2 = days( int(y2),int(m2),int(d2))
  year1 = days( int(y1), int(m1), int(d1) )
  return year2 - year1

if __name__ == "__main__":
  one = sys.argv[1]
  two = sys.argv[2]
  print diff(one,two)
Output:
python days-between.py 2019-01-01 2019-09-30
272

QB64

'Task
'Calculate the number of days between two dates.
'Date input should be of the form   YYYY-MM-DD.

from$ = "2000-05-29"
to$ = "2022-05-29"
Print NumberOfDays(from$, to$)

End

Function NumberOfDays (from$, to$)
    NumberOfDays = 0
    Dim As Integer Year(1 To 2), Mounth(1 To 2), Day(1 To 2)
    Dim As Integer NumberD, Index
    Year(1) = Val(Left$(from$, 4))
    Year(2) = Val(Left$(to$, 4))
    Mounth(1) = Val(Mid$(from$, 6, 2))
    Mounth(2) = Val(Mid$(to$, 6, 2))
    Day(1) = Val(Right$(from$, 2))
    Day(2) = Val(Right$(to$, 2))


    If Year(1) > Year(2) Then
        Swap Year(1), Year(2)
        Swap mount(1), mount(2)
        Swap Day(1), Day(2)
    End If

    If Day(1) > Day(2) Then
        Select Case Mounth(2) - 1
            Case 4, 6, 9, 11
                Day(2) = Day(2) + 30
            Case 1, 3, 5, 7, 8, 10, 12
                Day(2) = Day(2) + 31
            Case 2
                If (Index Mod 4 = 0) Or ((Index Mod 100 = 0) And (Index Mod 400 = 0)) Then Day(2) = Day(2) + 29 Else Day(2) = Day(2) + 28
        End Select
        Mounth(2) = Mounth(2) - 1
        If Mounth(2) = 0 Then Year(2) = Year(2) - 1: Mounth(2) = 12
    End If

    NumberD = (Day(2) - Day(1)) + 1

    For Index = Mounth(1) To Mounth(2) - 1 Step 1
        Select Case Index
            Case 4, 6, 9, 11
                NumberD = NumberD + 30
            Case 1, 3, 5, 7, 8, 10, 12
                NumberD = NumberD + 31
            Case 2
                If (Index Mod 4 = 0) Or ((Index Mod 100 = 0) And (Index Mod 400 = 0)) Then NumberD = NumberD + 29 Else NumberD = NumberD + 28
        End Select

    Next

    For Index = Year(1) To Year(2) - 1 Step 1
        If (Index Mod 4 = 0) Or ((Index Mod 100 = 0) And (Index Mod 400 = 0)) Then NumberD = NumberD + 366 Else NumberD = NumberD + 365
    Next Index
    NumberOfDays = NumberD
End Function

Raku

(formerly Perl 6) Dates are first class objects in Raku and may have arithmetic in days done directly on them.

say Date.new('2019-09-30') - Date.new('2019-01-01');

say Date.new('2019-03-01') - Date.new('2019-02-01');

say Date.new('2020-03-01') - Date.new('2020-02-01');

say Date.new('2029-03-29') - Date.new('2019-03-29');

say Date.new('2019-01-01') + 90;

say Date.new('2020-01-01') + 90;

say Date.new('2019-02-29') + 30;

CATCH { default { .message.say; exit; } };
272
28
29
3653
2019-04-01
2020-03-31
Day out of range. Is: 29, should be in 1..28

REBOL

REBOL accepts a multitude of lexically recognized date! formats. date! is a builtin datatype. Slashes or dashes, year-month-day, day/month/year, or dd/monthname/year, etc. Math on dates defaults to days.

prompt$ rebol -q
>> 2019-11-24 - 2000-01-01
== 7267
>> 2019-nov-24 - 01-jan-2000
== 7267

REXX

Works with: Regina REXX

bare bones version

Programming note:   the   B   (Base)   an option for the   date   BIF which indicates to compute the number of
days since the beginning of the Gregorian calendar,   and   I   which is the option that indicates the date is in
the   ISO   (International Standards Organization standard 8601:2004)   format.

