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我使用此功能（我在這個論壇上找到）來計算範圍之間的工作日天數
<?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)
?>
現在，我想稍微擴展一下這個功能。如果我從開始日期開始添加X個工作日，我想要生成日期。例如說我有保存價值
$workingdays = "20";
一個變量和$startdate是2012-06-01我想使這個函數計算的開始日期+ 20個工作日內將是2012-06-28。這可能嗎？
的功能你請解釋一下如何使用它？一般來說，我對PHP函數都很陌生。 – David
調用函數 - > addDays（strtotime（$ startDate），20，$ skipdays，$ skipdates = NULL）; 它確定如果您發送$ val skipDays和$ skipDays的空vals，在這種情況下通過默認星期六，太陽和沒有額外的日期將被跳過 –
如何將時間戳thingy轉換爲實際的Y-M-D日期？是的，我是一個新手.. – David