我需要一种在PHP中添加“工作日”的方法。例如,星期五12/5 + 3个工作日=星期三12/10。

我至少需要使用代码来理解周末,但理想情况下,它也应考虑美国联邦假日。我敢肯定,如有必要,我可以提出一种蛮力的解决方案,但我希望那里有一种更优雅的方法。有人吗?

谢谢。

评论

我为此创建了一个不错的库。 github.com/andrejsstepanovs/business-days-calculator它稳定且可以投入生产。

哦,我认为应该提到的是,如今,我们可以使用DateTime :: modify函数立即添加工作日:$ my_date = new \ DateTime(); $ my_date-> modify(“ + 7个工作日”);只会表现得很差劲。

详细博客:goo.gl/YOsfPX

一个更简单/更干净的答案:stackoverflow.com/questions/5532002/…

#1 楼

这是PHP手册中date()函数页面上用户注释的函数。它是对注释中早期功能的改进,增加了对leap年的支持。

输入开始日期和结束日期,以及可能在其间的任何假日数组,它会返回工作天为整数:

<?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
?>


评论


该函数期望开始和结束日期,如果您有开始日期,并且希望结果是给定日期x个工作日,该怎么办?

–mcgrailm
2010年5月18日在12:15

@mcgrailm:这是一个相似的想法,但是您可能想编写第二个函数,因为参数和返回值被交换了。就像是((X天%每周5天)*每个周末2天)+ X天+开始日期和结束日期的一周中的天数+假期)。

–flamingLogos
2010年5月20日下午3:32

@mcgrailm:刚刚找到了这个问题,它的答案可能会为您指明正确的方向:stackoverflow.com/questions/2681787/…。

–flamingLogos
2010年5月20日,3:46

此功能有一个错误。如果两个日期之间的时区发生变化怎么办?例如,CEST:echo getWorkingDays(“ 2012-01-01”,“ 2012-05-01”,$ holidays);不会给你整数。 stackoverflow.com/questions/12490521/…

–mnowotka
2012-09-19 7:58



那是做这件事的老方法。尽可能使用Glavić答案

–托马斯·鲁伊斯(Thomas Ruiz)
2014年6月30日10:42

#2 楼

获取两个日期之间没有假期的工作日数:

使用示例:

 echo number_of_working_days('2013-12-23', '2013-12-29');
 


输出:

 3
 


功能:

 function number_of_working_days($from, $to) {
    $workingDays = [1, 2, 3, 4, 5]; # date format = N (1 = Monday, ...)
    $holidayDays = ['*-12-25', '*-01-01', '2013-12-23']; # variable and fixed holidays

    $from = new DateTime($from);
    $to = new DateTime($to);
    $to->modify('+1 day');
    $interval = new DateInterval('P1D');
    $periods = new DatePeriod($from, $interval, $to);

    $days = 0;
    foreach ($periods as $period) {
        if (!in_array($period->format('N'), $workingDays)) continue;
        if (in_array($period->format('Y-m-d'), $holidayDays)) continue;
        if (in_array($period->format('*-m-d'), $holidayDays)) continue;
        $days++;
    }
    return $days;
}
 


评论


我投票赞成,因为它可以工作并且有一些很大的改进:)但是,您至少应该提到@Suresh Kamrushi,他在一个月前发布了大部分代码。 :)

– KOGI
15年1月20日在20:37

如果工作日总是星期一至星期五,则可以用DateInterval和workingdays数组代替$ interval = DateInterval :: createFromFormat('1 weekday');。我还建议使用$ holidays = array_flip($ holidays);在foreach和if isset之前($ holidays [$ period-> format('Y-m-d')]);减少每次迭代所需的处理时间。但是建议为假期创建一个自定义函数,以便能够处理相对的假期,例如11月的最后一个星期四的感恩节或9月的第一个星期一的劳动节。

– B
2015年2月25日在21:09



可以将其扩展为包括耶稣受难日(我似乎记得是3月21日之后的第一个满月之后的第一个星期日之前的星期五)吗?

–达米安·耶里克(Damian Yerrick)
15年7月21日在16:45

请注意不同的时区。在欧洲/柏林,这是行不通的。我现在只是在带有date('N')的时间戳上进行此操作,并添加1day及其时区独立。

–user3655829
19年7月10日在10:38

#3 楼

对于date()函数,有些参数应该会有所帮助。如果您选择date(“ w”),它将为您提供星期几的数字,从周日的0到周六的6。所以..也许像..

$busDays = 3;
$day = date("w");
if( $day > 2 && $day <= 5 ) { /* if between Wed and Fri */
  $day += 2; /* add 2 more days for weekend */
}
$day += $busDays;


这只是一种可能性的粗略示例。.

#4 楼

假期计算在每个州都是不标准的。我正在编写一个银行应用程序,我需要一些严格的业务规则,但仍然只能获得一个大致的标准。

/**
 * National American Holidays
 * @param string $year
 * @return array
 */
public static function getNationalAmericanHolidays($year) {


