PHP-确定未来的日期

我正在为一个班级写一份申请书，它要求我格式化一个预期的到达日期。理想情况下，预计到达日期为当天起5个工作日。我甚至不知道如何使它精确显示从当前日期起的5天，更不用说在没有周末的情况下这样做了。在此主题上的任何帮助都将不胜感激

我尝试过使用下面的函数，我在网站的其他地方看到过，但是我根本无法让这个函数运行，即使只是将它粘贴到我的代码中，而不在任何地方使用它，也会产生错误消息。我觉得它可能与strotime（）函数有关，因为在这个应用程序的其他地方我还不能使用它

此外，如果可能的话，我想避免使用下面的函数，因为我不了解它所涉及的一半内容。理想情况下，我希望使用一个我能理解的函数

<?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("2008-12-22","2009-01-02",$holidays)
// => will return 7
?>

---

您应该使用strotime（）函数

逻辑如下所述：


取当前日期并添加5天
如果该日期的天数为6（星期六）或0（星期日），则分别再增加2天或1天

我在XAMPP上测试了它，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

注：如果您觉得这个答案有用，请将其标记为正确答案。
我甚至不知道如何使其准确显示从当前日期起的5天。
非常简单<代码>$timestamp=strottime（“+5天”）或查看您是否需要更精确地了解“工作日”是什么？确切地说是星期一到星期五，还是必须考虑假期等等。