PHP "iş günü" eklenmesi için bir yöntem gerekir. Örneğin, Cuma 12/5 + 3 iş günü = Çarşamba 12/10.
En azından ben hafta sonları anlamak kodu gerekir, ama ideal aynı zamanda ABD federal tatil için dikkate alınmalıdır. Ben gerekirse ben kaba kuvvet tarafından bir çözüm ile gelip eminim, ama ben daha zarif bir yaklaşım orada umut ediyorum. Herkes?
Teşekkürler.
Burada PHP kılavuzda date () fonksiyonu sayfasında user comments bir fonksiyon. Bu artık yıl için destek ekler yorumların önceki bir fonksiyonun bir gelişme.
Arasında olabilecek herhangi bir tatil bir dizi ile birlikte başlangıç ve bitiş tarihleri girin ve bir tamsayı olarak iş günü döndürür:
<?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
?>
Yardımcı olmalıdır date() fonksiyonu için bazı argümanlar vardır. Eğer tarih ("w") kontrol edin eğer Cumartesi için 6 ile Pazar 0'dan, haftanın günü için size bir numara verecektir. Yani .. belki gibi bir şey ..
$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..
Tatil hesaplama her Devletin standart dışı olduğunu. Ben ancak hala sadece kaba bir standart alabilirsiniz için bazı zor iş kurallarını gereken bir banka uygulama yazıyorum.
/**
* 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 $bankHolidays;
}
İşte tarihe buisness gün ekleyerek için bir işlevdir
function add_business_days($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');
Başka sonrası (php.net yorumlardan ve buraya dahil) getWorkingDays bahseder ama bir Pazar günü başlar ve bir iş gününde bitirmek eğer kırar düşünüyorum.
Aşağıdakileri kullanarak (bir önceki görevinden getWorkingDays işlevini eklemeniz gerekir)
date_default_timezone_set('Europe\London');
//Example:
$holidays = array('2012-01-10');
$startDate = '2012-01-08';
$endDate = '2012-01-13';
echo getWorkingDays( $startDate,$endDate,$holidays);
5 değil 4 olarak sonuç verir
Sun, 08 Jan 2012 00:00:00 +0000 weekend
Mon, 09 Jan 2012 00:00:00 +0000
Tue, 10 Jan 2012 00:00:00 +0000 holiday
Wed, 11 Jan 2012 00:00:00 +0000
Thu, 12 Jan 2012 00:00:00 +0000
Fri, 13 Jan 2012 00:00:00 +0000
İzleyen fonksiyon, yukarıda üretmek için kullanıldı.
function get_working_days($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);
Tatillerde getir ...
Tatil için, () üretebilir tarih bazı biçiminde günlerde bir dizi yapmak. Örnek:
// I know, these aren't holidays
$holidays = array(
'Jan 2',
'Feb 3',
'Mar 5',
'Apr 7',
// ...
);
Ardından in_array() kullanmak ve zaman damgası bir tatil temsil ediyorsa date() fonksiyonları kontrol etmek için:
$day_of_year = date('M j', $timestamp);
$is_holiday = in_array($day_of_year, $holidays);
Belirli bir tarihten itibaren iş günü eklemek veya çıkarmak için bir fonksiyon, bu tatil için hesap değil.
function dateFromBusinessDays($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 $dateTime;
}
böylece gibi kullanabilirsiniz ...
echo date("m/d/Y", dateFromBusinessDays(-7));
echo date("m/d/Y", dateFromBusinessDays(3, time() + 3*60*60*24));
Ben bobin ilk örnek ile başlayan bu aynı ihtiyacı vardı ve bu ile sona erdi
function add_business_days($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 birisi
Varyant 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 $timestamp;
}
Varyant 2:
<?php
/*
* Does not count current day, the date returned is a business day
* following the last business day
* Requires PHP 5.1 (Using ISO-8601 week)
*/
function businessDays($timestamp = false, $bDays = 2) {
if($timestamp === false) $timestamp = time();
while ($bDays+1>0) {
$timestamp += 86400;
if (date('N', $timestamp)<6) $bDays--;
}
return $timestamp;
}
Varyant 3:
<?php
/*
* Does not count current day, the date returned is
* a date following the last business day (can be weekend or not.
