给定一个星期数,例如date -u +%W,您如何计算从星期一开始的那一周中的天数?
date -u +%W
第40周的rfc-3339输出示例:
2008-10-06 2008-10-07 2008-10-08 2008-10-09 2008-10-10 2008-10-11 2008-10-12
PHP
$week_number = 40; $year = 2008; for($day=1; $day<=7; $day++) { echo date('m/d/Y', strtotime($year."W".$week_number.$day))."\n"; }
下面的帖子是因为我是个白痴,没有正确阅读问题,但是会获得从星期一开始的一周中的日期,给出的是日期,而不是星期数。
在PHP中 ,改编自PHP日期手册页上的这篇文章
function week_from_monday($date) { // Assuming $date is in format DD-MM-YYYY list($day, $month, $year) = explode("-", $_REQUEST["date"]); // Get the weekday of the given date $wkday = date('l',mktime('0','0','0', $month, $day, $year)); switch($wkday) { case 'Monday': $numDaysToMon = 0; break; case 'Tuesday': $numDaysToMon = 1; break; case 'Wednesday': $numDaysToMon = 2; break; case 'Thursday': $numDaysToMon = 3; break; case 'Friday': $numDaysToMon = 4; break; case 'Saturday': $numDaysToMon = 5; break; case 'Sunday': $numDaysToMon = 6; break; } // Timestamp of the monday for that week $monday = mktime('0','0','0', $month, $day-$numDaysToMon, $year); $seconds_in_a_day = 86400; // Get date for 7 days from Monday (inclusive) for($i=0; $i<7; $i++) { $dates[$i] = date('Y-m-d',$monday+($seconds_in_a_day*$i)); } return $dates; }
来自的输出week_from_monday('07-10-2008'):
week_from_monday('07-10-2008')
Array ( [0] => 2008-10-06 [1] => 2008-10-07 [2] => 2008-10-08 [3] => 2008-10-09 [4] => 2008-10-10 [5] => 2008-10-11 [6] => 2008-10-12 )