/*REXX program computes the number of days between two dates in the form of  YYYY-MM-DD */
parse arg $1 $2 .                                /*get 2 arguments (dates) from the C.L.*/
say abs( date('B',$1,"I")  -  date('B',$2,"I") )   ' days between '    $1    " and "    $2
                                                 /*stick a fork in it,  we're all done. */
output   when using the inputs of:     2019-10-02   2000-01-01
7214  days between  2019-10-02  and  2000-01-01

supports more variations

This REXX version supports more variations in the date format   (allows a single digit month and/or day), as well as
allowing a single asterisk   (*)   to be used for a date   (which signifies that the current date is to be used).

The dates may be specified in the following formats:

       ,            (a comma)        indicates today's date
       *            (an asterisk)    indicates today's date 
       yyyy-mm-dd   where yyyy may be a 2- or 4-digit year, mm may be a 1- or 2-digit month, dd may be a 1- or 2-digit day of month
       mm/dd/yyyy   (as above)
       mm/dd        (as above),  but the current year is assumed
       dd\mm\yyyy   (as above)
       dd\mm        (as above),  but the current year is assumed



Commas   (,)   are inserted into numbers where appropriate.

Also, more informative error messages are generated.

/*REXX program computes the number of days between two dates in the form of  YYYY-MM-DD */
parse arg $.1 $.2 _ . 1 . . xtra                 /*obtain two arguments from the  C.L.  */
seps= '/-\';        yr.= .;   mon.= .;   dd.= .  /*define the defaults for both dates.  */
            do a=1  for 2                        /*process both of the specified dates. */
            if $.a=='' | $.a=="*" | $.a==','  then $.a= date("I")
               do s=1  for length(seps)          /*process a specified date by separator*/
               sep= substr(seps, s, 1)           /*obtain one of the supported sep char.*/
               if pos(sep, $.a)\==0  then call conv $.a         /*parse the date string.*/
               end   /*s*/
            end      /*a*/
?.1= '1st'
?.2= '2nd'
if _ \== ''              then call err "too many arguments specified: "   xtra
dy.= 31                                          /*default number of days for all months*/
parse value 30 with dy.4 1 dy.6 1 dy.9 1 dy.11   /*define 30─day months, Feb. is special*/
@notCorr= "isn't in a support date format: "     /*used for a literal for an error msg. */

  do j=1  for 2                                  /*examine both dates for correct format*/
  if $.j           ==''  then call err ?.j "date was not specified."
  if length(yr.j)==0     then call err ?.j "year"  @notCorr '(missing)'
  if isDec(yr.j)         then call err ?.j "year"  @notCorr '(has a non─decimal digit)'
  if yr.j<1 | yr.j>9999  then call err ?.j "year"  @notCorr '(not in the range 1──►9999)'
  if length(mon.j)==0    then call err ?.j "month" @notCorr '(missing)'
  if isDec(mon.j)        then call err ?.j "month" @notCorr '(has a non─decimal digit)'
  if mon.j<1 | mon.j>12  then call err ?.j "month" @notCorr '(not in the range 1──►12)'
  if length(dd.j)==0     then call err ?.j "day"   @notCorr '(missing)'
  if isDec(dd.j)         then call err ?.j "day"   @notCorr '(has a non─decimal digit)'
  mo= mon.j
  if leapYr(yr.j)  then dy.2= 29                 /*Is it a leapyear? Use 29 days for Feb*/
                   else dy.2= 28                 /*Isn't "     "      "  28   "   "   " */
  if dd.j<1 | dd.j>dy.mo then call err ?.j "day"   @notCorr '(day in month is invalid)'

    yr.j= right( yr.j  +0, 4, 0)                 /*force YYYY to be four decimal digits.*/
   mon.j= right(mon.j  +0, 2, 0)                 /*  "    MON  "  "  two    "       "   */
    dd.j= right( dd.j  +0, 2, 0)                 /*  "     DD  "  "   "     "       "   */
     $.j= yr.j'-'mon.j"-"dd.j                    /*reconstitute a date from above parts.*/
  end       /*j*/

