Question

我正在编写一个类的应用程序，它要求我格式化预期的到达日期。预计到达日期理想情况下是当天起的5个工作日。我甚至不知道如何让它在当天的5天内准确显示，更不用说在没有周末的情况下这样做了。任何有关该主题的帮助将不胜感激。

我尝试使用以下功能，我在网站的其他地方已经看过，但我根本无法使用该功能，即使只是在我的代码中粘贴它而不在任何地方使用它会产生错误信息。我觉得它可能与strtotime（）函数有关，因为我还没有能够在这个应用程序的其他地方使用它。

另外，如果可能的话，我想避免使用下面的功能，仅仅因为我不理解它涉及的一半。我理想地喜欢使用我能理解的功能。

<?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 1

你应该使用strtotime（）函数。

逻辑如下所述：

取当前日期并添加5天
如果该日期的日期编号为6（星期六）或0（星期日），则分别再加2或1天

我在XAMPP上进行了测试，效果非常好!!

function get_arrival_date($departure_date){

$arrival_date = date ('Y-m-d', strtotime ($departure_date. " +5 days"));

$day_number = date("w", strtotime($arrival_date));

switch($day_number)
    {
        case "6":
        $arrival_date = date ('Y-m-d', strtotime ($arrival_date. " +2 days"));
        break;

        case "0":
        $arrival_date = date ('Y-m-d', strtotime ($arrival_date. " +1 days"));
        break;
    }

return $arrival_date;

}

//End of function... Let's start with a little test

//First you set the departure date in a variable. 
//You can even supply this value directly as a string '2014-12-20' or 
//retrieve it from mysql database if you want

$date1 = date ('Y-m-d'); //Today (just and example)

//Then you call the function and assign its result to a variable

$date2= get_arrival_date($date1); //This variable holds the arrival date

echo $date2; //Let's print the arrival date

P.s.如果您觉得这个答案有用，请将其标记为正确答案。

PHP - 确定未来的日期

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