PHP计算某时间段内有几个周及某时间为今年第几周函数

date_default_timezone_set('PRC');
function count_weeks($startDate, $endDate = ''){//开始时间$startDate = date('Y-m-d', $startDate);//结束时间if(empty($endDate)) {$endDate = date('Y-m-d');}else{$endDate = date('Y-m-d', $endDate);}//跨越天数$n = (strtotime($endDate)-strtotime($startDate))/86400;//判断，跨度小于7天，可能是同一周，也可能是两周$endDate = date("Y-m-d",strtotime("$endDate +1 day"));if($n<7){//查开始时间 在 那周 的 位置$day            = date("w",strtotime($startDate))-1;//查开始时间  那周 的 周一$week_start        = date("Y-m-d",strtotime("$startDate -{$day} day"));//查开始时间  那周 的 周末$day            = 7-$day;$week_end        = date("Y-m-d",strtotime("$startDate +{$day} day"));//判断周末时间是否大于时间段的结束时间，如果大于，那就是时间段在同一周，否则时间段跨两周if($week_end>=$endDate){        $weekList[] =array('s'=>$startDate, 'e'=>date("Y-m-d",strtotime("$endDate -1 day")));}else{$weekList[] =array('s'=>$startDate, 'e'=>date("Y-m-d",strtotime("$week_end -1 day")));        $weekList[] =array('s'=>$week_end, 'e'=>date("Y-m-d",strtotime("$endDate -1 day")));    }}else{//如果跨度大于等于7天，可能是刚好1周或跨2周或跨N周，先找出开始时间 在 那周 的 位置和那周的周末时间$day         = date("w",strtotime($startDate))-1;$week_start  = date("Y-m-d",strtotime("$startDate -{$day} day"));$day         = 7-$day;$week_end    = date("Y-m-d",strtotime("$startDate +{$day} day"));//先把开始时间那周写入数组$weekList[]  =array('s'=>$startDate, 'e'=>date("Y-m-d",strtotime("$week_end -1 day"))); //判断周末是否大于等于结束时间，不管大于(2周)还是等于(1周)，结束时间都是时间段的结束时间。if($week_end >= $endDate){$weekList[] = array('s'=>$week_end, 'e'=>date("Y-m-d",strtotime("$endDate -1 day")));}else{//N周的情况用while循环一下，然后写入数组while($week_end <= $endDate){$start         = $week_end;$week_end    = date("Y-m-d",strtotime("$week_end +7 day"));if($week_end <= $endDate){$weekList[]  = array('s'=>$start, 'e'=>date("Y-m-d",strtotime("$week_end -1 day")));}else{$weekList[]  = array('s'=>$start, 'e'=>date("Y-m-d",strtotime("$endDate -1 day")));}}}}return $weekList;
}