営業日を計算する
PHPに「営業日」を追加する方法が必要です。たとえば、金曜日12/5 + 3営業日=水曜日12/10。
少なくとも週末を理解するためのコードが必要ですが、理想的には米国連邦の祝日も考慮する必要があります。必要に応じて総当たりで解決策を考え出すことができると確信していますが、もっと洗練されたアプローチがあることを望んでいます。誰でも？
ありがとう。
以下は、PHPマニュアルのdate（）関数ページの ユーザーコメント からの関数です。これは、リープのサポートを追加するコメントの以前の関数の改良です。年。
開始日と終了日、および間にある可能性のある休日の配列を入力すると、整数として稼働日が返されます。
<?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
?>
休日のない稼働日の2つの日付の間の数を取得します。
使用例：
echo number_of_working_days('2013-12-23', '2013-12-29');
出力：
3
関数：
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;
}
date（） 関数に役立ついくつかの引数があります。 date（ "w"）を確認すると、日曜日の0から土曜日の6までの曜日の番号が表示されます。それで…多分何か….
$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;
これは、1つの可能性の大まかな例です。
休日の計算は各州で非標準です。私はいくつかの厳しいビジネスルールを必要とする銀行のアプリケーションを書いていますが、それでも大まかな標準しか取得できません。
/**
* 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);
これは、日付にビジネス日を追加するための関数です
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');
別の投稿ではgetWorkingDaysについて言及しています（php.netのコメントから引用し、ここに含まれています）が、日曜日に開始して就業日に終了すると中断すると思います。
以下を使用して（前の投稿のgetWorkingDays関数を含める必要があります）
date_default_timezone_set('Europe\London');
//Example:
$holidays = array('2012-01-10');
$startDate = '2012-01-08';
$endDate = '2012-01-13';
echo getWorkingDays( $startDate,$endDate,$holidays);
結果を4ではなく5として返します
Sun, 08 Jan 2012 00:00:00 +0000 weekend
Mon, 09 Jan 2012 00:00:00 +0000
Tue, 10 Jan 2012 00:00:00 +0000 holiday
Wed, 11 Jan 2012 00:00:00 +0000
Thu, 12 Jan 2012 00:00:00 +0000
Fri, 13 Jan 2012 00:00:00 +0000
上記の生成には、次の関数が使用されました。
function get_working_days($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,$holidays);
休日を迎えましょう...
より簡単なこの機能を試すことができます。
function getWorkingDays($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;
}
}
以下に、毎日のforループのない別のソリューションを示します。
$from = new DateTime($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
労働時間を検出しようとする野teな試み-月曜日から金曜日の午前8時から午後4時まで：
if (date('N')<6 && date('G')>8 && date('G')<16) {
// we have a working time (or check for holidays)
}
以下は、特定の日付から営業日を計算するための作業コードです。
<?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, $holiday_date_array, 15);
?>
//output after adding 15 business working days in 2016-03-01 will be "2016-03-23"
レポートは3営業日以内にレビューする必要があるため、@ mcgrailmの作業に基づいた私のバージョンを調整しました。週末に提出した場合、カウントは次の月曜日に開始されます。
function business_days_add($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 $current_date;
}
そして、営業日に関して2つの日付の差を計算します。
function business_days_diff($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 $business_days;
}
注意として、86400または24 * 60 * 60を使用しているすべての人は、...忘れる時間は、1日が24時間ではない冬/夏時間から変わることに注意してください。 strtotime（ '+ 1 day'、$ timestamp）は少し遅くなりますが、はるかに信頼性が高くなります。
特定の日付から営業日を加算または減算する関数。これは休日を考慮しません。
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;
}
のように使用します...
echo date("m/d/Y", dateFromBusinessDays(-7));
echo date("m/d/Y", dateFromBusinessDays(3, time() + 3*60*60*24));
これは別のソリューションです。in_arrayで休日をチェックするよりもほぼ25％高速です。
/**
* 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));
}
休日の場合、date（）が生成できる形式で曜日の配列を作成します。例：
// I know, these aren't holidays
$holidays = array(
'Jan 2',
'Feb 3',
'Mar 5',
'Apr 7',
// ...
);
次に、 in_array（） および date（） 関数を使用して、タイムスタンプが休日を表すかどうかを確認します。
$day_of_year = date('M j', $timestamp);
$is_holiday = in_array($day_of_year, $holidays);
このコードスニペットは、週末と祝日がなくても営業日を計算するのが非常に簡単です。
function getWorkingDays($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,$holidays)
バリアント1：
<?php
/*
* Does not count current day, the date returned is the last business day
* Requires PHP 5.1 (Using ISO-8601 week)
*/
function businessDays($timestamp = false, $bDays = 2) {
if($timestamp === false) $timestamp = time();
while ($bDays>0) {
$timestamp += 86400;
if (date('N', $timestamp)<6) $bDays--;
}
return $timestamp;
}
バリアント2：
<?php
/*
* Does not count current day, the date returned is a business day
* following the last business day
* Requires PHP 5.1 (Using ISO-8601 week)
*/
function businessDays($timestamp = false, $bDays = 2) {
if($timestamp === false) $timestamp = time();
while ($bDays+1>0) {
$timestamp += 86400;
if (date('N', $timestamp)<6) $bDays--;
}
return $timestamp;
}
バリアント3：
<?php
/*
* Does not count current day, the date returned is
* a date following the last business day (can be weekend or not.
