blocks|key|2141220|text|如果限制为工作日，请使用字符串weekdays。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2141221|echo+date+(+'Y-m-j'+,+strtotime+(+'3+weekdays'+)+);|code-block|syntax|javascript|2141222|这应该会让你提前3个工作日，所以如果是周四，就会增加额外的周末时间。|2141223|来源：http://www.php.net/manual/en/datetime.formats.relative.php|offset|length|2141224|entityMap|0|LINK|mutability|MUTABLE|url|http://www.php.net/manual/en/datetime.formats.relative.php^0|0|0|0|3|1M|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|X|8|@]|9|@$K|Y|L|Z|1|10]]|A|$]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

If you are limiting to weekdays use the string weekdays.

<pre><code>echo date ( 'Y-m-j' , strtotime ( '3 weekdays' ) );
</code></pre>

This should jump you ahead by 3 weekdays, so if it is Thursday it will add the additional weekend time.

Source: <a href="http://www.php.net/manual/en/datetime.formats.relative.php" rel="noreferrer">http://www.php.net/manual/en/datetime.formats.relative.php</a>

blocks|key|2141250|text|当我需要更多的工作日时，我发现了这个问题。我正在寻找在本月1日之后的X个工作日。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2141251|一开始看起来很棒，直到添加了>5个工作日(类似于@zerkms的发现)。|2141252|事实证明，这对我来说更准确。|2141253|++++function+_getBusinessDayOfMonth(+$days+)+{+
++++++++$time+=+strtotime(date("m/1/Y+00:00"));+//finding+#+of+business+days+after+1st+of+the+month
++++++++$i+=+0;+//start+with+zero
++++++++while+($i+<+$days)+{+//loop+through+until+reached+the+amount+of+weekdays
++++++++++++$time+=+strtotime("%2B1+day",+$time);+//Increase+day+by+1
++++++++++++if+(date("N",+$time)+<+6)+{+//test+if+M-F
++++++++++++++++$i%2B%2B;+//Increase+by+1
++++++++++++}
++++++++}
++++++++echo+date("m/d/Y",+$time);
++++}|code-block|syntax|javascript|2141254|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|O|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|P|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I have found this buggy when needing a larger amount of weekdays. I was looking for X amount of business days after the 1st of the current month. 

Looked great at first until after adding > 5 business days (similar to what @zerkms found). 

This has proved more accurate for me. 

<pre><code> function _getBusinessDayOfMonth( $days ) { 
 $time = strtotime(date("m/1/Y 00:00")); //finding # of business days after 1st of the month
 $i = 0; //start with zero
 while ($i &lt; $days) { //loop through until reached the amount of weekdays
 $time = strtotime("+1 day", $time); //Increase day by 1
 if (date("N", $time) &lt; 6) { //test if M-F
 $i++; //Increase by 1
 }
 }
 echo date("m/d/Y", $time);
 }
</code></pre>

blocks|key|72677|text|我认为一个功能可以很容易地开发，你可以只导出当前日期的星期几，你可以添加两个，mod+5将很容易地给你工作日。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|72678|function+increaseWorkDay($workDayToProcess,+$dayToAdd){
++++if($workDayToProcess+>=+4+&&+$workDayToProcess+<=+6){
++++++++$workDayToProcess=+4;
++++}
++++$workDayToProcess%2B=+$dayToAdd;

++++return+$workDayToProcess+%25+5;
}|code-block|syntax|javascript|72679|您可以使用数组导出工作日名称，也可以使用此方法。|72680|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I think a function can be easily developed that you can just export current day no of the week and you can add two and mod of 5 will give you easily weekday.

<pre><code>function increaseWorkDay($workDayToProcess, $dayToAdd){
 if($workDayToProcess &gt;= 4 &amp;&amp; $workDayToProcess &lt;= 6){
 $workDayToProcess= 4;
 }
 $workDayToProcess+= $dayToAdd;

 return $workDayToProcess % 5;
}
</code></pre>

And you can export the weekday name by using an array, this method can alternatively be used.

blocks|key|2141333|text|我递归地做，对我很有效|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2141334|function+add_work_days($date,+$day){
++++if($day+==+0)
++++++++return+$date;

++++$date->add(new+DateInterval('P1D'));

++++if(!in_array($date->format('N'),+array('6',+'7')))
++++++++$day--;

++++return+add_work_days($date,+$day);
}

$date++=+add_work_days(new+DateTime(),+3);
echo+$date->format('d/m/Y');|code-block|syntax|javascript|2141335|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I do it recursively, worked for me

<pre><code>function add_work_days($date, $day){
 if($day == 0)
 return $date;

