Code Examples -> Function get_week_number which returns the week number of the given date according to ISO 8601:1988, also includes a function is_leap

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PHP Code Examples

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	Stefan Rצhrich
Date :	Jan 17th 1999
Grade :	2 of 5 (graded 3 times)
Viewed :	4662
File :	No file for this code example.
Images :	No Images for this code example.

Search :	More code by Stefan Rצhrich
Action :	Grade This Code Example
Tools :	My Examples List

Code

<?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 R�hrich, 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...