* See above for alternatives)
* Requires PHP 5.1 (Using ISO-8601 week)
*/
function businessDays($timestamp = false, $bDays = 2) {
if($timestamp === false) $timestamp = time();
while ($bDays>0) {
$timestamp += 86400;
if (date('N', $timestamp)<6) $bDays--;
}
return $timestamp += 86400;
}
Ek tatil hususlar aşağıdakileri yaparak yukarıdaki varyasyonları kullanılarak yapılabilir. Not! temin tüm damgaları gün (yani gece yarısı) aynı zaman vardır.
(Unixtimestamps gibi) tatil tarihleri bir dizi yapmak, yani:
$holidays = array_flip(strtotime('2011-01-01'),strtotime('2011-12-25'));
Satırı değiştirin:
if (date('N', $timestamp)<6) $bDays--;
olduğu:
if (date('N', $timestamp)<6 && !isset($holidays[$timestamp])) $bDays--;
Bitti!
<?php
/*
* Does not count current day, the date returned is the last business day
* Requires PHP 5.1 (Using ISO-8601 week)
*/
function businessDays($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 $timestamp;
}
: Benim versiyon @ mcgrailm ... gerekli raporu 3 iş günü içinde gözden geçirilmesi ve bir hafta sonu gönderilirse, sayma aşağıdaki Pazartesi günü başlayacak çünkü tweaked çalışmalarına dayalı
function business_days_add($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 $current_date;
}
Ve iş günü açısından iki tarih arasında fark çalışma dışarı:
function business_days_diff($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;
}
Bir not olarak, 86.400 kullanılarak, ya da 24 * 60 * 60 olan herkes, bir gün değil tam 24 saat yaz / kış zaman değil ... unutmadan zaman değişiklikleri yapın lütfen. Biraz yavaş strtotime ('+1 gün', $ timestamp) olsa da, çok daha güvenilir.
Burada bir özyinelemeli çözümdür. Kolayca, sadece takip ve son tarih dönmek için modifiye edilebilir.
// 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);
working days without holidays arasında iki tarih sayısını alın:
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 $days;
}
function get_business_days_forward_from_date($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 + $is_business_day; // increment if this is a business day
}
// if we get here, you are hosed
return ("ERROR");
}
Add_business_days küçük bir hata var. Mevcut fonksiyonu ile takip deneyin ve çıkış bir Cumartesi olacak.
Startdate = Friday Business days to add = 1 Holidays array = Add date for the following Monday.
Ben aşağıda benim işlevi bu sabit.
function add_business_days($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, $dayx);}
Ben sadece bana mükemmel çalıştı bazı şeyler ekleyerek, Bobin ve mcgrailm koduna dayalı çalışma benim fonksiyonu olsun.
function add_business_days($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.
Tüm Federal Tatil dahil etmek, ve (! Yukarıdaki 4 Haziran olarak hesaplanmıştır), ve ayrıca dizi anahtarı olarak tatil adını eklemek 4 Temmuz düzeltmek için, yukarıdaki James Makarna tarafından sunulan işlevine bir geliştirme ...
/**
* National American Holidays
* @param string $year
* @return array
*/
public static function getNationalAmericanHolidays($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 $bankHolidays;
}
Bobin, mcgrailm, Tony, James Pasta ve burada yayınlanan birkaç diğerleri için teşekkürler. Ben bir tarihte iş günü eklemek için kendi işlevi yazılı, ama burada bulunan bazı kod ile modifiye etmişti. Bu bir hafta sonu / tatil olmanın başlangıç tarihini idare edecek. Bu aynı zamanda iş saatlerini idare edecek. Ben bazı yorumlar eklenmiş ve daha kolay okunmasını sağlamak için kodunu kırmak.
<?php
function count_business_days($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);
?>
Şahsen ben bu daha temiz ve daha özlü bir çözüm olduğunu düşünüyorum:
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;
}
Sadece bu işlev için önerilen new tarihini gönderebilirsiniz.
Formula vardır:
number_of_days - math_round_down(10 * (number_of_days / (business_days_in_a_week * days_in_a_week)))
Tada! Eğer istersen bir de, bir hafta, bir ay içinde, iş günü sayıları hesaplamak.
math_round_down () is a hypothetical method, which implements a mathematical function that rounds down.
https://github.com/Arbitr108/useful_php İşte takvimine çalışma günü tahmin edebilecek bir sınıftır. Tahmin sadece ihtiyacınız tarihe takvim dönemi eklemek kadar sonra