 $date-&gt;add(new DateInterval('P1D'));

 if(!in_array($date-&gt;format('N'), array('6', '7')))
 $day--;

 return add_work_days($date, $day);
}

$date = add_work_days(new DateTime(), 3);
echo $date-&gt;format('d/m/Y');
</code></pre>

blocks|key|2141366|text|对于较早版本的PHP+<+5.3|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2141367|function+AddWorkingDays($startDate,+$adddays)
{
++$retdate+=+$startDate;
++$sign+=+"%2B";
++if($adddays+<+0){
++++$adddays+=+$adddays*-1;
++++$sign+=+"-";
++}
++while+($adddays+>+0)+{
++++$retdate+=+date+(+'Y-m-d'+,+strtotime+(+"$retdate+{$sign}1+day"+)+);

++++++$what_day+=+date("N",+strtotime($retdate));
++++++if+(+$what_day+!=+6+&&+$what_day+!=+7+)+//+6+and+7+are+weekend
++++++++++$adddays--;

++};

++return+$retdate;

}|code-block|syntax|javascript|2141368|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

For older versions of PHP &lt; 5.3

<pre><code>function AddWorkingDays($startDate, $adddays)
{
 $retdate = $startDate;
 $sign = "+";
 if($adddays &lt; 0){
 $adddays = $adddays*-1;
 $sign = "-";
 }
 while ($adddays &gt; 0) {
 $retdate = date ( 'Y-m-d' , strtotime ( "$retdate {$sign}1 day" ) );

 $what_day = date("N", strtotime($retdate));
 if ( $what_day != 6 &amp;&amp; $what_day != 7 ) // 6 and 7 are weekend
 $adddays--;

 };

 return $retdate;

}
</code></pre>

blocks|key|72757|text|对于PHP+>=+5.6|type|unstyled|depth|inlineStyleRanges|entityRanges|data|72758|public+function+addWorkingDays($date,+$day)
{

++++if+(!($date+instanceof+\DateTime)+%7C%7C+is_string($date))+{
++++++++$date+=+new+\DateTime($date);
++++}

++++if+($date+instanceof+\DateTime)+{
++++++++$newDate+=+clone+$date;
++++}

++++if+($day+==+0)+{
++++++++return+$newDate;
++++}

++++$i+=+1;

++++while+($i+<=+abs($day))+{

++++++++$newDate->modify(($day+>+0+?+'+%2B'+:+'+-')+.+'1+day');

++++++++$next_day_number+=+$newDate->format('N');

++++++++if+(!in_array($next_day_number,+[6,+7]))+{
++++++++++++$i%2B%2B;
++++++++}

++++}

++++return+$newDate;

}|code-block|syntax|javascript|72759|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

For PHP >= 5.6 

<pre><code>public function addWorkingDays($date, $day)
{

 if (!($date instanceof \DateTime) || is_string($date)) {
 $date = new \DateTime($date);
 }

 if ($date instanceof \DateTime) {
 $newDate = clone $date;
 }

 if ($day == 0) {
 return $newDate;
 }

 $i = 1;

 while ($i &lt;= abs($day)) {

 $newDate-&gt;modify(($day &gt; 0 ? ' +' : ' -') . '1 day');

 $next_day_number = $newDate-&gt;format('N');

 if (!in_array($next_day_number, [6, 7])) {
 $i++;
 }

 }

 return $newDate;

}
</code></pre>

blocks|key|1388147|text|你也可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1388148|echo+date_create()->modify("%2B3+weekdays")->format("Y-m-d");|code-block|syntax|javascript|1388149|请参见：|1388150|https://www.php.net/manual/en/function.date-create.php|offset|length|1388151|https://www.php.net/manual/en/datetime.modify.php|1388152|entityMap|0|LINK|mutability|MUTABLE|url|1^0|0|0|0|0|1I|0|0|0|1D|1|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|Z|8|@]|9|@$K|10|L|11|1|12]]|A|$]]|$1|M|3|N|5|6|7|13|8|@]|9|@$K|14|L|15|1|16]]|A|$]]|$1|O|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|J]]|V|$5|R|S|T|A|$U|N]]]]

You can do this too:

<pre><code>echo date_create()-&gt;modify("+3 weekdays")-&gt;format("Y-m-d");
</code></pre>

See:

<a href="https://www.php.net/manual/en/function.date-create.php" rel="nofollow noreferrer">https://www.php.net/manual/en/function.date-create.php</a>

<a href="https://www.php.net/manual/en/datetime.modify.php" rel="nofollow noreferrer">https://www.php.net/manual/en/datetime.modify.php</a>

Is there a way to use <code>strtotime</code> to add working days (Monday to Friday) to a date? Or some other method? What I want to do is:

<pre><code>date ( 'Y-m-j' , strtotime ( '+3 working days' ) )
</code></pre>

Working days (Mon-Fri) in PHP

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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有没有办法使用strtotime将工作日(星期一到星期五)添加到日期中？或者其他方法？我想做的是：date ( 'Y-m-j' , strtotime ( '+3 working days' ) )

问以PHP表示的工作日(星期一至星期五)
EN

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问以PHP表示的工作日(星期一至星期五)EN