between= abs( date('B', $.1, "I")  -  date('B', $.2, "I") )     /* # days between dates.*/
parse source . how .                                            /*determine how invoked.*/
if how='COMMAND'  then say commas(between)    ' days between '     $.1     " and "     $.2
exit between                                     /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: parse arg _; do c_=length(_)-3  to 1  by -3; _=insert(',', _, c_);  end;  return _
err:    say; say '***error*** ' arg(1);  exit 13 /*issue an error message (with text)   */
isDec:  return verify( arg(1), 1234567890) \== 0 /*insure argument is just decimal digs.*/
leapYr: arg _; ly=_//4==0; if ly==0  then return 0; ly=((_//100\==0)|_//400==0); return ly
serDAT: call err 'illegal date format with a separator= ['sep"],   date= "       @dat
/*──────────────────────────────────────────────────────────────────────────────────────*/
conv:   parse arg @dat                           /*obtain date that the user specified. */
        if sep=='-'  then parse var  @dat    yr.a  "-"  mon.a  '-'  dd.a    /* yy-mm-dd */
        if sep=='/'  then parse var  @dat   mon.a  "/"   dd.a  '/'  yr.a    /* mm/dd/yy */
        if sep=='\'  then parse var  @dat    dd.a  "\"  mon.a  '\'  yr.a    /* dd\mm\yy */
        if yr.a==''  then yr.a= right( date(), 4)                           /*omitted yy*/
        if length(yr.a)==2  then yr.a= left( date('S'), 2)yr.a              /*2 dig yy ?*/
        if yr.a==.  |  mon.a==.  |  dd.a==.  then call serDAT               /*validate. */
        return
output   when using the inputs of:     *   2000-1-1

Today   (indicated by the asterisk)   is   2019-10-2

7,214  days between  2019-10-02  and  2000-01-01

Ring

load "stdlib.ring"

DaysBetween = [["1995-11-21","1995-11-21"],
               ["2019-01-01","2019-01-02"],
               ["2019-01-02","2019-01-01"],
               ["2019-01-01","2019-03-01"],
               ["2020-01-01","2020-03-01"],
               ["1902-01-01","1968-12-25"],
               ["2090-01-01","2098-12-25"]]

for n = 1 to len(DaysBetween)
    date1 = DaysBetween[n][1]
    date2 = DaysBetween[n][2]
    date3 = substr(date1,9,2) + "/" + substr(date1,6,2) + "/" + substr(date1,1,4)
    date4 = substr(date2,9,2) + "/" + substr(date2,6,2) + "/" + substr(date2,1,4)
    ? "Days between " + DaysBetween[n][1] + " and " + DaysBetween[n][2] + ": " + diffdays(date4,date3)
next
Output:
Days between 1995-11-21 and 1995-11-21: 0
Days between 2019-01-01 and 2019-01-02: 1
Days between 2019-01-02 and 2019-01-01: -1
Days between 2019-01-01 and 2019-03-01: 59
Days between 2020-01-01 and 2020-03-01: 60
Days between 1902-01-01 and 1968-12-25: 0
Days between 2090-01-01 and 2098-12-25: 3280

RPL

Uses Python formula, in a forced binary calculation mode to avoid 'flooring' instructions

Works with: Halcyon Calc version 4.2.7
≪ → d m y 
  ≪ m 9 + 12 MOD
     y OVER #10d / - 
     DUP 365 * OVER #4d / + OVER #100d / - SWAP #400d / + 
     SWAP 306 * 5 + #10d / + d + 1 - B→R 
≫ ≫ 
'GREGN' STO

≪ SWAP 1 2 START
  → date 
  ≪ date 9 10 SUB STR→ date 6 7 SUB STR→ date 1 4 SUB STR→ 
     GREGN SWAP 
  ≫ 
  NEXT - 
≫
'NBDAYS' STO
"1902-01-01" "1968-12-25" NBDAYS
"2019-01-02" "2019-01-01" NBDAYS
"2019-01-01" "2019-03-01" NBDAYS
"2020-01-01" "2020-03-01" NBDAYS
Output:
4: 24465 
3: -1
2: 59
1: 60

