Мне нужен метод добавления "рабочих дней" в PHP. Например, пятница 12/5 + 3 рабочих дня = среда 12/10.
Как минимум мне нужен код для понимания выходных, но в идеале он должен учитывать и федеральные праздники США. Я уверен, что при необходимости придумаю решение грубой силой, но я надеюсь, что там будет более элегантный подход. Кто-нибудь?
Спасибо.
Здесь функция из комментариев пользователя на странице функции date() в руководстве по PHP. Это улучшение более ранней функции в комментариях, которая добавляет поддержку високосных лет.
Введите даты начала и окончания вместе с массивом любых праздничных дней, которые могут находиться между ними, и возвращает рабочие дни как целое число:
<?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 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
?>
Получите количество рабочих дней без отпуска между двумя датами:
echo number_of_working_days('2013-12-23', '2013-12-29');
3
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;
}
Есть несколько аргументов для функции date(), которая должна помочь. Если вы проверите дату ( "w" ), он даст вам номер для дня недели, от 0 до воскресенья до 6 в субботу. Так.. может что-то вроде..
$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;
Это лишь примерный пример одной возможности.
Праздничный расчет нестандартен в каждом государстве. Я пишу банковское приложение, для которого мне нужны какие-то жесткие бизнес-правила, но все же может получить только приблизительный стандарт.
/**
* 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;
}
$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);
Вот функция добавления дней работы к дате
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');
В другой публикации упоминается getWorkingDays (из комментариев php.net и включено здесь), но я думаю, что он ломается, если вы начинаете в воскресенье и заканчиваете рабочий день.
Используя следующее (вам нужно включить функцию getWorkingDays из предыдущего сообщения)
date_default_timezone_set('Europe\London');
//Example:
$holidays = array('2012-01-10');
$startDate = '2012-01-08';
$endDate = '2012-01-13';
echo getWorkingDays( $startDate,$endDate,$holidays);
Дает результат как 5 не 4
Sun, 08 Jan 2012 00:00:00 +0000 weekend
Mon, 09 Jan 2012 00:00:00 +0000
Tue, 10 Jan 2012 00:00:00 +0000 holiday
Wed, 11 Jan 2012 00:00:00 +0000
Thu, 12 Jan 2012 00:00:00 +0000
Fri, 13 Jan 2012 00:00:00 +0000
Для генерации вышеприведенной функции использовалась следующая функция.
function get_working_days($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,$holidays);
Принесите праздничные дни...
Вы можете попробовать эту функцию, которая проще.
function getWorkingDays($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 $no_days;
}
}
Функция добавления или вычитания рабочих дней с определенной даты, это не учитывает праздники.
function dateFromBusinessDays($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 $dateTime;
}
используйте так...
echo date("m/d/Y", dateFromBusinessDays(-7));
echo date("m/d/Y", dateFromBusinessDays(3, time() + 3*60*60*24));
Моя версия, основанная на работе @mcgrailm... изменилась, потому что отчет нужно было пересмотреть в течение 3 рабочих дней, и если он был отправлен в выходные дни, подсчет начнется в следующий понедельник:
function business_days_add($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 $current_date;
}
И разработав разницу двух дат в рабочие дни:
function business_days_diff($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;
}
В качестве примечания, каждый, кто использует 86400, или 24 * 60 * 60, пожалуйста, не... ваше время забывания меняется с зимнего/летнего времени, где в день это не ровно 24 часа. Хотя он немного медленнее strtotime ('+ 1 день', $timestamp), он намного надежнее.
Грубая попытка определить рабочее время - с понедельника по пятницу с 8:00 до 16:00:
if (date('N')<6 && date('G')>8 && date('G')<16) {
// we have a working time (or check for holidays)
}
Я создал для этого приличную библиотеку.
https://github.com/andrejsstepanovs/business-days-calculator
Он стабилен и готов к производству.
Ниже приведен рабочий код для расчета рабочих рабочих дней с определенной даты.
<?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"
Вот еще одно решение без цикла для каждого дня.
$from = new DateTime($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
В праздничные дни создайте массив дней в каком-то формате, который может дать date(). Пример:
// I know, these aren't holidays
$holidays = array(
'Jan 2',
'Feb 3',
'Mar 5',
'Apr 7',
// ...
);
Затем используйте in_array() и date(), чтобы проверить, является ли отметка времени праздником:
$day_of_year = date('M j', $timestamp);
$is_holiday = in_array($day_of_year, $holidays);
У меня была такая же потребность, я начал с первого примера бобины и закончил с этим
function add_business_days($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);
}
hth someone
<?php
// $today is the UNIX timestamp for today 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', $shipdate) . "</font></strong>";
?>
Вариант 1:
<?php
/*
* Does not count current day, the date returned is the last business day
* Requires PHP 5.1 (Using ISO-8601 week)
*/
function businessDays($timestamp = false, $bDays = 2) {
if($timestamp === false) $timestamp = time();
while ($bDays>0) {
$timestamp += 86400;
if (date('N', $timestamp)<6) $bDays--;
}
return $timestamp;
}
Вариант 2:
<?php
/*
* Does not count current day, the date returned is a business day
* following the last business day
* Requires PHP 5.1 (Using ISO-8601 week)
*/
function businessDays($timestamp = false, $bDays = 2) {
if($timestamp === false) $timestamp = time();
while ($bDays+1>0) {
$timestamp += 86400;
if (date('N', $timestamp)<6) $bDays--;
}
return $timestamp;
}
Вариант 3:
<?php
/*
* Does not count current day, the date returned is
* a date following the last business day (can be weekend or not.
