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Day of the week: Difference between revisions

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Day of the week (view source)

Revision as of 22:22, 30 July 2010

1,307 bytes added ,  13 years ago
Undo revision 87053 by Underscore (Talk)
(Perl 5: Removed all but the most concise solution. Perl 6: Added.)
(Undo revision 87053 by Underscore (Talk))
Line 813:
=={{header|Perl}}==
 
<lang perl>#! /usr/bin/perl -w
In one line using grep (read from right to left):▼
 
use Time::Local;
<lang perl>use DateTime;▼
use strict;
 
foreach my $i (2008 .. 2121)
print join " ", grep { DateTime->new(year => $_, month => 12, day => 25)->day_of_week == 7 } (2008 .. 2121);</lang>▼
{
my $time = timelocal(0,0,0,25,11,$i);
my ($s,$m,$h,$md,$mon,$y,$wd,$yd,$is) = localtime($time);
if ( $wd == 0 )
{
print "25 Dec $i is Sunday\n";
}
}
 
exit 0;</lang>
 
Output:
<pre>2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118</pre>▼
 
<pre>
=={{header|Perl 6}}==
25 Dec 2011 is Sunday
25 Dec 2016 is Sunday
25 Dec 2022 is Sunday
25 Dec 2033 is Sunday
Day too big - 25195 > 24855
Sec too small - 25195 < 78352
Sec too big - 25195 > 15247
Cannot handle date (0, 0, 0, 25, 11, 2038) at ./ydate.pl line 8
</pre>
 
Using the DateTime module from CPAN:
{{works with|Rakudo|2010.07}}
<lang perl>#! /usr/bin/perl -w
 
▲<lang perl>use DateTime;
As Perl 5, except <code>DateTime</code> is built-in, so you don't need to download a module of that name:
use strict;
 
foreach my $i (2008 .. 2121)
<lang perl6>say join ' ', grep { Date.new($_, 12, 25).day-of-week == 7 }, 2008 .. 2121;</lang>
{
my $dt = DateTime->new( year => $i,
month => 12,
day => 25
);
if ( $dt->day_of_week == 7 )
{
print "25 Dec $i is Sunday\n";
}
}
 
exit 0;</lang>
or shorter:
<lang perl>#! /usr/bin/perl -w
 
use DateTime;
use strict;
for (2008 .. 2121) {
print "25 Dec $_ is Sunday\n"
if DateTime->new(year => $_, month => 12, day => 25)->day_of_week == 7;
}
 
exit 0;</lang>
Output:
 
<pre>
25 Dec 2011 is Sunday
25 Dec 2016 is Sunday
25 Dec 2022 is Sunday
25 Dec 2033 is Sunday
25 Dec 2039 is Sunday
25 Dec 2044 is Sunday
25 Dec 2050 is Sunday
25 Dec 2061 is Sunday
25 Dec 2067 is Sunday
25 Dec 2072 is Sunday
25 Dec 2078 is Sunday
25 Dec 2089 is Sunday
25 Dec 2095 is Sunday
25 Dec 2101 is Sunday
25 Dec 2107 is Sunday
25 Dec 2112 is Sunday
25 Dec 2118 is Sunday
</pre>
 
▲InAlternatively in one line using grep (read from right to left):
<lang perl>#! /usr/bin/perl -w
 
use DateTime;
use strict;
 
▲print join " ", grep { DateTime->new(year => $_, month => 12, day => 25)->day_of_week == 7 } (2008 .. 2121);</lang>
 
0;</lang>
Output:
▲<pre>2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118</pre>
 
=={{header|PHP}}==
Anonymous user
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