PHP根据函数计算工作日的日期

我正在使用此函数（我在本论坛上找到）计算以下范围之间的工作日数：

<?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("$startdate","$enddate",$holidays)
?>

---

而

$startdate

是

2012-06-01

我想让这个函数计算startdate+20个工作日将是2012-06-28。这可能吗？

不久前，我使用下面的函数做了类似的事情。这里的关键是跳过周末，你也可以扩展到跳过假期

例如：

调用函数->

addDays（strotime（$startDate），20，$skipdays，$skipdates=array（））

---

[编辑]：刚刚阅读了一篇关于Netuts的精彩文章，希望这能有所帮助

如果有人感兴趣，我正在使用此功能为日期添加X个工作日。函数接受时间戳并返回时间戳。可以通过数组指定假日（如果在美国，可以使用

usBankHolidays（）

）

目前，假设周六和周日不是工作日，但这很容易改变

代码：

function addBusinessDays($date, $days, $holidays = array()) {
    $output = new DateTime();
    $output->setTimestamp($date);
    while ($days > 0) {
        $weekDay = $output->format('N');

        // Skip Saturday and Sunday
        if ($weekDay == 6 || $weekDay == 7) {
            $output = $output->add(new DateInterval('P1D'));
            continue;
        }

        // Skip holidays
        $strDate = $output->format('Y-m-d');
        foreach ($holidays as $s) {
            if ($s == $strDate) {
                $output = $output->add(new DateInterval('P1D'));
                continue 2;
            }
        }

        $days--;
        $output = $output->add(new DateInterval('P1D'));
    }
    return $output->getTimestamp();
}

function usBankHolidays($format = 'datesonly') {
    $output = array(
        array('2015-05-25', 'Memorial Day'),
        array('2015-07-03', 'Independence Day'),
        array('2015-09-07', 'Labor Day'),
        array('2015-10-12', 'Columbus Day'),
        array('2015-11-11', 'Veterans Day'),
        array('2015-11-26', 'Thanksgiving Day'),
        array('2015-12-25', 'Christmas Day'),
        array('2016-01-01', 'New Year Day'),
        array('2016-01-18', 'Martin Luther King Jr. Day'),
        array('2016-02-15', 'Presidents Day (Washingtons Birthday)'),
        array('2016-05-30', 'Memorial Day'),
        array('2016-07-04', 'Independence Day'),
        array('2016-09-05', 'Labor Day'),
        array('2016-10-10', 'Columbus Day'),
        array('2016-11-11', 'Veterans Day'),
        array('2016-11-24', 'Thanksgiving Day'),
        array('2016-12-25', 'Christmas Day'),
        array('2017-01-02', 'New Year Day'),
        array('2017-01-16', 'Martin Luther King Jr. Day'),
        array('2017-02-20', 'Presidents Day (Washingtons Birthday)'),
        array('2017-05-29', 'Memorial Day'),
        array('2017-07-04', 'Independence Day'),
        array('2017-09-04', 'Labor Day'),
        array('2017-10-09', 'Columbus Day'),
        array('2017-11-10', 'Veterans Day'),
        array('2017-11-23', 'Thanksgiving Day'),
        array('2017-12-25', 'Christmas Day'),
        array('2018-01-01', 'New Year Day'),
        array('2018-01-15', 'Martin Luther King Jr. Day'),
        array('2018-02-19', 'Presidents Day (Washingtons Birthday)'),
        array('2018-05-28', 'Memorial Day'),
        array('2018-07-04', 'Independence Day'),
        array('2018-09-03', 'Labor Day'),
        array('2018-10-08', 'Columbus Day'),
        array('2018-11-12', 'Veterans Day'),
        array('2018-11-22', 'Thanksgiving Day'),
        array('2018-12-25', 'Christmas Day'),
        array('2019-01-01', 'New Year Day'),
        array('2019-01-21', 'Martin Luther King Jr. Day'),
        array('2019-02-18', 'Presidents Day (Washingtons Birthday)'),
        array('2019-05-27', 'Memorial Day'),
        array('2019-07-04', 'Independence Day'),
        array('2019-09-02', 'Labor Day'),
        array('2019-10-14', 'Columbus Day'),
        array('2019-11-11', 'Veterans Day'),
        array('2019-11-28', 'Thanksgiving Day'),
        array('2019-12-25', 'Christmas Day'),
        array('2020-01-01', 'New Year Day'),
        array('2020-01-20', 'Martin Luther King Jr. Day'),
        array('2020-02-17', 'Presidents Day (Washingtons Birthday)'),
        array('2020-05-25', 'Memorial Day'),
        array('2020-07-03', 'Independence Day'),
        array('2020-09-07', 'Labor Day'),
        array('2020-10-12', 'Columbus Day'),
        array('2020-11-11', 'Veterans Day'),
        array('2020-11-26', 'Thanksgiving Day'),
        array('2020-12-25', 'Christmas Day '),
    );

