Question

我在线搜索了这段代码。这是关于获得本月的工作日（减去周末假期的数量（仅限星期日）和当月的公众假期数量。）

例如，我的工作日是星期一到星期六，但本月有3个公众假期，分别是6月24日（星期六），25日（星期日）和26日（星期一）。据说我的工作日是24.但是，使用此代码，

{<?php 
function 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 = $no_full_weeks * 6;
    if ($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:
$holidays=array("2017-06-25","2017-06-24","2017-06-26");
echo getWorkingDays("2017-06-01","2017-06-30",$holidays)
// => will return 7
?>}

我得到额外的工作日 - ＆gt; （24 + 1）天。我认为6月24日（星期六）的公众假期被视为工作日，计划不会将其视为假期。

感谢您的帮助，非常感谢。

Answer 1

如果你想跳过星期六，那么只需将它从假日计数条件中删除

<?php
//If the holiday doesn't fall in weekend 
if ($startDate <= $time_stamp && $time_stamp <= $endDate && date("N",$time_stamp) != 7 )
    $workingDays--;

PHP中的周末假期和公共假期问题

1 个答案: