PHPで「営業日」を追加する方法が必要です。たとえば、12/5 金曜日 + 3 営業日 = 12/10 水曜日。
少なくとも、週末を理解するコードが必要ですが、理想的には、米国連邦の祝日も考慮に入れる必要があります。必要に応じて力ずくで解決策を思いつくことができると確信していますが、もっとエレガントなアプローチがあることを願っています。誰？
ありがとう。
AdamTheHutt
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141000 次
37 に答える
109
これは、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、うるう年のサポートを追加するコメントの以前の機能の改善です。
開始日と終了日を入力し、その間にある可能性のある休日の配列を入力すると、営業日が整数として返されます。
<?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
?>
于 2008-12-03T03:56:28.000 に答える
14
date()関数には役立つ引数がいくつかあります。date("w") をチェックすると、日曜日の 0 から土曜日の 6 までの曜日の数字が表示されます。だから..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にすぎません。
于 2008-12-03T03:35:42.207 に答える
11
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が必要な銀行のアプリケーションを書いていますが、まだ大まかな基準しか得られません。
/**
* 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;
}
于 2012-11-29T20:01:52.477 に答える
11
$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);
于 2014-09-04T05:54:23.390 に答える
7
これは、日付に営業日を追加する関数です
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 getWorkingDays について言及されています (php.net のコメントから、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壊れると思います。
以下を使用します (以前の投稿の 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,$holidays);
4 ではなく 5 として結果を返します
Sun, 08 Jan 2012 00:00:00 +0000 weekend
Mon, 09 Jan 2012 00:00:00 +0000
Tue, 10 Jan 2012 00:00:00 +0000 holiday
Wed, 11 Jan 2012 00:00:00 +0000
Thu, 12 Jan 2012 00:00:00 +0000
Fri, 13 Jan 2012 00:00:00 +0000
上記の生成には、次の関数が使用されました。
function get_working_days($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,$holidays);
休日を迎えて...
于 2009-12-02T01:59:43.150 に答える
7
よりシンプルなこの機能を試すことができます。
function getWorkingDays($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 $no_days;
}
}
于 2015-09-08T13:53:20.610 に答える
3
@mcgrailm の作業に基づく私のバージョンは、レポートを 3 営業日以内にレビューする必要があり、週末に提出された場合、カウントは次の月曜日に開始されるため、微調整されています。
function business_days_add($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 $current_date;
}
そして、2 つの日付の差を営業日で計算します。
function business_days_diff($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、忘れないでください... あなたの忘却時間は、1 日が正確に 24 時間ではない冬時間/夏時間から変わります。strtotime('+1 day', $timestamp) は少し遅くなりますが、はるかに信頼性が高くなります。
于 2011-10-14T10:34:27.047 に答える
2
ブルートは労働時間を検出しようとします-月曜日から金曜日の午前8時から午後4時：
if (date('N')<6 && date('G')>8 && date('G')<16) {
// we have a working time (or check for holidays)
}
于 2011-12-07T23:37:27.507 に答える
2
特定の日付から営業日を加算または減算する関数で、休日は考慮されません。
function dateFromBusinessDays($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 $dateTime;
}
みたいに使って…
echo date("m/d/Y", dateFromBusinessDays(-7));
echo date("m/d/Y", dateFromBusinessDays(3, time() + 3*60*60*24));
于 2010-06-19T16:23:17.743 に答える
1
ボビンの最初の例から始めて、これで終わったのと同じニーズがありました
function add_business_days($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);
}
hth誰か
于 2010-06-29T15:36:01.510 に答える
1
休日の場合は、date() が生成できる形式で日付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している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);
于 2008-12-03T03:49:06.563 に答える
1
休日とカスタムの稼働日を含む 2 つの日付間の稼働日を計算する
答えはそれほど簡単ではありません。したがって、私の提案は、単純化された関数に依存する (または固定のロケールと文化を想定する) 以上の構成が可能なクラスを使用することです。特定の営業日後の日付を取得するには、次のようにします。
- 勤務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することができます。
- 正確な祝日（国と州）を考慮する必要があることを知っておく必要があります
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;
于 2015-05-20T09:47:49.843 に答える
1
バリエーション 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 $timestamp;
}
バリエーション 3:
<?php
/*
* Does not count current day, the date returned is
* a date following the last business day (can be weekend or not.
* See above for alternatives)
* Requires PHP 5.1 (Using ISO-8601 week)
*/
function businessDays($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。ノート！すべてのタイムスタンプが同じ時刻 (真夜中など) であることを確認してください。
休日の日付の配列を (unixtimestamps として) 作成します。つまり:
$holidays = array_flip(strtotime('2011-01-01'),strtotime('2011-12-25'));
行を修正:
if (date('N', $timestamp)<6) $bDays--;
することが ：
if (date('N', $timestamp)<6 && !isset($holidays[$timestamp])) $bDays--;
終わり！
<?php
/*
* Does not count current day, the date returned is the last business day
* Requires PHP 5.1 (Using ISO-8601 week)
*/
function businessDays($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;
}
于 2011-03-02T19:02:01.717 に答える
1
<?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);
}
?>
于 2011-04-07T13:24:03.063 に答える
1
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";
于 2012-05-31T16:17:17.967 に答える
1
<?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>";
?>
于 2010-10-07T17:20:04.580 に答える
1
これは別の解決策です。in_array で休日をチェックするよりも約 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));
}
于 2016-11-18T10:33:52.353 に答える
1
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);
于 2012-03-21T21:50:15.893 に答える
0
ボビンとmcgrailm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.