Ruby

require "date"

d1, d2 = Date.parse("2019-1-1"), Date.parse("2019-10-19")

p (d1 - d2).to_i  # => -291
p (d2 - d1).to_i  # => 291

Rust

// [dependencies]
// chrono = "0.4"

use chrono::NaiveDate;

fn main() {
    let args: Vec<String> = std::env::args().collect();
    if args.len() != 3 {
        eprintln!("usage: {} start-date end-date", args[0]);
        std::process::exit(1);
    }
    if let Ok(start_date) = NaiveDate::parse_from_str(&args[1], "%F") {
        if let Ok(end_date) = NaiveDate::parse_from_str(&args[2], "%F") {
            let d = end_date.signed_duration_since(start_date);
            println!("{}", d.num_days());
        } else {
            eprintln!("Can't parse end date");
            std::process::exit(1);
        }        
    } else {
        eprintln!("Can't parse start date");
        std::process::exit(1);
    }
}
Output:
days_between_dates 2020-01-01 2020-09-06
249

S-BASIC

Error checking of entered dates is omitted in order to focus on the stated task but would obviously have to be included in production code.

comment
    Given a month, day, and year in the Gregorian calendar,
    return a numeric date which is equal to the number of 
    days since the start of the Common era.
end
function serial_date(da, mo, yr = integer) = real
    var n = real
    n = 365 * yr + da + 31 * (mo - 1)
    if mo >= 3 then 
        n = n - int(.4 * mo + 2.3) 
    else
        yr = yr - 1
    n = n + int(yr/4) - int(.75 * (int(yr/100) + 1))
end = n

comment
   Read a date in YYYY-MM-DD format from the console and 
   return its serial date equivalent.
end
function get_date = real
    var date = string : 20
    var y, m, d = integer
    input2 date 
    y = val(mid(date,1,4))
    m = val(mid(date,6,2))
    d = val(mid(date,9,2))
end = serial_date(d, m, y)

rem  --  main program begins here

var date1, date2 = real
var another = char

repeat
    begin
        print "First date   : ";
        date1 = get_date
        print "Second date  : ";
        date2 = get_date
        print "Elapsed days = "; date2 - date1
        input "Do another (y/n)"; another
    end
until not another

end
Output:

Test dates taken from Delphi example

First date   : 1970-01-01
Second date  : 2019-10-18
Elapsed days =  18187
Do another (y/n)? n

Scala

object DaysBetweenDates {

  /*Inspired by the Python version of the algorithm and the discussion here
   https://stackoverflow.com/questions/12862226/the-implementation-of-calculating-the-number-of-days-between-2-dates.*/

  /**Transform a date into a day number in the Gregorian Calendar*/
  def dateToDays(year : Int, month : Int, day : Int ) : Int = {
    val m = (month+ 9) % 12
    val y = year - m/10
    val d = day
    365*y + y/4 - y/100 + y/400 + (m*306 + 5)/10 + (d - 1)
  }

  /**Compute the difference of days between both input dates*/
  def daysDifference(firstDate : String, secondDate : String) : Int = {
    val firstDateTuple = firstDate.split('-') match { case Array(a, b, c) => (a, b, c) }
    val secondDateTuple = secondDate.split('-') match { case Array(a, b, c) => (a, b, c) }

    val firstYear = dateToDays( firstDateTuple._1.toInt, firstDateTuple._2.toInt, firstDateTuple._3.toInt)
    val secondYear = dateToDays( secondDateTuple._1.toInt, secondDateTuple._2.toInt, secondDateTuple._3.toInt )

    return secondYear - firstYear
  }

  def main(args: Array[String]): Unit = {
    println(daysDifference("2019-01-01", "2019-09-30"))
    println(daysDifference("1995-11-21", "1995-11-21"))
    println(daysDifference("2019-01-01", "2019-01-02"))
    println(daysDifference("2019-01-02", "2019-01-01"))
    println(daysDifference("2019-01-01", "2019-03-01"))
    println(daysDifference("2020-01-01", "2020-03-01"))
    println(daysDifference("1902-01-01", "1968-12-25"))
    println(daysDifference("2090-01-01", "2098-12-25"))
    println(daysDifference("1902-01-01", "2098-12-25"))
  }

}
Output:
272
0
1
-1
59
60
24465
3280
71947

SenseTalk

SenseTalk supports date and time calculations in many forms, and many different units of time (such as 'days' in this example). Rounding is needed due to daylight savings time changes.