    //  January 1 - New Year’s Day (Observed)
    //  Calc Last Monday in May - Memorial Day  strtotime("last Monday of May 2011");
    //  July 4 Independence Day
    //  First monday in september - Labor Day strtotime("first Monday of September 2011")
    //  November 11 - Veterans’ Day (Observed)
    //  Fourth Thursday in November Thanksgiving strtotime("fourth Thursday of November 2011");
    //  December 25 - Christmas Day        
    $bankHolidays = array(
          $year . "-01-01" // New Years
        , "". date("Y-m-d",strtotime("last Monday of May " . $year) ) // Memorial Day
        , $year . "-07-04" // Independence Day (corrected)
        , "". date("Y-m-d",strtotime("first Monday of September " . $year) ) // Labor Day
        , $year . "-11-11" // Veterans Day
        , "". date("Y-m-d",strtotime("fourth Thursday of November " . $year) ) // Thanksgiving
        , $year . "-12-25" // XMAS
        );

    return $bankHolidays;
}


评论


独立日是$ year。'-07-04'(7月4日),而不是6月4日

–solepixel
13年4月22日在19:13

周末的假期在前一天(如果是星期六)或后一天(如果是星期日)进行观察。我能够使用您的数组来获取实际的观测天数。谢谢。

–BrookeAH
17年11月30日在16:20

#5 楼

$startDate = new DateTime( '2013-04-01' );    //intialize start date
$endDate = new DateTime( '2013-04-30' );    //initialize end date
$holiday = array('2013-04-11','2013-04-25');  //this is assumed list of holiday
$interval = new DateInterval('P1D');    // set the interval as 1 day
$daterange = new DatePeriod($startDate, $interval ,$endDate);
foreach($daterange as $date){
if($date->format("N") <6 AND !in_array($date->format("Y-m-d"),$holiday))
$result[] = $date->format("Y-m-d");
}
echo "<pre>";print_r($result);


#6 楼

这是一个为日期添加商务日的功能

 function add_business_days($startdate,$buisnessdays,$holidays,$dateformat){
  $i=1;
  $dayx = strtotime($startdate);
  while($i < $buisnessdays){
   $day = date('N',$dayx);
   $date = date('Y-m-d',$dayx);
   if($day < 6 && !in_array($date,$holidays))$i++;
   $dayx = strtotime($date.' +1 day');
  }
  return date($dateformat,$dayx);
 }

 //Example
 date_default_timezone_set('Europe\London');
 $startdate = '2012-01-08';
 $holidays=array("2012-01-10");
 echo '<p>Start date: '.date('r',strtotime( $startdate));
 echo '<p>'.add_business_days($startdate,7,$holidays,'r');


另一篇文章提到getWorkingDays(来自php.net注释,并包含在此处),但我认为如果您开始,它会中断

使用以下内容(您需要包括上一篇文章的getWorkingDays函数)

 date_default_timezone_set('Europe\London');
 //Example:
 $holidays = array('2012-01-10');
 $startDate = '2012-01-08';
 $endDate = '2012-01-13';
 echo getWorkingDays( $startDate,$endDate,$holidays);


给出结果为5而不是4

Sun, 08 Jan 2012 00:00:00 +0000 weekend
Mon, 09 Jan 2012 00:00:00 +0000
Tue, 10 Jan 2012 00:00:00 +0000 holiday
Wed, 11 Jan 2012 00:00:00 +0000
Thu, 12 Jan 2012 00:00:00 +0000
Fri, 13 Jan 2012 00:00:00 +0000 


以下函数用于生成上述结果。

     function get_working_days($startDate,$endDate,$holidays){
      $debug = true;
      $work = 0;
      $nowork = 0;
      $dayx = strtotime($startDate);
      $endx = strtotime($endDate);
      if($debug){
       echo '<h1>get_working_days</h1>';
       echo 'startDate: '.date('r',strtotime( $startDate)).'<br>';
       echo 'endDate: '.date('r',strtotime( $endDate)).'<br>';
       var_dump($holidays);
       echo '<p>Go to work...';
      }
      while($dayx <= $endx){
       $day = date('N',$dayx);
       $date = date('Y-m-d',$dayx);
       if($debug)echo '<br />'.date('r',$dayx).' ';
       if($day > 5 || in_array($date,$holidays)){
        $nowork++;
     if($debug){
      if($day > 5)echo 'weekend';
      else echo 'holiday';
     }
       } else $work++;
       $dayx = strtotime($date.' +1 day');
      }
      if($debug){
      echo '<p>No work: '.$nowork.'<br>';
      echo 'Work: '.$work.'<br>';
      echo 'Work + no work: '.($nowork+$work).'<br>';
      echo 'All seconds / seconds in a day: '.floatval(strtotime($endDate)-strtotime($startDate))/floatval(24*60*60);
      }
      return $work;
     }

    date_default_timezone_set('Europe\London');
     //Example:
     $holidays=array("2012-01-10");
     $startDate = '2012-01-08';
     $endDate = '2012-01-13';
//broken
     echo getWorkingDays( $startDate,$endDate,$holidays);
//works
     echo get_working_days( $startDate,$endDate,$holidays);


放假...

#7 楼

您可以尝试更简单的此功能。

function getWorkingDays($startDate, $endDate)
{
    $begin = strtotime($startDate);
    $end   = strtotime($endDate);
    if ($begin > $end) {

        return 0;
    } else {
        $no_days  = 0;
        while ($begin <= $end) {
            $what_day = date("N", $begin);
            if (!in_array($what_day, [6,7]) ) // 6 and 7 are weekend
                $no_days++;
            $begin += 86400; // +1 day
        };

        return $no_days;
    }
}


#8 楼

从给定日期增加或减少工作日的功能,这不考虑假期。

function dateFromBusinessDays($days, $dateTime=null) {
  $dateTime = is_null($dateTime) ? time() : $dateTime;
  $_day = 0;
  $_direction = $days == 0 ? 0 : intval($days/abs($days));
  $_day_value = (60 * 60 * 24);

  while($_day !== $days) {
    $dateTime += $_direction * $_day_value;

    $_day_w = date("w", $dateTime);
    if ($_day_w > 0 && $_day_w < 6) {
      $_day += $_direction * 1; 
    }
  }

  return $dateTime;
}


像这样使用...

echo date("m/d/Y", dateFromBusinessDays(-7));
echo date("m/d/Y", dateFromBusinessDays(3, time() + 3*60*60*24));


