php PHP中给定日期的下一个工作日
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Next business day of given date in PHP
提问by Marko
Does anyone have a PHP snippet to calculate the next business day for a given date? How does, for example, YYYY-MM-DD need to be converted to find out the next business day?
有没有人有一个 PHP 代码段来计算给定日期的下一个工作日？例如，需要如何转换 YYYY-MM-DD 才能找出下一个工作日？
Example: For 03.04.2011 (DD-MM-YYYY) the next business day is 04.04.2011. For 08.04.2011 the next business day is 11.04.2011.
示例：对于 03.04.2011 (DD-MM-YYYY)，下一个工作日是 04.04.2011。对于 08.04.2011，下一个工作日是 11.04.2011。
This is the variable containing the date I need to know the next business day for
这是包含我需要知道下一个工作日的日期的变量
$cubeTime['time'];
Variable contains: 2011-04-01 result of the snippet should be: 2011-04-04
变量包含：2011-04-01 片段的结果应该是：2011-04-04
回答by Dave
Next Weekday
下一个工作日
This finds the next weekday from a specific date (not including Saturday or Sunday):
这将查找特定日期（不包括星期六或星期日）的下一个工作日：
echo date('Y-m-d', strtotime('2011-04-05 +1 Weekday'));
You could also do it with a date variable of course:
当然，您也可以使用日期变量来做到这一点：
$myDate = '2011-04-05';
echo date('Y-m-d', strtotime($myDate . ' +1 Weekday'));
UPDATE:Or, if you have access to PHP's DateTime class(very likely):
更新：或者，如果您有权访问PHP 的 DateTime 类（很可能）：
$date = new DateTime('2018-01-27');
$date->modify('+7 weekday');
echo $date->format('Y-m-d');
Want to Skip Holidays?:
想跳过假期？：
Although the original poster mentioned "I don't need to consider holidays", if you DO happen to want to ignore holidays, just remember - "Holidays" is just an array of whatever dates you don't want to include and differs by country, region, company, person...etc.
尽管原始海报提到“我不需要考虑假期”，但如果您碰巧想忽略假期，请记住-“假期”只是您不想包含的任何日期的数组，并且因国家/地区而异，地区，公司，个人...等。
Simply put the above code into a function that excludes/loops past the dates you don't want included. Something like this:
只需将上述代码放入一个函数中，该函数会排除/循环超过您不希望包含的日期。像这样的东西：
$tmpDate = '2015-06-22';
$holidays = ['2015-07-04', '2015-10-31', '2015-12-25'];
$i = 1;
$nextBusinessDay = date('Y-m-d', strtotime($tmpDate . ' +' . $i . ' Weekday'));
while (in_array($nextBusinessDay, $holidays)) {
$i++;
$nextBusinessDay = date('Y-m-d', strtotime($tmpDate . ' +' . $i . ' Weekday'));
}
I'm sure the above code can be simplified or shortened if you want. I tried to write it in an easy-to-understand way.
如果你愿意，我相信上面的代码可以被简化或缩短。我试图以易于理解的方式编写它。
回答by user2412642
For UK holidays you can use
对于英国假期，您可以使用
https://www.gov.uk/bank-holidays#england-and-wales
https://www.gov.uk/bank-holidays#england-and-wales
The ICS format data is easy to parse. My suggestion is...
ICS 格式数据易于解析。我的建议是...
# $date must be in YYYY-MM-DD format
# You can pass in either an array of holidays in YYYYMMDD format
# OR a URL for a .ics file containing holidays
# this defaults to the UK government holiday data for England and Wales
function addBusinessDays($date,$numDays=1,$holidays='') {
if ($holidays==='') $holidays = 'https://www.gov.uk/bank-holidays/england-and-wales.ics';
if (!is_array($holidays)) {
$ch = curl_init($holidays);
curl_setopt($ch,CURLOPT_RETURNTRANSFER,true);
$ics = curl_exec($ch);
curl_close($ch);
$ics = explode("\n",$ics);
$ics = preg_grep('/^DTSTART;/',$ics);
$holidays = preg_replace('/^DTSTART;VALUE=DATE:(\d{4})(\d{2})(\d{2}).*/s','--',$ics);
}
$addDay = 0;
while ($numDays--) {
while (true) {
$addDay++;
$newDate = date('Y-m-d', strtotime("$date +$addDay Days"));
$newDayOfWeek = date('w', strtotime($newDate));
if ( $newDayOfWeek>0 && $newDayOfWeek<6 && !in_array($newDate,$holidays)) break;
}
}
return $newDate;
}
回答by mhitza
function next_business_day($date) {
$add_day = 0;
do {
$add_day++;
$new_date = date('Y-m-d', strtotime("$date +$add_day Days"));
$new_day_of_week = date('w', strtotime($new_date));
} while($new_day_of_week == 6 || $new_day_of_week == 0);
return $new_date;
}
This function should ignore weekends (6 = Saturday and 0 = Sunday).
此函数应忽略周末（6 = 星期六和 0 = 星期日）。
回答by John Parker
What you need to do is:
你需要做的是：
Convert the provided date into a timestamp.
Use this along with the or
worNformatters for PHP's datecommand to tell you what day of the week it is.If it isn't a "business day", you can then increment the timestamp by a day (86400 seconds) and check again until you hit a business day.
将提供的日期转换为时间戳。
将它与PHP 的date命令的 or
w或Nformatters一起使用来告诉您它是一周中的哪一天。如果它不是“工作日”，您可以将时间戳增加一天（86400 秒）并再次检查，直到您到达工作日。
N.B.: For this is reallywork, you'd also need to exclude any bank or public holidays, etc.