于 2011-01-25T03:17:16.230 に答える
0
function get_business_days_forward_from_date($num_days, $start_date='', $rtn_fmt='Ymd') {
// $start_date will default to today
if ($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");
}
于 2010-04-02T15:36:31.447 に答える
0
営業日を操作するために使用できる API の作成が完了しました (これらのソリューションはどれも、私の状況ではまったく機能しませんでした :-); 他の誰かが役に立つと思う場合に備えて、ここにリンクします。
〜ネイト
于 2013-06-09T03:17:28.527 に答える
0
個人的には、これはよりクリーン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 $d;
}
提案されたnew日付をこの関数に送信するだけです。
于 2013-12-19T15:36:24.500 に答える
0
この関数を作成したところ、非常にうまく機能しているようです。
function getBusinessDays($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");
于 2014-08-14T13:32:05.833 に答える
0
上記の James Pasta が提供する機能の拡張。すべての連邦の祝日を含め、7 月 4 日を修正し (上記では 6 月 4 日として計算されました!)、配列キーとして祝日名も含めます...
/**
* アメリカの祝日
* @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 $bankHolidays;
}
于 2013-01-14T13:08:46.103 に答える
0
この質問には、執筆時点で 35 を超える回答がありますが、この問題に対するより良い解決策があります。
@flamingLogosは、数学に基づいた、ループを含まない回答を提供しました。その解は の時間計算量を提供しますΘ(1)。ただし、その背後にある計算、特にうるう年の処理はかなり複雑です。
@Glavićは、最小限でエレガントな優れたソリューションを提供しました。ただし、一定時間内に計算を実行しないため、1 日間隔でループするため、10 年または 100 年のような長い期間で使用すると、サービス拒否 (DOS) 攻撃または少なくともタイムアウトが発生する可能性があります。
そこで私は、時間が一定でありながら非常に読みやすい数学的アプローチを提案します。
アイデアは、完全な週に何日かかるかを数えることです。
<?php
function getWorkingHours($start_date, $end_date) {
// validate input
if(!validateDate($start_date) || !validateDate($end_date)) return ['error' => 'Invalid Date'];
if($end_date < $start_date) return ['error' => 'End date must be greater than or equal Start date'];
//We save timezone and switch to UTC to prevent issues
$old_timezone = date_default_timezone_get();
date_default_timezone_set("UTC");
$startDate = strtotime($start_date);
$endDate = strtotime($end_date);
//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 include both dates in the interval.
$days = ($endDate - $startDate) / 86400 + 1;
$no_full_weeks = ceil($days / 7);
//we get only missing days count to complete full weeks
//we take modulo 7 in case it was already full weeks
$no_of_missing_days = (7 - ($days % 7)) % 7;
$workingDays = $no_full_weeks * 5;
//Next we remove the working days we added, this loop will have max of 6 iterations.
for ($i = 1; $i <= $no_of_missing_days; $i++){
if(date('N', $endDate + $i * 86400) < 6) $workingDays--;
}
$holidays = getHolidays(date('Y', $startDate), date('Y', $endDate));
//We subtract the holidays
foreach($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)
$workingDays--;
}
date_default_timezone_set($old_timezone);
return ['days' => $workingDays];
}
関数への入力はY-m-d、php またはyyyy-mm-dd一般的な日付形式の形式です。
get holiday 関数は、開始年から終了年までの休日の日付の配列を返します。
于 2020-10-20T21:25:55.423 に答える
0
ここに投稿してくれた Bobbin、mcgrailm、Tony、James Pasta、その他数名に感謝します。日付に営業日を追加する独自の関数を作成しましたが、ここで見つけたコードで変更しました。これにより、週末/休日の開始日が処理されます。営業時間もこれに準じます。コメントを追加し、コードを分割して読みやすくしました。
<?php
function count_business_days($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, $holidays);
?>
于 2013-07-09T22:50:15.390 に答える
0
add_business_days には小さなバグがあります。既存の関数で次を試すと、出力は土曜日になります。
Startdate = Friday 追加する営業日 = 1 Holidays 配列 = 次の月曜日の日付を追加します。
以下の関数でそれを修正しました。
function add_business_days($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);}
于 2011-02-04T08:20:45.897 に答える
-1
式があります：
number_of_days - math_round_down(10 * (number_of_days / (business_days_in_a_week * days_in_a_week)))
多田！1 か月、1 週間、任意の営業日数を計算します。
math_round_down () は、切り捨てを行う数学関数を実装する仮想メソッドです。
于 2012-07-26T19:54:18.073 に答える
-1
https://github.com/Arbitr108/useful_php稼働日をカレンダーに換算できるクラスです。したがって、見積もりの後、必要な日付にカレンダー期間を追加するだけです
于 2013-05-12T17:03:47.040 に答える