PHP-確定將來的日期

Question

我正在為一個班級寫一個應用程序，它要求我格式化一個預計到達日期。 理想的到達日期是從當天算起的5個工作日。 我什至不知道如何從當前日期開始准確顯示5天，更不用說沒有周末了。 任何有關該主題的幫助將不勝感激。

我嘗試使用以下功能，該功能在網站上的其他地方也可以看到，但我根本無法使該功能正常運行，即使只是將其粘貼到代碼中而不在任何地方使用都會產生錯誤消息。 我覺得它可能與strtotime（）函數有關，因為我無法在該應用程序的其他地方使用它。

另外，我只想避免使用下面的函數，這僅僅是因為我不了解其中涉及的一半。 理想情況下，我想使用我可以理解的功能。

<?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

您應該使用strtotime（）函數。

邏輯將如下所述：

• 以當前日期加上5天
• 如果該日期的天數是6（星期六）或0（星期日），則分別增加2天或1天

我在XAMPP上進行了測試，效果很好！

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

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