PHP根据工作日的函数计算日期

时间:2012-06-21 01:24:32

标签: php function date

我正在使用此功能(我在此论坛上找到)来计算一个范围之间的工作天数:

<?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)
?>

现在我想扩展一下这个功能。如果我从开始日期开始添加X个工作日,我想生成什么日期。比方说,我有一个保持值 20

的变量
$workingdays = "20";

$startdate2012-06-01我想让这个函数计算起始日期+20个工作日将是2012-06-28。这有可能吗?

6 个答案:

答案 0 :(得分:8)

我使用下面的功能做了类似的事情。这里的关键是跳过周末,你可以延长这个以跳过假期。

示例:

调用函数 - &gt; addDays(strtotime($startDate), 20, $skipdays,$skipdates = array())

 <?php
    function addDays($timestamp, $days, $skipdays = array("Saturday", "Sunday"), $skipdates = NULL) {
        // $skipdays: array (Monday-Sunday) eg. array("Saturday","Sunday")
        // $skipdates: array (YYYY-mm-dd) eg. array("2012-05-02","2015-08-01");
       //timestamp is strtotime of ur $startDate
        $i = 1;

        while ($days >= $i) {
            $timestamp = strtotime("+1 day", $timestamp);
            if ( (in_array(date("l", $timestamp), $skipdays)) || (in_array(date("Y-m-d", $timestamp), $skipdates)) )
            {
                $days++;
            }
            $i++;
        }

        return $timestamp;
        //return date("m/d/Y",$timestamp);
    }
    ?>

[编辑]:刚刚阅读了一篇关于nettuts的精彩文章,希望这有助于http://net.tutsplus.com/tutorials/php/dates-and-time-the-oop-way/

答案 1 :(得分:2)

如果有人感兴趣,我可以使用此功能将X个工作日添加到某个日期。该函数接受时间戳并返回时间戳。可以通过数组指定假期(如果在美国,您可以使用usBankHolidays())。

目前,它假设周六和周日不是工作日,但可以轻松更改。

<强>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;
}

<强>用法:

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

答案 2 :(得分:1)

来自@ this.lau_回答一个重写更简单,更合理的算法,动态(固定)瞻礼

public function addBusinessDays($date, $days) {

    $output = new DateTime();
    $output->setTimestamp($date);

    while ($days > 0) {

        $output = $output->add(new DateInterval('P1D'));
        $weekDay = $output->format('N');
        $strDate = $output->format('Y-m-d');

        // Skip Saturday and Sunday
        if ($weekDay == 6 || $weekDay == 7) {

            continue;

        }

        // Skip holidays
        $holidays = $this->_getHolidays();            

        foreach ($holidays as $holiday_date => $holiday_name) {

            if ($holiday_date == $strDate) {

                continue 2;

            }

        }

        $days--;

    }

    return $output->getTimestamp();

}



public function _getHolidays() {

    $feste = array(
        date("Y") . "-01-01" => "Capodanno", 
        date("Y") . "-01-06" => "Epifania", 
        date("Y") . "-04-25" => "Liberazione", 
        date("Y") . "-05-01" => "Festa Lavoratori", 
        date("Y") . "-06-02" => "Festa della Repubblica", 
        date("Y") . "-08-15" => "Ferragosto", 
        date("Y") . "-11-01" => "Tutti Santi", 
        date("Y") . "-12-08" => "Immacolata", 
        date("Y") . "-12-25" => "Natale", 
        date("Y") . "-12-26" => "St. Stefano"
    );

    return $feste;

}

使用

调用该函数
$deliveryDate = addBusinessDays(time(), 7);

答案 3 :(得分:0)

使用this.lau_ function,对于减法日期,我无法撤销它,以便使用到期日期。希望这有助于某人:

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();
        }

答案 4 :(得分:0)

在函数上只有两个参数的情况下,我有一个更好的解决方案。

只需调用函数

addDays($currentdate, $WordkingDatestoadd);

然后,使用以下功能

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;
}

答案 5 :(得分:0)

这里我动态地使用HolidayAPI 来获取假期列表。HolidayAPI 可以免费实现。我是孟加拉国人,这就是我使用国家/地区代码 BD 的原因。 API 链接 - https://holidayapi.com/v1/holidays?pretty&key=your_key&country=your_country_code&year=2020

// 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;
        }
    }
}

输入:2020-02-26 输出:2020-03-05