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php基础,该如何解决
- WBOYOriginal
- 2016-06-13 11:51:55802browse
php基础
编写一个函数，使它可以根据传入的指定年周号字符串(如201205)和偏移周数(如25表示后25周，或-25表示前25周)来返回新的年周号字符串.
------解决方案--------------------
date_default_timezone_set('Asia/Shanghai');
$timestamp = strtotime('2014 + 10 week -20 week');
echo '2014年第10周减20周的结果是：'.date('Y年W周',$timestamp);
echo '，timestamp是'.$timestamp;
?>
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