我正在为一个类编写应用程序，它要求我格式化预期到达日期。理想情况下，预计到达日期为当天起5个工作日。我甚至不知道如何让它准确地显示当前日期后的5天，更不用说在没有周末的情况下这样做了。如有任何关于该主题的帮助，我们将不胜感激。
我尝试过使用以下函数，我在网站上的其他地方也看到过，但我根本无法使该函数运行，即使只是将其粘贴在代码中，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strtotime（）函数有关，因为我还没能在这个应用程序的其他地方使用它。
此外，如果可能的话，我想尽量避免使用下面的函数，因为我不明白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
- 如果该日期的天数是6（星期六）或0（星期日），则分别增加2天或1天
我在XAMPP上测试了它，它运行得很完美！！
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
附言：如果你觉得这个答案有用，请把它标记为正确的答案。