如何计算一个月没有周六周日的天数

Question

我正在使用它来动态查找一个月中的总天数

$count = cal_days_in_month(CAL_GREGORIAN, $_POST['PayMonth'], $_POST['PayYear']);

但是我如何计算那个特定月份离开周六和周日的天数？

Answer 1

function countDays($year, $month, $ignore) {
$count = 0;
$counter = mktime(0, 0, 0, $month, 1, $year);
while (date("n", $counter) == $month) {
    if (in_array(date("w", $counter), $ignore) == false) {
        $count++;
    }
    $counter = strtotime("+1 day", $counter);
}
return $count;  }echo countDays(2013, 1, array(0, 6)); // 23

参考：link1 和 link2

Answer 2

这是用户 cmets 在 PHP 手册中 date() 函数页面上的一个函数。这是对 cme​​ts 中早期功能的改进，增加了对闰年的支持。

输入开始日期和结束日期，以及可能介于两者之间的任何假期的数组，并将工作日作为整数返回：

<?php
//The function returns the no. of business days between two dates and it skips the holidays
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 = $no_full_weeks * 5;
    if ($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
?>

Answer 3

这是我用来计算一个月没有周末的天数：

$nbdayinmonth = cal_days_in_month(CAL_GREGORIAN, $_POST['PayMonth'], $_POST['PayYear']);

$nbday = 0;
for ($i = 1; $i <= $nbdayinmonth; $i++) {
    $weekday = (int)date("w", strtotime($i . "-" . $_POST['PayMonth'] . "-" . $_POST['PayYear']));
    if ($weekday > 0 && $weekday < 6)
        $nbday++;
}
echo $nbday;