#9 楼

我的版本基于@mcgrailm的工作...进行了调整,因为该报告需要在3个工作日内进行审核,并且如果在周末提交,则从下周一开始计数:

function business_days_add($start_date, $business_days, $holidays = array()) {
    $current_date = strtotime($start_date);
    $business_days = intval($business_days); // Decrement does not work on strings
    while ($business_days > 0) {
        if (date('N', $current_date) < 6 && !in_array(date('Y-m-d', $current_date), $holidays)) {
            $business_days--;
        }
        if ($business_days > 0) {
            $current_date = strtotime('+1 day', $current_date);
        }
    }
    return $current_date;
}


并计算两个日期在工作日方面的差额:

function business_days_diff($start_date, $end_date, $holidays = array()) {
    $business_days = 0;
    $current_date = strtotime($start_date);
    $end_date = strtotime($end_date);
    while ($current_date <= $end_date) {
        if (date('N', $current_date) < 6 && !in_array(date('Y-m-d', $current_date), $holidays)) {
            $business_days++;
        }
        if ($current_date <= $end_date) {
            $current_date = strtotime('+1 day', $current_date);
        }
    }
    return $business_days;
}


请注意,每个使用86400的人,或者24 * 60 * 60,请不要...您忘记的时间会随着冬/夏时间而变化,一天中的时间并非24小时。虽然strtotime(“ + 1天”,$ timestamp)要慢一些,但它的可靠性要高得多。

#10 楼

蛮力尝试检测工作时间-周一至周五上午8点至下午4点:

if (date('N')<6 && date('G')>8 && date('G')<16) {
   // we have a working time (or check for holidays)
}


#11 楼

以下是从给定日期计算工作日的工作代码。

<?php
$holiday_date_array = array("2016-01-26", "2016-03-07", "2016-03-24", "2016-03-25", "2016-04-15", "2016-08-15", "2016-09-12", "2016-10-11", "2016-10-31");
$date_required = "2016-03-01";

function increase_date($date_required, $holiday_date_array=array(), $days = 15){
    if(!empty($date_required)){
        $counter_1=0;
        $incremented_date = '';
        for($i=1; $i <= $days; $i++){
            $date = strtotime("+$i day", strtotime($date_required));
            $day_name = date("D", $date);
            $incremented_date = date("Y-m-d", $date);
            if($day_name=='Sat'||$day_name=='Sun'|| in_array($incremented_date ,$holiday_date_array)==true){
                $counter_1+=1;
            }
        }
        if($counter_1 > 0){
            return increase_date($incremented_date, $holiday_date_array, $counter_1);
        }else{
            return $incremented_date;
        }
    }else{
        return 'invalid';
    }
}

echo increase_date($date_required, $holiday_date_array, 15);
?>

//output after adding 15 business working days in 2016-03-01 will be "2016-03-23"


#12 楼

这是每天都没有for循环的另一种解决方案。

$from = new DateTime($first_date);
$to = new DateTime($second_date);

$to->modify('+1 day');
$interval = $from->diff($to);
$days = $interval->format('%a');

$extra_days = fmod($days, 7);
$workdays = ( ( $days - $extra_days ) / 7 ) * 5;

$first_day = date('N', strtotime($first_date));
$last_day = date('N', strtotime("1 day", strtotime($second_date)));
$extra = 0;
if($first_day > $last_day) {
   if($first_day == 7) {
       $first_day = 6;
   }

   $extra = (6 - $first_day) + ($last_day - 1);
   if($extra < 0) {
       $extra = $extra * -1;
   }
}
if($last_day > $first_day) {
    $extra = $last_day - $first_day;
}
$days = $workdays + $extra


#13 楼

在假期中,以date()可以产生的某种格式排列天的数组。示例:

// I know, these aren't holidays
$holidays = array(
    'Jan 2',
    'Feb 3',
    'Mar 5',
    'Apr 7',
    // ...
);


然后使用in_array()和date()函数检查时间戳是否表示假期:

$day_of_year = date('M j', $timestamp);
$is_holiday = in_array($day_of_year, $holidays);


#14 楼

我也有同样的需要,我从梭芯的第一个例子开始,最后得到了这个

  function add_business_days($startdate,$buisnessdays,$holidays=array(),$dateformat){
    $enddate = strtotime($startdate);
    $day = date('N',$enddate);
    while($buisnessdays > 1){
        $enddate = strtotime(date('Y-m-d',$enddate).' +1 day');
        $day = date('N',$enddate);
        if($day < 6 && !in_array($enddate,$holidays))$buisnessdays--;
    }
    return date($dateformat,$enddate);
  }


还有一个人

评论


抱歉,但是您的mcgrailm不能正常工作。...它没有考虑$ enddate中的日期是否是假期...它只在添加过程中关注假期,除非我错过了一些东西

–user605865
2011年2月7日在2:35

@理查德,我想我明白你的意思。它不会检查开始日期,以查看它是假日还是周末,而是在开始日期之后计算营业日。如果您想在支票中加上开始日期,可以选择+1天

–mcgrailm
2011-2-7 14:09

我认为这取决于语法以及您要达到的目标。例如,我正在尝试计算何时需要审核报告,但是如果报告是在周末提交的,并且需要在3个业务中完成,则计数从星期一开始(考虑这不是假日)...我已经发布了我的版本,它们都是基于您的代码的,但是做了一些微调。

–克雷格·弗朗西斯(Craig Francis)
2011年10月14日上午10:35

非常有用-一件事,如果使用像Bobbins一样传递的假数组,我必须将!in_array($ enddate,$ holidays)更改为!in_array(date('Ym-d',$ enddate),$ holidays)。 holiday = array('2013-06-16','2013-07-12','2013-08-05');否则,您将检查一个充满日期的数组以获取时间戳,该时间戳始终返回false。

– McNab
2013年6月10日14:08

#15 楼

变体1:

<?php
/*
 * Does not count current day, the date returned is the last business day
 * Requires PHP 5.1 (Using ISO-8601 week)
 */

function businessDays($timestamp = false, $bDays = 2) {
    if($timestamp === false) $timestamp = time();
    while ($bDays>0) {
        $timestamp += 86400;
        if (date('N', $timestamp)<6) $bDays--;
    }
    return $timestamp;
}


变体2:

<?php
/*
 * Does not count current day, the date returned is a business day 
 * following the last business day
 * Requires PHP 5.1 (Using ISO-8601 week)
 */

function businessDays($timestamp = false, $bDays = 2) {
    if($timestamp === false) $timestamp = time();
    while ($bDays+1>0) {
        $timestamp += 86400;
        if (date('N', $timestamp)<6) $bDays--;
    }
    return $timestamp;
}


变体3:

<?php
/*
 * Does not count current day, the date returned is 
 * a date following the last business day (can be weekend or not. 
 * See above for alternatives)
 * Requires PHP 5.1 (Using ISO-8601 week)
 */

function businessDays($timestamp = false, $bDays = 2) {
    if($timestamp === false) $timestamp = time();
    while ($bDays>0) {
        $timestamp += 86400;
        if (date('N', $timestamp)<6) $bDays--;
    }
    return $timestamp += 86400;
}


通过执行以下操作,可以使用上述变化形式做出其他假期考虑。注意!确保所有时间戳都是一天中的同一时间(即午夜)。

制作一系列假日日期(作为unixtimestamps),即:

$holidays = array_flip(strtotime('2011-01-01'),strtotime('2011-12-25'));


修改行:

if (date('N', $timestamp)<6) $bDays--;


要成为:

if (date('N', $timestamp)<6 && !isset($holidays[$timestamp])) $bDays--;


完成!

<?php
/*
 * Does not count current day, the date returned is the last business day
 * Requires PHP 5.1 (Using ISO-8601 week)
 */

function businessDays($timestamp = false, $bDays = 2) {
    if($timestamp === false) $timestamp = strtotime(date('Y-m-d',time()));
    $holidays = array_flip(strtotime('2011-01-01'),strtotime('2011-12-25'));
    while ($bDays>0) {
        $timestamp += 86400;
        if (date('N', $timestamp)<6 && !isset($holidays[$timestamp])) $bDays--;
    }
    return $timestamp;
}


#16 楼

<?php 
function AddWorkDays(){
$i = 0;
$d = 5; // Number of days to add

    while($i <= $d) {
    $i++;
        if(date('N', mktime(0, 0, 0, date(m), date(d)+$i, date(Y))) < 5) {
            $d++;
        }
    }
    return date(Y).','.date(m).','.(date(d)+$d);
}
?>


#17 楼

这是一个递归解决方案。可以轻松地对其进行修改以仅跟踪并返回最新日期。

//  Returns a $numBusDays-sized array of all business dates, 
//  starting from and including $currentDate. 
//  Any date in $holidays will be skipped over.

function getWorkingDays($currentDate, $numBusDays, $holidays = array(), 
  $resultDates = array())
{
  //  exit when we have collected the required number of business days
  if ($numBusDays === 0) {
    return $resultDates;
  }

  //  add current date to return array, if not a weekend or holiday
  $date = date("w", strtotime($currentDate));
  if ( $date != 0  &&  $date != 6  &&  !in_array($currentDate, $holidays) ) {
    $resultDates[] = $currentDate;
    $numBusDays -= 1;
  }

  //  set up the next date to test
  $currentDate = new DateTime("$currentDate + 1 day");
  $currentDate = $currentDate->format('Y-m-d');

  return getWorkingDays($currentDate, $numBusDays, $holidays, $resultDates);
}

//  test
$days = getWorkingDays('2008-12-05', 4);
print_r($days);


#18 楼

date_default_timezone_set('America/New_York');


/** Given a number days out, what day is that when counting by 'business' days
  * get the next business day. by default it looks for next business day
  * ie calling  $date = get_next_busines_day(); on monday will return tuesday
  *             $date = get_next_busines_day(2); on monday will return wednesday
  *             $date = get_next_busines_day(2); on friday will return tuesday
  *
  * @param $number_of_business_days (integer)       how many business days out do you want
  * @param $start_date (string)                     strtotime parseable time value
  * @param $ignore_holidays (boolean)               true/false to ignore holidays
  * @param $return_format (string)                  as specified in php.net/date
 */
function get_next_business_day($number_of_business_days=1,$start_date='today',$ignore_holidays=false,$return_format='m/d/y') {

    // get the start date as a string to time
    $result = strtotime($start_date);

    // now keep adding to today's date until number of business days is 0 and we land on a business day
    while ($number_of_business_days > 0) {
        // add one day to the start date
        $result = strtotime(date('Y-m-d',$result) . " + 1 day");

        // this day counts if it's a weekend and not a holiday, or if we choose to ignore holidays
        if (is_weekday(date('Y-m-d',$result)) && (!(is_holiday(date('Y-m-d',$result))) || $ignore_holidays) ) 
            $number_of_business_days--;

    }

    // when my $number of business days is exausted I have my final date

    return(date($return_format,$result));
}

    function is_weekend($date) {
    // return if this is a weekend date or not.
    return (date('N', strtotime($date)) >= 6);
}

function is_weekday($date) {
    // return if this is a weekend date or not.
    return (date('N', strtotime($date)) < 6);
}

function is_holiday($date) {
    // return if this is a holiday or not.