set startDate to "2020-03-13"
set endDate to "2021-07-14"

put (endDate - startDate) rounded to nearest day
Output:
488 days

Sidef

require('Date::Calc')

func days_diff(a,b) {
    %S<Date::Calc>.Delta_Days(a.split('-')..., b.split('-')...)
}

var date1 = "1970-01-01"
var date2 = "2019-10-02"

say "Date 1: #{date1}"
say "Date 2: #{date2}"

var days = days_diff(date1, date2)

say "There are #{days} days between these dates"
Output:
Date 1: 1970-01-01
Date 2: 2019-10-02
There are 18171 days between these dates

Swift

import Foundation

func daysFromTimeInterval(_ interval: Double) -> Int {
  return Int(interval) / 86400
}

let formatter = DateFormatter()

formatter.dateFormat = "yyyy-MM-dd"

print("Enter date one (yyyy-MM-dd): ", terminator: "")

guard let date1Str = readLine(strippingNewline: true), let date1 = formatter.date(from: date1Str) else {
  fatalError("Invalid date two")
}

print("Enter date two (yyyy-MM-dd): ", terminator: "")

guard let date2Str = readLine(strippingNewline: true), let date2 = formatter.date(from: date2Str) else {
  fatalError("Invalid date two")
}

let (start, end) = date1 > date2 ? (date2, date1) : (date1, date2)
let days = daysFromTimeInterval(DateInterval(start: start, end: end).duration)

print("There are \(days) days between \(start) and \(end)")
Output:
Enter date one (yyyy-MM-dd): 2019-01-01
Enter date two (yyyy-MM-dd): 2019-12-02
There are 335 days between 2019-01-01 05:00:00 +0000 and 2019-12-02 05:00:00 +0000

UNIX Shell

Works with: Bourne Again Shell
# return true if year is leap in Gregorian calendar
leap() {
  local -i year
  year=$1
  if (( year % 4 )); then return 1; fi
  if (( year % 100 )); then return 0; fi
  ! (( year % 400 ))
}

# convert date to Gregorian day count (Rata Die), where RD 1 = January 1, 1 CE
rd() {
   local year month day
   IFS=- read year month day <<<"$1"
   local -i elapsed_years=year-1
   (( day += elapsed_years * 365 ))
   (( day += elapsed_years/4 ))
   (( day -= elapsed_years/100 ))
   (( day += elapsed_years/400 ))
   local month_lengths=(31 28 31 30 31 30 31 31 30 31 30 31)
   if leap "$year"; then let month_lengths[1]+=1; fi
   local m
   for (( m=0; m<month-1; ++m)); do
     (( day += month_lengths[m] ))
   done
   printf '%d\n' "$day"
}

days_between() {
  local -i date1 date2
  date1=$(rd "$1")
  date2=$(rd "$2")
  printf '%d\n' $(( date2 - date1 ))
}

days_between 1970-01-01 2019-12-04
Output:
18234

Visual Basic .NET

Translation of: C#
Imports System.Globalization

Module Module1

    Function DateDiff(d1 As String, d2 As String) As Integer
        Dim a = DateTime.ParseExact(d1, "yyyy-MM-dd", CultureInfo.InvariantCulture)
        Dim b = DateTime.ParseExact(d2, "yyyy-MM-dd", CultureInfo.InvariantCulture)
        Return (b - a).TotalDays
    End Function

    Sub Main()
        Console.WriteLine(DateDiff("1970-01-01", "2019-10-18"))
    End Sub

End Module
Output:
18187

V (Vlang)

Translation of: go
import time

fn days_between(d1 string, d2 string) ?int {
    t1 := time.parse('$d1 01:01:01')?
    t2 := time.parse('$d2 01:01:01')?
    days := int((t2-t1).hours()/24)
    return days
}
fn main(){
    mut date1,mut date2 := "2019-01-01", "2019-09-30"
    mut days := days_between(date1, date2)?
    println("There are $days days between $date1 and $date2")
 
    date1, date2 = "2015-12-31", "2016-09-30"
    days = days_between(date1, date2)?
    println("There are $days days between $date1 and $date2")
}
Output:
There are 272 days between 2019-01-01 and 2019-09-30
There are 274 days between 2015-12-31 and 2016-09-30