* See above for alternatives)
* Requires PHP 5.1 (Using ISO-8601 week)
*/
function businessDays($timestamp = false, $bDays = 2) {
if($timestamp === false) $timestamp = time();
while ($bDays>0) {
$timestamp += 86400;
if (date('N', $timestamp)<6) $bDays--;
}
return $timestamp += 86400;
}
追加の休日の考慮事項は、以下を実行することにより、上記のバリエーションを使用して行うことができます。注意！すべてのタイムスタンプが同じ時刻（つまり、真夜中）であることを確認します。
休日の日付の配列を作成します（unixtimestampsとして）、つまり：
$holidays = array_flip(strtotime('2011-01-01'),strtotime('2011-12-25'));
行を変更します。
if (date('N', $timestamp)<6) $bDays--;
することが ：
if (date('N', $timestamp)<6 && !isset($holidays[$timestamp])) $bDays--;
できた！
<?php
/*
* Does not count current day, the date returned is the last business day
* Requires PHP 5.1 (Using ISO-8601 week)
*/
function businessDays($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 $timestamp;
}
私はボビンの最初の例から始めてこれで終わったこの同じニーズを持っていました
function add_business_days($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);
}
誰か
<?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>";
?>
休日とカスタム就業週を含む2つの日付間の就業日を計算します
答えはささいなことではありません-したがって、私の提案は、単純な機能に依存する（または固定のロケールとカルチャを想定する）以上のことを構成できるクラスを使用することです。特定の就業日数後に日付を取得するには：
- 作業する曜日を指定する必要があります（デフォルトはMON-FRI）-クラスでは、各曜日を個別に有効または無効にできます。
- 祝日（国と州）を正確に考慮する必要があることを知っておく必要があります。
- 例えば https://github.com/khatfield/php-HolidayLibrary/blob/master/Holidays.class.php
- または、データをハードコーディングします。 from http://www.feiertagskalender.ch/?hl=en
- またはデータAPIの支払い http://www.timeanddate.com/services/api/holiday-api.html
機能的アプローチ
/**
* @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;
これが再帰的な解決策です。最新の日付のみを追跡して返すように簡単に変更できます。
// 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);
私はパーティーに遅れていることは知っていますが、休日と営業日を把握するために、Marcos J. Montesのこの古い機能セットを使用しています。彼はイースターのために1876年からアルゴリズムを追加するのに時間をかけ、米国の主要な祝日をすべて追加しました。これは他の国でも簡単に更新できます。
//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;
}
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";
<?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);
}
?>
Bobbin、mcgrailm、Tony、James Pasta、およびここに投稿した他の数人に感謝します。日付に営業日を追加する独自の関数を作成しましたが、ここで見つけたいくつかのコードで修正しました。これにより、週末/休日の開始日が処理されます。これも営業時間を処理します。いくつかのコメントを追加し、コードを分割して読みやすくしました。
<?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, $holidays);
?>
function get_business_days_forward_from_date（$ 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");
}
個人的には、これはより簡潔で簡潔なソリューションだと思います。
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 $d;
}
提案されたnew日付をこの関数に送信するだけです。
PHPClassesには、この名前の PHP Working Days という名前のNiceクラスがあります。このクラスを確認できます。
この関数を作成しましたが、非常にうまく機能しているようです：
function getBusinessDays($date1, $date2){
if(!is_numeric($date1)){
$date1 = strtotime($date1);
}
if(!is_numeric($date2)){
$date2 = strtotime($date2);
}
if($date2 < $date1){
$temp_date = $date1;
$date1 = $date2;
$date2 = $temp_date;
unset($temp_date);
}
$diff = $date2 - $date1;
$days_diff = intval($diff / (3600 * 24));
$current_day_of_week = intval(date("N", $date1));
$business_days = 0;
for($i = 1; $i <= $days_diff; $i++){
if(!in_array($current_day_of_week, array("Sunday" => 1, "Saturday" => 7))){
$business_days++;
}
$current_day_of_week++;
if($current_day_of_week > 7){
$current_day_of_week = 1;
}
}
return $business_days;
}
echo "Business days: " . getBusinessDays("8/15/2014", "8/8/2014");
Add_business_daysには小さなバグがあります。既存の関数で次を試してください。出力は土曜日になります。
開始日=金曜日追加する営業日= 1休日配列=次の月曜日の日付を追加します。
以下の関数で修正しました。
function add_business_days($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);}
2つの日付の間の時間を取得する必要がある場合は、 https://github.com/maximnara/business-days-counter を使用できます。簡単に機能しますが、laravel now $diffInSeconds = $this->datesCounter->getDifferenceInSeconds(Carbon::create(2019, 1, 1), Carbon::now(), DateCounter::COUNTRY_FR);
祝日と週末はカウントされません。たとえば、起動時間を9時間から18時間に設定したり、起動時間を設定したりすることはできません。
または、週末だけが必要で、Carbonを使用する場合は、組み込み関数を使用できます。
$date1 = Carbon::create(2019, 1, 1)->endOfDay();
$date2 = $dt->copy()->startOfDay();
$diff = $date1->diffFiltered(CarbonInterval::minute(), function(Carbon $date) {
return !$date->isWeekend();
}, $date2, true);
しかし、それは間隔で毎分foreachします、ビット間隔のために、それは時間がかかることができます。
Bobbinとmcgrailmのコードに基づいて関数を機能させ、完璧に機能するものをいくつか追加しました。
function add_business_days($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.
上記のジェームズパスタが提供する機能の強化、すべての連邦休日を含める、7月4日を修正する（上記の6月4日として計算された！）、配列キーとして休日名も含める...
/ **
*国民の祝日
* @param string $ year
* @return配列
* /
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;
}
営業日を操作するために使用できるAPIの記述を終了しました（これらのソリューションはどれも私の状況ではうまくいきませんでした:-)。他の誰かが便利だと思った場合にここにリンクします。
〜ネイト