    // what are my holidays for this year
    $holidays = array("New Year's Day 2011" => "12/31/10",
                        "Good Friday" => "04/06/12",
                        "Memorial Day" => "05/28/12",
                        "Independence Day" => "07/04/12",
                        "Floating Holiday" => "12/31/12",
                        "Labor Day" => "09/03/12",
                        "Thanksgiving Day" => "11/22/12",
                        "Day After Thanksgiving Day" => "11/23/12",
                        "Christmas Eve" => "12/24/12",
                        "Christmas Day" => "12/25/12",
                        "New Year's Day 2012" => "01/02/12",
                        "New Year's Day 2013" => "01/01/13"
                        );

    return(in_array(date('m/d/y', strtotime($date)),$holidays));
}


print get_next_business_day(1) . "\n";


#19 楼

<?php
// $today is the UNIX timestamp for today's date
$today = time();
echo "<strong>Today is (ORDER DATE): " . '<font color="red">' . date('l, F j, Y', $today) . "</font></strong><br/><br/>";

//The numerical representation for day of week (Ex. 01 for Monday .... 07 for Sunday
$today_numerical = date("N",$today);

//leadtime_days holds the numeric value for the number of business days 
$leadtime_days = $_POST["leadtime"];

//leadtime is the adjusted date for shipdate
$shipdate = time();

while ($leadtime_days > 0) 
{
 if ($today_numerical != 5 && $today_numerical != 6)
 {
  $shipdate = $shipdate + (60*60*24);
  $today_numerical = date("N",$shipdate);
  $leadtime_days --;
 }
 else
  $shipdate = $shipdate + (60*60*24);
  $today_numerical = date("N",$shipdate);
}

echo '<strong>Estimated Ship date: ' . '<font color="green">' . date('l, F j, Y', $shipdate) . "</font></strong>";
?>


#20 楼

计算两个日期(包括假期和自定义工作周)之间的工作日

答案不是那么简单-因此,我的建议是使用一个类,在该类中,您可以配置更多的内容而不是依赖于简单函数(或假设使用固定的语言环境)和文化)。要在一定数量的工作日后获取日期,您将:


需要指定要工作的工作日(默认为MON-FRI)-该类允许您启用或分别禁用每个工作日。
需要知道您需要考虑公共假期(国家和州)的准确性

例如https://github.com/khatfield/php-HolidayLibrary/blob/master/Holidays.class.php

或对数据进行硬编码:例如从http://www.feiertagskalender.ch/?hl=zh_CN

或为数据API付费http://www.timeanddate.com/services/api/holiday-api.html



功能方法

/**
 * @param days, int
 * @param $format, string: dateformat (if format defined OTHERWISE int: timestamp) 
 * @param start, int: timestamp (mktime) default: time() //now
 * @param $wk, bit[]: flags for each workday (0=SUN, 6=SAT) 1=workday, 0=day off
 * @param $holiday, string[]: list of dates, YYYY-MM-DD, MM-DD 
 */
function working_days($days, $format='', $start=null, $week=[0,1,1,1,1,1,0], $holiday=[])
{
    if(is_null($start)) $start = time();
    if($days <= 0) return $start;
    if(count($week) != 7) trigger_error('workweek must contain bit-flags for 7 days');
    if(array_sum($week) == 0) trigger_error('workweek must contain at least one workday');
    $wd = date('w', $start);//0=sun, 6=sat
    $time = $start;
    while($days)
    {
        if(
        $week[$wd]
        && !in_array(date('Y-m-d', $time), $holiday)
        && !in_array(date('m-d', $time), $holiday)
        ) --$days; //decrement on workdays
        $wd = date('w', $time += 86400); //add one day in seconds
    }
    $time -= 86400;//include today
    return $format ? date($format, $time): $time;
}

//simple usage
$ten_days = working_days(10, 'D F d Y');
echo '<br>ten workingdays (MON-FRI) disregarding holidays: ',$ten_days;

//work on saturdays and add new years day as holiday
$ten_days = working_days(10, 'D F d Y', null, [0,1,1,1,1,1,1], ['01-01']);
echo '<br>ten workingdays (MON-SAT) disregarding holidays: ',$ten_days;


#21 楼

这是另一种解决方案,比使用in_array检查假期快了将近25%:

/**
 * Function to calculate the working days between two days, considering holidays.
 * @param string $startDate -- Start date of the range (included), formatted as Y-m-d.
 * @param string $endDate -- End date of the range (included), formatted as Y-m-d.
 * @param array(string) $holidayDates -- OPTIONAL. Array of holidays dates, formatted as Y-m-d. (e.g. array("2016-08-15", "2016-12-25"))
 * @return int -- Number of working days.
 */
function getWorkingDays($startDate, $endDate, $holidayDates=array()){
    $dateRange = new DatePeriod(new DateTime($startDate), new DateInterval('P1D'), (new DateTime($endDate))->modify("+1day"));
    foreach ($dateRange as $dr) { if($dr->format("N")<6){$workingDays[]=$dr->format("Y-m-d");} }
    return count(array_diff($workingDays, $holidayDates));
}


#22 楼

此代码段非常容易计算没有周末和节假日的工作日:

function getWorkingDays($startDate,$endDate,$offdays,$holidays){
$endDate = strtotime($endDate);
$startDate = strtotime($startDate);
$days = ($endDate - $startDate) / 86400 + 1;
$counter=0;
for ($i = 1; $i <= $days; $i++) {
    $the_first_day_of_week = date("N", $startDate);
    $startDate+=86400;
if (!in_array($the_first_day_of_week, $offdays) && !in_array(date("Y-m-
d",$startDate), $holidays)) {
$counter++;
}

}   
return $counter;
}
//example to use
$holidays=array("2017-07-03","2017-07-20");
$offdays=array(5,6);//weekend days Monday=1 .... Sunday=7
echo getWorkingDays("2017-01-01","2017-12-31",$offdays,$holidays)


#23 楼

我知道我参加晚会很晚,但是我使用Marcos J. Montes的这套旧功能来确定假期和工作日。他花时间为复活节添加了1876年以来的算法,并添加了美国所有主要的假期。可以轻松地在其他国家/地区进行更新。

