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Title : Function get_week_number which returns the week number of the given date according to ISO 8601:1988, also includes a function is_leap_year.
Categories : Date Time, PHP Update Picture
Stefan Rצhrich
Date : Jan 17th 1999
Grade : 2 of 5 (graded 3 times)
Viewed : 5181
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Search : More code by Stefan Rצhrich
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<?php
/* weeknumber.php3

(int) get_week_number($timestamp)

Week number of year with Monday as first day of the week
(1...53). If the week containing January 1 has four or more days
in the new year, then it is considered week 1; otherwise, it is
week 53 of the previous year, and the next week is week 1. (See the
ISO 8601: 1988 standard.)

Adapted from GNU sh-utils for PHP3 by Stefan Rhrich, sr@linux.de,
http://home.pages.de/~sr/.

*/

function is_leap_year($year) {
if ((($year % 4) == 0 and ($year % 100)!=0) or ($year % 400)==0) {
return 1;
} else {
return 0;
}
}

/*
#define ISO_WEEK_START_WDAY 1 // Monday
#define ISO_WEEK1_WDAY 4 // Thursday
#define YDAY_MINIMUM (-366)
int big_enough_multiple_of_7 = (-YDAY_MINIMUM / 7 + 2) * 7;
return (yday
- (yday - wday + ISO_WEEK1_WDAY + big_enough_multiple_of_7) % 7
+ ISO_WEEK1_WDAY - ISO_WEEK_START_WDAY);
*/

function iso_week_days($yday, $wday) {
return $yday - (($yday - $wday + 382) % 7) + 3;
}

function get_week_number($timestamp) {

$d = getdate($timestamp);

$days = iso_week_days($d[ "yday"], $d[ "wday"]);

if ($days < 0) {
$d[ "yday"] += 365 + is_leap_year(--$d[ "year"]);
$days = iso_week_days($d[ "yday"], $d[ "wday"]);
} else {
$d[ "yday"] -= 365 + is_leap_year($d[ "year"]);
$d2 = iso_week_days($d[ "yday"], $d[ "wday"]);
if (0 <= $d2) {
/* $d["year"]++; */
$days = $d2;
}
}

return (int)($days / 7) + 1;
}
?>



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User Comments
 Thomas Kunth wrote :766
You say that the result would be 53 if week containing January 1st doesn`t have at least four days in the new year.

This may not be correct:
E.g. January 1st, 2000 is week 52 of 1999.

As you see here it would be cool if you provide a function the related year...