Datum Kalenderwoche (KW)

[Sebastian Becker]

Hi, Melanie,

für die Wochennummer gibt es den ISO-Standard 8601. Dazu habe ich mal folgende Funktion gefunden (aber nicht getestet) ...

http://px.sklar.com/code.html?id=28

Grüße,

Sebastian

<?php
/* weeknumber.php

(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 Rohrich, [e-mail],
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;
}
?>