//Usage
$days = 30;
$next_working_date = nextWorkingDay($days, $somedate);

//add date function
function DateAdd($interval, $number, $date) {

    $date_time_array = getdate($date);
    //die(print_r($date_time_array));

    $hours = $date_time_array["hours"];
    $minutes = $date_time_array["minutes"];
    $seconds = $date_time_array["seconds"];
    $month = $date_time_array["mon"];
    $day = $date_time_array["mday"];
    $year = $date_time_array["year"];

    switch ($interval) {

        case "yyyy":
            $year+=$number;
            break;
        case "q":
            $year+=($number*3);
            break;
        case "m":
            $month+=$number;
            break;
        case "y":
        case "d":
        case "w":
            $day+=$number;
            break;
        case "ww":
            $day+=($number*7);
            break;
        case "h":
            $hours+=$number;
            break;
        case "n":
            $minutes+=$number;
            break;
        case "s":
            $seconds+=$number; 
            break;            
    }
    //      echo "day:" . $day;
    $timestamp= mktime($hours,$minutes,$seconds,$month,$day,$year);
    return $timestamp;
}

// the following function get_holiday() is based on the work done by
// Marcos J. Montes
function get_holiday($year, $month, $day_of_week, $week="") {
    if ( (($week != "") && (($week > 5) || ($week < 1))) || ($day_of_week > 6) || ($day_of_week < 0) ) {
        // $day_of_week must be between 0 and 6 (Sun=0, ... Sat=6); $week must be between 1 and 5
        return FALSE;
    } else {
        if (!$week || ($week == "")) {
            $lastday = date("t", mktime(0,0,0,$month,1,$year));
            $temp = (date("w",mktime(0,0,0,$month,$lastday,$year)) - $day_of_week) % 7;
        } else {
            $temp = ($day_of_week - date("w",mktime(0,0,0,$month,1,$year))) % 7;
        }

        if ($temp < 0) {
            $temp += 7;
        }

        if (!$week || ($week == "")) {
            $day = $lastday - $temp;
        } else {
            $day = (7 * $week) - 6 + $temp;
        }
        //echo $year.", ".$month.", ".$day . "<br><br>";
        return format_date($year, $month, $day);
    }
}

function observed_day($year, $month, $day) {
    // sat -> fri & sun -> mon, any exceptions?
    //
    // should check $lastday for bumping forward and $firstday for bumping back,
    // although New Year's & Easter look to be the only holidays that potentially
    // move to a different month, and both are accounted for.

    $dow = date("w", mktime(0, 0, 0, $month, $day, $year));

    if ($dow == 0) {
        $dow = $day + 1;
    } elseif ($dow == 6) {
        if (($month == 1) && ($day == 1)) {    // New Year's on a Saturday
            $year--;
            $month = 12;
            $dow = 31;
        } else {
            $dow = $day - 1;
        }
    } else {
        $dow = $day;
    }

    return format_date($year, $month, $dow);
}

function calculate_easter($y) {
    // In the text below, 'intval($var1/$var2)' represents an integer division neglecting
    // the remainder, while % is division keeping only the remainder. So 30/7=4, and 30%7=2
//
    // This algorithm is from Practical Astronomy With Your Calculator, 2nd Edition by Peter
    // Duffett-Smith. It was originally from Butcher's Ecclesiastical Calendar, published in
    // 1876. This algorithm has also been published in the 1922 book General Astronomy by
    // Spencer Jones; in The Journal of the British Astronomical Association (Vol.88, page
    // 91, December 1977); and in Astronomical Algorithms (1991) by Jean Meeus. 

    $a = $y%19;
    $b = intval($y/100);
    $c = $y%100;
    $d = intval($b/4);
    $e = $b%4;
    $f = intval(($b+8)/25);
    $g = intval(($b-$f+1)/3);
    $h = (19*$a+$b-$d-$g+15)%30;
    $i = intval($c/4);
    $k = $c%4;
    $l = (32+2*$e+2*$i-$h-$k)%7;
    $m = intval(($a+11*$h+22*$l)/451);
    $p = ($h+$l-7*$m+114)%31;
    $EasterMonth = intval(($h+$l-7*$m+114)/31);    // [3 = March, 4 = April]
    $EasterDay = $p+1;    // (day in Easter Month)

    return format_date($y, $EasterMonth, $EasterDay);
}


function nextWorkingDay($number_days, $start_date = "") {
    $day_counter = 0;
    $intCounter = 0;    

    if ($start_date=="") {
        $today  = mktime(0, 0, 0, date("m")  , date("d"), date("Y"));
    } else {
        $start_time = strtotime($start_date);
        $today  = mktime(0, 0, 0, date("m", $start_time)  , date("d", $start_time), date("Y", $start_time));
    }

    while($day_counter < $number_days) {
        $working_time = DateAdd("d", 1, $today);
        $working_date = date("Y-m-d", $working_date);
        if (!isWeekend($working_date) && !confirm_holiday(date("Y-m-d", strtotime($working_date))) ) {
            $day_counter++;
        }
        $intCounter++;
        $today  = $working_time;
        if ($intCounter > 1000) {
            //just in case out of control?
            break;
        }
    }

