PHP根据工作日的函数计算日期

Question

我正在使用此功能(我在此论坛上找到)来计算范围之间的工作天数:

<?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("$startdate","$enddate",$holidays)
?>

现在我想扩展一下这个功能.如果我从开始日期开始添加X个工作日,我想生成什么日期.比方说,我有一个保持值20的变量

$workingdays = "20";

而$startdate就是2012-06-01我想使这个函数计算的开始日期+ 20个工作日内将是2012-06-28.这可能吗？

Answer 1

我使用下面的功能做了类似的事情.这里的关键是跳过周末,你可以延长这个以跳过假期.

例:

调用函数 - > addDays(strtotime($startDate), 20, $skipdays,$skipdates = array())

 <?php
    function addDays($timestamp, $days, $skipdays = array("Saturday", "Sunday"), $skipdates = NULL) {
        // $skipdays: array (Monday-Sunday) eg. array("Saturday","Sunday")
        // $skipdates: array (YYYY-mm-dd) eg. array("2012-05-02","2015-08-01");
       //timestamp is strtotime of ur $startDate
        $i = 1;

        while ($days >= $i) {
            $timestamp = strtotime("+1 day", $timestamp);
            if ( (in_array(date("l", $timestamp), $skipdays)) || (in_array(date("Y-m-d", $timestamp), $skipdates)) )
            {
                $days++;
            }
            $i++;
        }

        return $timestamp;
        //return date("m/d/Y",$timestamp);
    }
    ?>

[编辑]:刚刚阅读了一篇关于nettuts的精彩文章,希望这有助于http://net.tutsplus.com/tutorials/php/dates-and-time-the-oop-way/