php 计算工作日
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Calculate business days
提问by AdamTheHutt
I need a method for adding "business days" in PHP. For example, Friday 12/5 + 3 business days = Wednesday 12/10.
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。
At a minimum I need the code to understand weekends, but ideally it should account for US federal holidays as well. I'm sure I could come up with a solution by brute force if necessary, but I'm hoping there's a more elegant approach out there. Anyone?
至少我需要代码来理解周末，但理想情况下它也应该考虑美国联邦假期。我敢肯定，如有必要，我可以通过蛮力提出解决方案，但我希望有更优雅的方法。任何人？
Thanks.
谢谢。
采纳答案by flamingLogos
Here's a function from the user commentson the date() function page in the PHP manual. It's an improvement of an earlier function in the comments that adds support for leap years.
这是PHP 手册中 date() 函数页面上用户评论的一个函数。这是对评论中较早功能的改进，增加了对闰年的支持。
Enter the starting and ending dates, along with an array of any holidays that might be in between, and it returns the working days as an integer:
输入开始日期和结束日期，以及可能介于两者之间的任何假期的数组，它会以整数形式返回工作日：
<?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
?>
回答by Glavi?
Get the number of working days without holidaysbetween two dates :
获取两个日期之间没有假期的工作日数：
Use example:
使用示例：
echo number_of_working_days('2013-12-23', '2013-12-29');
Output:
输出：
3
Function:
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;
}
回答by Tim
There are some args for the date()function that should help. If you check date("w") it will give you a number for the day of the week, from 0 for Sunday through 6 for Saturday. So.. maybe something like..
date()函数有一些参数应该会有所帮助。如果你检查 date("w") 它会给你一个星期几的数字，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;
This is just a rough example of one possibility..
这只是一种可能性的粗略示例。
回答by James Pasta
Holiday calculation is non-standard in each State. I am writing a bank application which I need some hard business rules for but can still only get a rough standard.
每个州的假期计算都是非标准的。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，但仍然只能得到一个粗略的标准。
/**
* 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;
}
回答by Suresh Kamrushi
$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);
回答by Bobbin
Here is a function for adding buisness days to a date
这是一个将营业日添加到日期的函数
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');
Another post mentions getWorkingDays (from php.net comments and included here) but I think it breaks if you start on a Sunday and finish on a work day.
另一篇文章提到了 getWorkingDays（来自 php.net 评论并包含在此处），但我认为如果您在周日开始并在工作日结束，它会中断。
Using the following (you'll need to include the getWorkingDays function from previous post)
使用以下内容（您需要包含上一篇文章中的 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,$holidays);
Gives the result as 5 not 4
结果为 5 而不是 4
Sun, 08 Jan 2012 00:00:00 +0000 weekend
Mon, 09 Jan 2012 00:00:00 +0000
Tue, 10 Jan 2012 00:00:00 +0000 holiday
Wed, 11 Jan 2012 00:00:00 +0000
Thu, 12 Jan 2012 00:00:00 +0000
Fri, 13 Jan 2012 00:00:00 +0000
The following function was used to generate the above.
以下函数用于生成上述内容。
function get_working_days($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,$holidays);
Bring on the holidays...
带上假期...
回答by George John
You can try this function which is more simple.
你可以试试这个更简单的功能。
function getWorkingDays($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;
}
}
回答by easycodingclub
Below is the working code to calculate working business days from a given date.
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"
回答by Laird
Here is another solution without for loop for each day.
这是每天没有 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 + $extra
回答by Alex
A function to add or subtract business days from a given date, this doesn't account for holidays.
从给定日期添加或减去工作日的功能，这不考虑假期。
function dateFromBusinessDays($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;
}
use like so...
像这样使用...
echo date("m/d/Y", dateFromBusinessDays(-7));
echo date("m/d/Y", dateFromBusinessDays(3, time() + 3*60*60*24));