注意：因为这确实是工作，您还需要排除任何银行或公共假期等。
回答by Angus
This function will calculate the business day in the future or past. Arguments are number of days, forward (1) or backwards(0), and a date. If no date is supplied todays date will be used:
此函数将计算未来或过去的营业日。参数是天数、向前 (1) 或向后 (0) 和日期。如果没有提供日期，将使用今天的日期：
// returned $date Y/m/d
function work_days_from_date($days, $forward, $date=NULL)
{
if(!$date)
{
$date = date('Y-m-d'); // if no date given, use todays date
}
while ($days != 0)
{
$forward == 1 ? $day = strtotime($date.' +1 day') : $day = strtotime($date.' -1 day');
$date = date('Y-m-d',$day);
if( date('N', strtotime($date)) <= 5) // if it's a weekday
{
$days--;
}
}
return $date;
}
回答by GeniusDesign
I stumbled apon this thread when I was working on a Danish website where I needed to code a "Next day delivery" PHP script.
当我在一个丹麦网站上工作时，我偶然发现了这个线程，我需要编写一个“次日交付”PHP 脚本。
Here is what I came up with (This will display the name of the next working day in Danish, and the next working + 1 if current time is more than a given limit)
这是我想出的（这将用丹麦语显示下一个工作日的名称，如果当前时间超过给定的限制，则下一个工作+1）
$day["Mon"] = "Mandag";
$day["Tue"] = "Tirsdag";
$day["Wed"] = "Onsdag";
$day["Thu"] = "Torsdag";
$day["Fri"] = "Fredag";
$day["Sat"] = "L?rdag";
$day["Sun"] = "S?ndag";
date_default_timezone_set('Europe/Copenhagen');
$date = date('l');
$checkTime = '1400';
$date2 = date(strtotime($date.' +1 Weekday'));
if( date( 'Hi' ) >= $checkTime) {
$date2 = date(strtotime($date.' +2 Weekday'));
}
if (date('l') == 'Saturday'){
$date2 = date(strtotime($date.' +2 Weekday'));
}
if (date('l') == 'Sunday') {
$date2 = date(strtotime($date.' +2 Weekday'));
}
echo '<p>N?ste levering: <span>'.$day[date("D", $date2)].'</span></p>';
As you can see in the sample code $checkTime is where I set the time limit which determines if the next day delivery will be +1 working day or +2 working days.
正如您在示例代码中看到的，$checkTime 是我设置时间限制的地方，该时间限制决定了第二天交货是 +1 个工作日还是 +2 个工作日。
'1400' = 14:00 hours
'1400' = 14:00 小时
I know that the if statements can be made more compressed, but I show my code for people to easily understand the way it works.
我知道可以更压缩 if 语句，但我向人们展示了我的代码，以便人们轻松理解它的工作方式。
I hope someone out there can use this little snippet.
我希望有人可以使用这个小片段。
回答by Suresh Kamrushi
See the example below:
请参阅下面的示例：
$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);
For more info: http://goo.gl/YOsfPX
欲了解更多信息：http: //goo.gl/YOsfPX
回答by Ishwor Khadka
You could do something like this.
你可以做这样的事情。
/**
* @param string $date
* @param DateTimeZone|null|null $DateTimeZone
* @return \NavigableDate\NavigableDateInterface
*/
function getNextBusinessDay(string $date, ? DateTimeZone $DateTimeZone = null):\NavigableDate\NavigableDateInterface
{
$Date = \NavigableDate\NavigableDateFacade::create($date, $DateTimeZone);
$NextDay = $Date->nextDay();
while(true)
{
$nextDayIndexInTheWeek = (int) $NextDay->format('N');
// check if the day is between Monday and Friday. In DateTime class php, Monday is 1 and Friday is 5
if ($nextDayIndexInTheWeek >= 1 && $nextDayIndexInTheWeek <= 5)
{
break;
}
$NextDay = $NextDay->nextDay();
}
return $NextDay;
}
$date = '2017-02-24';
$NextBussinessDay = getNextBusinessDay($date);
var_dump($NextBussinessDay->format('Y-m-d'));
Output:
输出：
string(10) "2017-02-27"
\NavigableDate\NavigableDateFacade::create($date, $DateTimeZone), is provided by php library available at https://packagist.org/packages/ishworkh/navigable-date. You need to first include this library in your project with composer or direct download.
\NavigableDate\NavigableDateFacade::create($date, $DateTimeZone)，由https://packagist.org/packages/ishworkh/navigable-date 上的php 库提供。您需要首先使用 composer 或直接下载将这个库包含在您的项目中。
回答by Nishant Bhatt
I used below methods in PHP, strtotime()does not work specially in leap yearFebruarymonth.
我在 PHP 中使用了以下方法，strtotime()在闰年2月不起作用。
public static function nextWorkingDay($date, $addDays = 1)
{
if (strlen(trim($date)) <= 10) {
$date = trim($date)." 09:00:00";
}
$date = new DateTime($date);
//Add days
$date->add(new DateInterval('P'.$addDays.'D'));
while ($date->format('N') >= 5)
{
$date->add(new DateInterval('P1D'));
}
return $date->format('Y-m-d H:i:s');
}
This solution for 5 working days (you can change if you required for 6 or 4 days working). if you want to exclude more days like holidays then just check another condition in while loop.
此解决方案 5 个工作日（如果需要 6 或 4 天的工作，您可以更改）。如果你想排除更多的日子，比如假期，那么只需在 while 循环中检查另一个条件。
//
while ($date->format('N') >= 5 && !in_array($date->format('Y-m-d'), self::holidayArray()))
