我想到的笨方法:$a = strtotime("19 Dec 2004");//19 Dec 2004可以改为任意一个已过去的星期的开始时间
$t = ceil((time() - $a) / 7 / 24 / 3600) - 1;
$starttime = $a + $t * 7 * 24 * 3600;
$endtime = $starttime + 7 * 24 * 3600;
$t = ceil((time() - $a) / 7 / 24 / 3600) - 1;
$starttime = $a + $t * 7 * 24 * 3600;
$endtime = $starttime + 7 * 24 * 3600;
$d = time(); //待处理的日期$w = date("w",$d); //这天是星期几
$d0 = date("Y-m-d",strtotime("-$w day",$d)); //周开始
$d6 = date("Y-m-d",strtotime((6-$w)." day",$d)); //周结束
?>
$d = time(); //待处理的日期
$w = date("w",$d); //这天是星期几 $d0 = mktime (0,0,0,date("m"),date("d") - $w,date("Y"));//周开始
$d0 = mktime (0,0,0,date("m"),date("d") - $w + 6,date("Y"));//周结束
?>
$d0 = date("Y-m-d",strtotime("-$w day",$date)); //周开始
$d6 = date("Y-m-d",strtotime((6-$w)." day",$date)); //周结束
$d0 = date("Y-m-d",strtotime("-$w day",$date)); //周开始
$d6 = date("Y-m-d",strtotime((6-$w)." day",$date)); //周结束
$date2 = date("Y-m-d",strtotime("Saturday")); //周结束
$date2 = date("Y-m-d",strtotime("last Saturday")); //上周结束