    return $working_date;
}
function isWeekend($check_date) {
    return (date("N",  strtotime($check_date)) > 5);
}
function confirm_holiday($somedate="") {
    if ($somedate=="") {
        $somedate = date("Y-m-d");
    }
    $year = date("Y", strtotime($somedate));
    $blnHoliday = false;
    //newyears
    if ($somedate == observed_day($year, 1, 1)) {
        $blnHoliday = true;
    }
    if ($somedate == format_date($year, 1, 1)) {
        $blnHoliday = true;
    }
    if ($somedate == format_date($year, 12, 31)) {
        $blnHoliday = true;
    }
    //Martin Luther King
    if ($somedate == get_holiday($year, 1, 1, 3)) {
        $blnHoliday = true;
    }
    //President's
    if ($somedate == get_holiday($year, 2, 1, 3)) {
        $blnHoliday = true;
    }
    //easter
    if ($somedate == calculate_easter($year)) {
        $blnHoliday = true;
    }
    //Memorial
    if ($somedate == get_holiday($year, 5, 1)) {
        $blnHoliday = true;
    }
    //july4
    if ($somedate == observed_day($year, 7, 4)) {
        $blnHoliday = true;
    }
    //labor
    if ($somedate == get_holiday($year, 9, 1, 1)) {
        $blnHoliday = true;
    }
    //columbus
    if ($somedate == get_holiday($year, 10, 1, 2)) {
        $blnHoliday = true;
    }
    //thanks
    if ($somedate == get_holiday($year, 11, 4, 4)) {
        $blnHoliday = true;
    }
    //xmas
    if ($somedate == format_date($year, 12, 24)) {
        $blnHoliday = true;
    }
    if ($somedate == format_date($year, 12, 25)) {
        $blnHoliday = true;
    }
    return $blnHoliday;
}


#24 楼

函数get_business_days_forward_from_date($ num_days,$ start_date ='',$ rtn_fmt ='Y-m-d')
{

// $start_date will default to today    

if ($start_date=='') { $start_date = date("Y-m-d"); }

$business_day_ct = 0;

$max_days = 10000 + $num_days;  // to avoid any possibility of an infinite loop


// define holidays, this currently only goes to 2012 because, well, you know... ;-)
// if the world is still here after that, you can find more at
// http://www.opm.gov/Operating_Status_Schedules/fedhol/2013.asp
// always add holidays in order, because the iteration will stop when the holiday is > date being tested

$fed_holidays=array(
    "2010-01-01",
    "2010-01-18",
    "2010-02-15",
    "2010-05-31",
    "2010-07-05",
    "2010-09-06",
    "2010-10-11",
    "2010-11-11",
    "2010-11-25",
    "2010-12-24",

    "2010-12-31",
    "2011-01-17",
    "2011-02-21",
    "2011-05-30",
    "2011-07-04",
    "2011-09-05",
    "2011-10-10",
    "2011-11-11",
    "2011-11-24",
    "2011-12-26",

    "2012-01-02",
    "2012-01-16",
    "2012-02-20",
    "2012-05-28",
    "2012-07-04",
    "2012-09-03",
    "2012-10-08",
    "2012-11-12",
    "2012-11-22",
    "2012-12-25",
    );

$curr_date_ymd = date('Y-m-d', strtotime($start_date));    

for ($x=1;$x<$max_days;$x++)
{
    if (intval($num_days)==intval($business_day_ct)) { return(date($rtn_fmt, strtotime($curr_date_ymd))); }  // date found - return

    // get next day to check

    $curr_date_ymd = date('Y-m-d', (strtotime($start_date)+($x * 86400)));   // add 1 day to the current date

    $is_business_day = 1;

    // check if this is a weekend   1 (for Monday) through 7 (for Sunday)

    if ( intval(date("N",strtotime($curr_date_ymd))) > 5) { $is_business_day = 0; }

    //check for holiday
    foreach($fed_holidays as $holiday)
    {
        if (strtotime($holiday)==strtotime($curr_date_ymd))  // holiday found
        {
            $is_business_day = 0;
            break 1;
        }

        if (strtotime($holiday)>strtotime($curr_date_ymd)) { break 1; }  // past date, stop searching (always add holidays in order)


    }

    $business_day_ct = $business_day_ct + $is_business_day;  // increment if this is a business day

} 

// if we get here, you are hosed
return ("ERROR");


}

#25 楼

add_business_days有一个小错误。尝试使用现有功能执行以下操作,输出将为星期六。

开始日期=星期五
要添加的工作日= 1
假日数组=添加下一个星期一的日期。

我已经在下面的功能中修复了该问题。

function add_business_days($startdate, $buisnessdays, $holidays = array(), $dateformat = 'Y-m-d'){
$i= 1;
$dayx= strtotime($startdate);
$buisnessdays= ceil($buisnessdays);

while($i < $buisnessdays)
{
    $day= date('N',$dayx);

    $date= date('Y-m-d',$dayx);
    if($day < 6 && !in_array($date,$holidays))
        $i++;

    $dayx= strtotime($date.' +1 day');
}

## If the calculated day falls on a weekend or is a holiday, then add days to the next business day
$day= date('N',$dayx);
$date= date('Y-m-d',$dayx);

while($day >= 6 || in_array($date,$holidays))
{
    $dayx= strtotime($date.' +1 day');
    $day= date('N',$dayx);
    $date= date('Y-m-d',$dayx);
}

return date($dateformat, $dayx);}


评论


我假设这是基于Bobbin的代码,我相信我也解决了这个问题

–mcgrailm
2011-2-4在16:20

#26 楼

我只是基于Bobbin和mcgrailm代码使函数工作,并添加了一些对我来说很完美的东西。

function add_business_days($startdate,$buisnessdays,$holidays,$dateformat){
    $enddate = strtotime($startdate);
    $day = date('N',$enddate);
    while($buisnessdays > 0){ // compatible with 1 businessday if I'll need it
        $enddate = strtotime(date('Y-m-d',$enddate).' +1 day');
        $day = date('N',$enddate);
        if($day < 6 && !in_array(date('Y-m-d',$enddate),$holidays))$buisnessdays--;
    }
    return date($dateformat,$enddate);
}

// as a parameter in in_array function we should use endate formated to 
// compare correctly with the holidays array.


