Php将5个工作日添加到当前日期，不包括周末（周六-周日）和（多个）假日

为了交付我们的网店，我们需要从php中的当前日期算起5个工作日

我们的工作日是从周一到周五，我们有几天的休息日，假期也不包括在内

我找到了这个脚本，但不包括假期

<?php 

    $_POST['startdate'] = date("Y-m-d");
    $_POST['numberofdays'] = 5;

    $d = new DateTime( $_POST['startdate'] );
    $t = $d->getTimestamp();

    // loop for X days
    for($i=0; $i<$_POST['numberofdays']; $i++){

        // add 1 day to timestamp
        $addDay = 86400;

        // get what day it is next day
        $nextDay = date('w', ($t+$addDay));

        // if it's Saturday or Sunday get $i-1
        if($nextDay == 0 || $nextDay == 6) {
            $i--;
        }

        // modify timestamp, add 1 day
        $t = $t+$addDay;
    }

    $d->setTimestamp($t);

    echo $d->format('Y-m-d'). "\n";

?>

---

您可以使用while语句，循环直到获得足够的5天。每次循环get&check，第二天是否在假日列表中

以下是示例：

$holidayDates = array(
    '2016-03-26',
    '2016-03-27',
    '2016-03-28',
    '2016-03-29',
    '2016-04-05',
);

$count5WD = 0;
$temp = strtotime("2016-03-25 00:00:00"); //example as today is 2016-03-25
while($count5WD<5){
    $next1WD = strtotime('+1 weekday', $temp);
    $next1WDDate = date('Y-m-d', $next1WD);
    if(!in_array($next1WDDate, $holidayDates)){
        $count5WD++;
    }
    $temp = $next1WD;
}

$next5WD = date("Y-m-d", $temp);

echo $next5WD; //if today is 2016-03-25 then it will return 2016-04-06 as many days between are holidays

---

基于Tinh Dang答案的函数：

function getFutureBusinessDay($num_business_days, $today_ymd = null, $holiday_dates_ymd = []) {
    $num_business_days = min($num_business_days, 1000);
    $business_day_count = 0;
    $current_timestamp = empty($today_ymd) ? time() : strtotime($today_ymd);
    while ($business_day_count < $num_business_days) {
        $next1WD = strtotime('+1 weekday', $current_timestamp);
        $next1WDDate = date('Y-m-d', $next1WD);        
        if (!in_array($next1WDDate, $holiday_dates_ymd)) {
            $business_day_count++;
        }
        $current_timestamp = $next1WD;
    }
    return date('Y-m-d', $current_timestamp);
}

---

我将循环限制为1000个工作日。如果需要，没有限制。

根据卢克的回答：

function getFutureBusinessDay($num_business_days, $today_ymd = null, $holiday_dates_ymd = []) {
    $num_business_days = min($num_business_days, 1000);
    $business_day_count = 0;
    $current_timestamp = empty($today_ymd) ? time() : strtotime($today_ymd);
    while ($business_day_count < $num_business_days) {
        $next1WD = strtotime('+1 weekday', $current_timestamp);
        $next1WDDate = date('Y-m-d', $next1WD);        
        if (!in_array($next1WDDate, $holiday_dates_ymd)) {
            $business_day_count++;
        }
        $current_timestamp = $next1WD;
    }
    return date('Y-m-d', $current_timestamp);
}

不同的是，这一项每年都会产生假期

<?php
class DateHelper
{
    //change at will
    const HOLIDAY_DATES = [
        ['day' => 25, 'month' => 12],//christimas
        ['day' => 1, 'month' => 1],//new year
        ['day' => 13, 'month' => 4]//easter
    ];
    /**
     * @param int $numBusinessDays
     * @param \DateTimeInterface $date
     * @return \DateTime
     */
    public static function getFutureBusinessDay(int $numBusinessDays, \DateTimeInterface $date)
    {
        $numBusinessDays = min($numBusinessDays, 1000);
        $businessDayCount = 0;
        $currentTimestamp = strtotime($date->format('Y-m-d'));
        $holidayDates = self::getHolidayDates();
        while ($businessDayCount < $numBusinessDays) {
            $next1WD = strtotime('+1 weekday', $currentTimestamp);
            $next1WDDate = date('Y-m-d', $next1WD);
            if (!in_array($next1WDDate, $holidayDates)) {
                $businessDayCount++;
            }
            $currentTimestamp = $next1WD;
        }
        return (new \DateTime())->setTimestamp($currentTimestamp);
    }

    /**
     * @return array
     */
    private static function getHolidayDates()
    {
        $holidays = [];
        foreach (self::HOLIDAY_DATES as $holidayDate) {
            $date = new \DateTime();
            $date->setDate($date->format('Y'), $holidayDate['month'], $holidayDate['day']);
            $holidays[] = $date->format('Y-m-d');
        }
        return $holidays;
    }
}

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如果不尝试编码，您将无法获得。可能会重复感谢。我将试着把它放在一起，昨天我必须学习基本的php；我们的开发者病了…，希望它能工作：我不能让它工作；我必须做一个循环吗？非常感谢！解决了我的问题：