    if ($format == 'datesonly') {
        $temp = array();
        foreach ($output as $item) {
            $temp[] = $item[0];
        }
        $output = $temp;
    }

    return $output;
}

用法：

$deliveryDate = addBusinessDays(time(), 7, usBankHolidays());

使用this.lau，我被缩写为反转它，用于减法日期，用于使用到期日期。希望这对某人有所帮助：

function subBusinessDays( $date, $days, $holidays = array() ) {
            $output = new DateTime();
            $output->setTimestamp( $date );

            while ( $days > 0 ) {

                $output = $output->sub( new DateInterval( 'P1D' ) );

                // Skip holidays
                $strDate = $output->format( 'Y-m-d' );
                if ( in_array( $strDate, $holidays ) ) {
                    // Skip Saturday and Sunday
                    $output = $output->sub( new DateInterval( 'P1D' ) );
                    continue;
                }

                $weekDay = $output->format( 'N' );
                if ($weekDay <= 5 ) {
                    $days --;
                }

            }

            return $output->getTimestamp();
        }

函数subBusinessDays（$date，$days，$holidays=array（））{
$output=新的日期时间（）；
$output->setTimestamp（$date）；
而（$days>0）{
$output=$output->sub（新的日期间隔（'P1D'））；
//跳过假期
$strDate=$output->format（'Y-m-d'）；
if（在数组中（$strDate，$holidays））{
//跳过周六和周日
$output=$output->sub（新的日期间隔（'P1D'））；
继续；
}
$weekDay=$output->format（'N'）；
if（$weekDay getTimestamp（）；
}

来自@this.lau uuu回答一个重写更简单、更逻辑的算法，带有动态（固定）holyday

使用

$deliveryDate = addBusinessDays(time(), 7);

---

我有一个更好的解决方案，函数上只有两个参数

只需调用函数

addDays($currentdate, $WordkingDatestoadd);

然后，使用下面的函数

function addDays($timestamp, $days) {

$workingDays = [1, 2, 3, 4, 5]; # date format = N (1 = Monday, ...)
$holidayDays = ['*-12-25', '*-01-01']; # variable and fixed holidays

$timestamp = new DateTime($timestamp);
$days = $days;

for($i=1 ; $i <= $days; $i++){
    $timestamp = $timestamp->modify('+1 day');
    if ((!in_array($timestamp->format('N'), $workingDays)) && (!in_array($timestamp->format('*-m-d'), $holidayDays))){
        $i--;
    }
}

$timestamp = $timestamp->format('Y-m-d');
return $timestamp;
}

function addDays（$timestamp，$days）{
$workingDays=[1,2,3,4,5]；#日期格式=N（1=周一，…）
$holidayDays=['*-12-25'，'*-01-01']；#可变假日和固定假日
$timestamp=新日期时间（$timestamp）；
$days=$days；
对于（$i=1；$i修改（“+1天”）；
如果（！in_数组（$timestamp->format（'N'），$workingDays））&（！in_数组（$timestamp->format（'*-m-d'），$holidayDays）））{
$i--；
}
}
$timestamp=$timestamp->format（'Y-m-d'）；
返回$timestamp；
}

