这是PHP手册中date（）函数页面上用户注释的函数。它是对注释中早期功能的改进，增加了对leap年的支持。
输入开始和结束日期，以及可能在其间的任何假日的数组，它以整数形式返回工作日：
<?php//The function returns the no. of business days between two dates and it skips the holidaysfunction getWorkingDays($startDate,$endDate,$holidays){ // do strtotime calculations just once $endDate = strtotime($endDate); $startDate = strtotime($startDate); //The total number of days between the two dates. We compute the no. of seconds and divide it to 60*60*24 //We add one to inlude both dates in the interval. $days = ($endDate - $startDate) / 86400 + 1; $no_full_weeks = floor($days / 7); $no_remaining_days = fmod($days, 7); //It will return 1 if it's Monday,.. ,7 for Sunday $the_first_day_of_week = date("N", $startDate); $the_last_day_of_week = date("N", $endDate); //---->The two can be equal in leap years when february has 29 days, the equal sign is added here //In the first case the whole interval is within a week, in the second case the interval falls in two weeks. if ($the_first_day_of_week <= $the_last_day_of_week) { if ($the_first_day_of_week <= 6 && 6 <= $the_last_day_of_week) $no_remaining_days--; if ($the_first_day_of_week <= 7 && 7 <= $the_last_day_of_week) $no_remaining_days--; } else { // (edit by Tokes to fix an edge case where the start day was a Sunday // and the end day was NOT a Saturday) // the day of the week for start is later than the day of the week for end if ($the_first_day_of_week == 7) { // if the start date is a Sunday, then we definitely subtract 1 day $no_remaining_days--; if ($the_last_day_of_week == 6) { // if the end date is a Saturday, then we subtract another day $no_remaining_days--; } } else { // the start date was a Saturday (or earlier), and the end date was (Mon..Fri) // so we skip an entire weekend and subtract 2 days $no_remaining_days -= 2; } } //The no. of business days is: (number of weeks between the two dates) * (5 working days) + the remainder//---->february in none leap years gave a remainder of 0 but still calculated weekends between first and last day, this is one way to fix it $workingDays = $no_full_weeks * 5; if ($no_remaining_days > 0 ) { $workingDays += $no_remaining_days; } //We subtract the holidays foreach($holidays as $holiday){ $time_stamp=strtotime($holiday); //If the holiday doesn't fall in weekend if ($startDate <= $time_stamp && $time_stamp <= $endDate && date("N",$time_stamp) != 6 && date("N",$time_stamp) != 7) $workingDays--; } return $workingDays;}//Example:$holidays=array("2008-12-25","2008-12-26","2009-01-01");echo getWorkingDays("2008-12-22","2009-01-02",$holidays)// => will return 7?>