Wren

Library: Wren-date
import "./date" for Date

var datePairs = [
    ["1995-11-21", "1995-11-21"],
    ["2019-01-01", "2019-01-02"],
    ["2019-01-02", "2019-01-01"],
    ["2019-01-01", "2019-03-01"],
    ["2020-01-01", "2020-03-01"],
    ["1902-01-01", "1968-12-25"],
    ["2090-01-01", "2098-12-25"],
    ["1902-01-01", "2098-12-25"],
    ["1970-01-01", "2019-10-18"],
    ["2019-03-29", "2029-03-29"],
    ["2020-02-29", "2020-03-01"]
]
Date.default = Date.isoDate
for (dates in datePairs) {
    var date1 = Date.parse(dates[0])
    var date2 = Date.parse(dates[1])
    var days = (date2 - date1).days
    System.print("Days between %(date1) and %(date2) = %(days)")
}
Output:
Days between 1995-11-21 and 1995-11-21 = 0
Days between 2019-01-01 and 2019-01-02 = 1
Days between 2019-01-02 and 2019-01-01 = -1
Days between 2019-01-01 and 2019-03-01 = 59
Days between 2020-01-01 and 2020-03-01 = 60
Days between 1902-01-01 and 1968-12-25 = 24465
Days between 2090-01-01 and 2098-12-25 = 3280
Days between 1902-01-01 and 2098-12-25 = 71947
Days between 1970-01-01 and 2019-10-18 = 18187
Days between 2019-03-29 and 2029-03-29 = 3653
Days between 2020-02-29 and 2020-03-01 = 1

XPL0

Translation of: FreeBASIC
func Gregorian(Y, M, D);        \Return Gregorian day given date
int  Y, M, D;
int  N, W;
[N:= M + 9 - (M+9)/12*12;
W:= Y - N/10;
return 365*W + W/4 - W/100 + W/400 + (N*306+5)/10 + D - 1;
];

int Dates, N, Y, M, D, G0, G1;
[Dates:= [
    ["2019-01-01", "2019-01-02"],
    ["2019-01-02", "2019-01-01"],
    ["2019-01-01", "2019-03-01"],
    ["2020-01-01", "2020-03-01"],
    ["1995-11-21", "1995-11-21"],
    ["2090-01-01", "2098-12-25"] ];
OpenO(8);  OpenI(8);
for N:= 0 to 6-1 do
    [Text(8, Dates(N,0));
    Y:= IntIn(8);  M:= IntIn(8);  D:= IntIn(8);
    G0:= Gregorian(Y, M, D);
    Text(8, Dates(N,1));
    Y:= IntIn(8);  M:= IntIn(8);  D:= IntIn(8);
    G1:= Gregorian(Y, M, D);
    Text(0, "Number of days between ");  Text(0, Dates(N,0));  Text(0, " and ");
    Text(0, Dates(N,1));  Text(0, " is ");  IntOut(0, abs(G1-G0));  CrLf(0);
    ];
]
Output:
Number of days between 2019-01-01 and 2019-01-02 is 1
Number of days between 2019-01-02 and 2019-01-01 is 1
Number of days between 2019-01-01 and 2019-03-01 is 59
Number of days between 2020-01-01 and 2020-03-01 is 60
Number of days between 1995-11-21 and 1995-11-21 is 0
Number of days between 2090-01-01 and 2098-12-25 is 3280

zkl

var [const] TD=Time.Date;
today:=TD.parseDate("--");  // "yyyy-mm-dd" and variations --> (y,m,d)
// or Time.Clock.UTC  --> (y,m,d,h,m,s)
then:=TD.parseDate("2018-9-30");
diff:=TD.deltaDays(then,today.xplode());  // ( (y,m,d), y,m,d )
println("Number of days between %s and %s: %d".fmt(then,today,diff));
println("Number of days between %s and %s: %d".fmt(
   TD.toYMDString(today.xplode()),	// to(y,m,d) not to((y,m,d))
   TD.toYMDString(then.xplode()),
   TD.deltaDays(today,then.xplode())));
Output:
Number of days between L(2018,9,30) and L(2019,9,30): 365
Number of days between 2019-09-30 and 2018-09-30: -365