在这里，我将HolidayAPI动态用于假日列表。HolidayAPI可以自由实现。我是孟加拉国人，这就是我使用国家代码BD的原因。 API链接-

//卷曲开始
$cURLConnection=curl_init（）；
curl_setopt（$cURLConnection，CURLOPT_URL，“您的api”）；
curl_setopt（$cURLConnection，CURLOPT_RETURNTRANSFER，true）；
$phoneList=curl\u exec（$cURLConnection）；
卷曲关闭（$卷曲连接）；
$dateResponse=json_decode（$phoneList）；
//卷曲端
$holidays=array（）；//清除假日数组`在此处输入代码`
对于（$i=0；$iholidays）；$i++）{
$holidays[]=$dateResponse->holidays[$i]->date；//从api在数组中插入假日
}
$startDate=“2020-02-26”；//插入您喜欢的值
$timestamp=strottime（$startDate）；//将日期转换为时间
$weekends=数组（“星期五”、“星期六”）；
$days=0；//初始化天数计数器
而第（1）款{//
$timestamp=STROTIME（“+1天，$timestamp）；//天增量
if（（在_数组中（日期（$l），时间戳，$weekends））|（在_数组中（日期（$Y-m-d，时间戳，$holidays））//检查周末和假日
{
继续；
}否则{
$days++``
如果（$days>=5）{//interval-5天后的工作日是多少
回显日期（“Y-m-d”，$timestamp）；//打印预期输出
打破
}
}
}

输入：2020-02-26

---

输出：2020-03-05

你能再解释一下如何使用它吗？我对PHP函数基本上是新手。调用函数->addDays（strotime（$startDate），20，$skipdays，$skipdates=NULL）；如果您发送$skipDays和$skipDates的空VAL就可以了，在这种情况下，由defauld Sat、sun发送，不会跳过任何额外的日期。我如何将时间戳内容Y转换为实际的Y-M-D日期？是的，我是一个新手。$startdate作为变量似乎有一些问题。。如果我手动使用

StrotTime（2012-06-01），它会起作用

但是

strottime（$startdate）

不要这样做。即使我

echo$startdate；

它输出正确的日期…杂耍问题？你能做一个快速测试吗，你能做类型转换吗？do:$startdate=（字符串）$startdate；echo strottime（$startdate）；尝试使所有假期都动态：）foreach$holidays中有一个bug。continue在foreach范围内，而不是while范围内，您必须使用continue 2。不客气，我喜欢您的算法，我对它进行了一些小的修改。谢谢。这看起来是一个学习TDD的完美例子：）

$deliveryDate = addBusinessDays(time(), 7);

addDays($currentdate, $WordkingDatestoadd);

function addDays($timestamp, $days) {

$workingDays = [1, 2, 3, 4, 5]; # date format = N (1 = Monday, ...)
$holidayDays = ['*-12-25', '*-01-01']; # variable and fixed holidays

$timestamp = new DateTime($timestamp);
$days = $days;

for($i=1 ; $i <= $days; $i++){
    $timestamp = $timestamp->modify('+1 day');
    if ((!in_array($timestamp->format('N'), $workingDays)) && (!in_array($timestamp->format('*-m-d'), $holidayDays))){
        $i--;
    }
}

$timestamp = $timestamp->format('Y-m-d');
return $timestamp;
}

// cURL start
$cURLConnection = curl_init();
curl_setopt($cURLConnection, CURLOPT_URL, 'your api');
curl_setopt($cURLConnection, CURLOPT_RETURNTRANSFER, true);
$phoneList = curl_exec($cURLConnection);
curl_close($cURLConnection);
$dateResponse = json_decode($phoneList);
// cURL end

$holidays = array(); //declear holidays array`enter code here`
for ($i=0;$i<count($dateResponse->holidays);$i++){
    $holidays[] = $dateResponse->holidays[$i]->date;  //inserting holidays in array from api
}

$startDate = "2020-02-26";  //insert value thats you prefer
$timestamp  = strtotime($startDate);  // convert date to time
$weekends = array("Friday", "Saturday");
$days = 0;  //initialization days counter

while (1){  //
    $timestamp = strtotime("+1 day", $timestamp); //day increment
    if ( (in_array(date("l", $timestamp), $weekends)) || (in_array(date("Y-m-d", $timestamp), $holidays)) ) //checking weekends and holidays
    {
        continue;
    }else{
        $days++;``
        if($days >= 5){  //interval - What will be the working days after 5 days
            echo date("Y-m-d", $timestamp); // print expected output
            break;
        }
    }
}