* See above for alternatives)
* Requires PHP 5.1 (Using ISO-8601 week)
*/
function businessDays($timestamp = false, $bDays = 2) {
if($timestamp === false) $timestamp = time();
while ($bDays>0) {
$timestamp += 86400;
if (date('N', $timestamp)<6) $bDays--;
}
return $timestamp += 86400;
}
Дополнительные варианты отдыха могут быть сделаны с использованием вариантов вышеизложенного, выполнив следующее. Заметка! убедитесь, что все отметки времени - это одно и то же время дня (то есть полночь).
Сделайте массив праздничных дат (как unixtimestamps), т.е.:
$holidays = array_flip(strtotime('2011-01-01'),strtotime('2011-12-25'));
Изменить строку:
if (date('N', $timestamp)<6) $bDays--;
:
if (date('N', $timestamp)<6 && !isset($holidays[$timestamp])) $bDays--;
Готово!
<?php
/*
* Does not count current day, the date returned is the last business day
* Requires PHP 5.1 (Using ISO-8601 week)
*/
function businessDays($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;
}
<?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);
}
?>
Вот рекурсивное решение. Его можно легко изменить, чтобы отслеживать и возвращать последнюю дату.
// 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);
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 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 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
$holidays = array("New Year Day 2011" => "12/31/10",
"Good Friday" => "04/06/12",
"Memorial Day" => "05/28/12",
"Independence Day" => "07/04/12",
"Floating Holiday" => "12/31/12",
"Labor Day" => "09/03/12",
"Thanksgiving Day" => "11/22/12",
"Day After Thanksgiving Day" => "11/23/12",
"Christmas Eve" => "12/24/12",
"Christmas Day" => "12/25/12",
"New Year Day 2012" => "01/02/12",
"New Year Day 2013" => "01/01/13"
);
return(in_array(date('m/d/y', strtotime($date)),$holidays));
}
print get_next_business_day(1) . "\n";
рассчитать рабочие дни между двумя датами, включая праздники и пользовательскую рабочую неделю
Ответ не такой тривиальный, поэтому моим предложением было бы использовать класс, в котором вы можете настраивать больше, чем полагаться на упрощенную функцию (или предполагать фиксированную локаль и культуру). Чтобы получить дату после определенного количества рабочих дней, вы:
Функциональный подход
/**
* @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;
Это другое решение, оно почти на 25% быстрее, чем проверка праздников с помощью in_array:
/**
* 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));
}
Я знаю, что опаздываю на вечеринку, но я использую этот старый набор функций Маркоса Монтеса для определения выходных и рабочих дней. Он взял время, чтобы добавить алгоритм с 1876 года для Пасхи, и он добавил все основные американские праздники. Это может быть легко обновлено для других стран.
//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 & 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 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 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 $blnHoliday;
}
Этот фрагмент кода очень легко вычислить рабочий день без окончания недели и праздников:
function getWorkingDays($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,$holidays)
function get_business_days_forward_from_date ($ 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 + $is_business_day; // increment if this is a business day
}
// if we get here, you are hosed
return ("ERROR");
}
Я просто использую свою функцию, основанную на коде Bobbin и mcgrailm, добавляя некоторые вещи, которые идеально подходят для меня.
function add_business_days($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.
В add_business_days есть небольшая ошибка. Попробуйте следующее с существующей функцией, и выход будет в субботу.
Начало = пятница Рабочие дни для добавления = 1 Holidays array = Добавить дату для следующего понедельника.
Я исправил это в моей функции ниже.
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);}
Улучшение функции, предложенной Джеймсом Паста выше, включить все федеральные праздники и исправить 4 июля (было рассчитано как 4 июня выше!), а также включить название праздника в качестве ключа массива...
/**
* Национальные американские праздники
* @param string $year
* @return array
*/
public static function getNationalAmericanHolidays ($ year) {
// January 1 - New Year Day (Observed)
// Third Monday in January - Birthday of Martin Luther King, Jr.
// Third Monday in February - Washington’s Birthday / President 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 $bankHolidays;
}
Только что закончил писать API, который можно использовать для управления рабочими днями (ни одно из этих решений не сработало для моей ситуации:-); ссылаясь на него здесь, если кто-либо найдет это полезным.
~ Nate
Благодаря Bobbin, mcgrailm, Tony, James Pasta и нескольким другим, которые разместили здесь. Я написал свою собственную функцию, чтобы добавить рабочие дни к дате, но изменил ее с помощью некоторого кода, который я нашел здесь. Это будет означать, что дата начала будет в выходные/праздничные дни. Это также будет обрабатывать часы работы. Я добавил несколько комментариев и разбил код, чтобы упростить его чтение.
<?php
function count_business_days($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 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);
?>