评论


我和你的唯一的区别是我的是基于你的是零。您在这里真的什么都没做。我很高兴这段代码对您有所帮助。但是,我不认为这是“基于”代码,更像是您试图将我的代码声明为您自己的代码

–mcgrailm
2011年1月25日13:55

#27 楼

上面的James Pasta提供的功能的增强,包括所有联邦假日,并更正了7月4日(计算为上面的6月4日!),还包括假日名称作为数组键...

/ **
*美国国定假日
* @param string $ year
* @return array
* /
公共静态函数getNationalAmericanHolidays($ year ){

//  January 1 - New Year's Day (Observed)
//  Third Monday in January - Birthday of Martin Luther King, Jr.
//  Third Monday in February - Washington’s Birthday / President's Day
//  Last Monday in May - Memorial Day
//  July 4 - Independence Day
//  First Monday in September - Labor Day
//  Second Monday in October - Columbus Day
//  November 11 - Veterans’ Day (Observed)
//  Fourth Thursday in November Thanksgiving Day
//  December 25 - Christmas Day
$bankHolidays = array(
    ['New Years Day'] => $year . "-01-01",
    ['Martin Luther King Jr Birthday'] => "". date("Y-m-d",strtotime("third Monday of January " . $year) ),
    ['Washingtons Birthday'] => "". date("Y-m-d",strtotime("third Monday of February " . $year) ),
    ['Memorial Day'] => "". date("Y-m-d",strtotime("last Monday of May " . $year) ),
    ['Independance Day'] => $year . "-07-04",
    ['Labor Day'] => "". date("Y-m-d",strtotime("first Monday of September " . $year) ),
    ['Columbus Day'] => "". date("Y-m-d",strtotime("second Monday of October " . $year) ),
    ['Veterans Day'] => $year . "-11-11",
    ['Thanksgiving Day'] => "". date("Y-m-d",strtotime("fourth Thursday of November " . $year) ),
    ['Christmas Day'] => $year . "-12-25"
);

return $bankHolidays;


}

#28 楼

刚写完一个可用于处理工作日的API(对于我的情况,这些解决方案都无法奏效:-);如果有人发现它有用,请在此处链接。

〜Nate

用于计算营业日的PHP类

#29 楼

感谢Bobbin,mcgrailm,Tony,James Pasta和其他一些在此处发布的人。我已经编写了自己的函数来向日期添加工作日,但是使用在这里找到的一些代码对其进行了修改。这将在周末/节假日处理开始日期。这也将处理营业时间。我添加了一些注释并分解了代码,使其更易于阅读。

<?php
function count_business_days($date, $days, $holidays) {
    $date = strtotime($date);

    for ($i = 1; $i <= intval($days); $i++) { //Loops each day count

        //First, find the next available weekday because this might be a weekend/holiday
        while (date('N', $date) >= 6 || in_array(date('Y-m-d', $date), $holidays)){
            $date = strtotime(date('Y-m-d',$date).' +1 day');
        }

        //Now that we know we have a business day, add 1 day to it
        $date = strtotime(date('Y-m-d',$date).' +1 day');

        //If this day that was previously added falls on a weekend/holiday, then find the next business day
        while (date('N', $date) >= 6 || in_array(date('Y-m-d', $date), $holidays)){
            $date = strtotime(date('Y-m-d',$date).' +1 day');
        }
    }
    return date('Y-m-d', $date);
}

//Also add in the code from Tony and James Pasta to handle holidays...

function getNationalAmericanHolidays($year) {
$bankHolidays = array(
    'New Years Day' => $year . "-01-01",
    'Martin Luther King Jr Birthday' => "". date("Y-m-d",strtotime("third Monday of January " . $year) ),
    'Washingtons Birthday' => "". date("Y-m-d",strtotime("third Monday of February " . $year) ),
    'Memorial Day' => "". date("Y-m-d",strtotime("last Monday of May " . $year) ),
    'Independance Day' => $year . "-07-04",
    'Labor Day' => "". date("Y-m-d",strtotime("first Monday of September " . $year) ),
    'Columbus Day' => "". date("Y-m-d",strtotime("second Monday of October " . $year) ),
    'Veterans Day' => $year . "-11-11",
    'Thanksgiving Day' => "". date("Y-m-d",strtotime("fourth Thursday of November " . $year) ),
    'Christmas Day' => $year . "-12-25"
);
return $bankHolidays;

}

//Now to call it... since we're working with business days, we should
//also be working with business hours so check if it's after 5 PM
//and go to the next day if necessary.

//Go to next day if after 5 pm (5 pm = 17)
if (date(G) >= 17) {
    $start_date = date("Y-m-d", strtotime("+ 1 day")); //Tomorrow
} else {
    $start_date = date("Y-m-d"); //Today
}

//Get the holidays for the current year and also for the next year
$this_year = getNationalAmericanHolidays(date('Y'));
$next_year = getNationalAmericanHolidays(date('Y', strtotime("+12 months")));
$holidays = array_merge($this_year, $next_year);

//The number of days to count
$days_count = 10;

echo count_business_days($start_date, $days_count, $holidays);

?>


#30 楼

就我个人而言,我认为这是一个更简洁,更简洁的解决方案:

function onlyWorkDays( $d ) {
    $holidays = array('2013-12-25','2013-12-31','2014-01-01','2014-01-20','2014-02-17','2014-05-26','2014-07-04','2014-09-01','2014-10-13','2014-11-11','2014-11-27','2014-12-25','2014-12-31');
    while (in_array($d->format("Y-m-d"), $holidays)) { // HOLIDAYS
        $d->sub(new DateInterval("P1D"));
    }
    if ($d->format("w") == 6) { // SATURDAY
        $d->sub(new DateInterval("P1D"));
    }
    if ($d->format("w") == 0) { // SUNDAY
        $d->sub(new DateInterval("P2D"));
    }
    return $d;
}


只需将拟议的new日期发送到此函数。