blocks|key|1142702|text|如果您使用的是PHP5.3，则可以使用DateTime对象及其add方法：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1142703|$Date1+=+'2010-09-17';
$date+=+new+DateTime($Date1);
$date->add(new+DateInterval('P1D'));+//+P1D+means+a+period+of+1+day
$Date2+=+$date->format('Y-m-d');|code-block|syntax|javascript|1142704|看看DateInterval+constructor手册页，看看如何构造其他时间段添加到您的日期中(2天是'P2D'，3天是'P3D'，依此类推)。|1142705|没有PHP5.3，你应该能够像以前那样使用strtotime+(我已经测试过了，它在5.1.6和5.2.10中都能工作)：|1142706|$Date1+=+'2010-09-17';
$Date2+=+date('Y-m-d',+strtotime($Date1+.+"+%2B+1+day"));
//+var_dump($Date2)+returns+"2010-09-18"|1142707|entityMap|0|LINK|mutability|MUTABLE|url|http://ca.php.net/manual/en/class.datetime.php|1|http://ca.php.net/manual/en/datetime.add.php|2|http://ca.php.net/manual/en/dateinterval.construct.php^0|J|8|V|3|J|8|0|V|3|1|0|0|2|C|1H|5|1Q|5|2|O|2|0|L|9|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@$9|17|A|18|1|19]|$9|1A|A|1B|1|1C]]|E|$]]|$1|F|3|G|5|H|7|1D|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1E|8|@$9|1F|A|1G|B|C]|$9|1H|A|1I|B|C]|$9|1J|A|1K|B|C]]|D|@$9|1L|A|1M|1|1N]]|E|$]]|$1|M|3|N|5|6|7|1O|8|@$9|1P|A|1Q|B|C]]|D|@]|E|$]]|$1|O|3|P|5|H|7|1R|8|@]|D|@]|E|$I|J]]|$1|Q|3|-4|5|6|7|1S|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]|Y|$5|T|U|V|E|$W|Z]]|10|$5|T|U|V|E|$W|11]]]]

If you're using PHP 5.3, you can use a <a href="http://ca.php.net/manual/en/class.datetime.php" rel="noreferrer"><code>DateTime</code></a> object and its <a href="http://ca.php.net/manual/en/datetime.add.php" rel="noreferrer"><code>add</code></a> method:

<pre><code>$Date1 = '2010-09-17';
$date = new DateTime($Date1);
$date-&gt;add(new DateInterval('P1D')); // P1D means a period of 1 day
$Date2 = $date-&gt;format('Y-m-d');
</code></pre>

Take a look at the <a href="http://ca.php.net/manual/en/dateinterval.construct.php" rel="noreferrer"><code>DateInterval</code> constructor</a> manual page to see how to construct other periods to add to your date (2 days would be <code>'P2D'</code>, 3 would be <code>'P3D'</code>, and so on).

Without PHP 5.3, you should be able to use <code>strtotime</code> the way you did it (I've tested it and it works in both 5.1.6 and 5.2.10):

<pre><code>$Date1 = '2010-09-17';
$Date2 = date('Y-m-d', strtotime($Date1 . " + 1 day"));
// var_dump($Date2) returns "2010-09-18"
</code></pre>

blocks|key|928255|text|您所要做的就是使用days代替day，如下所示：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|928256|<?php
$Date+=+"2010-09-17";
echo+date('Y-m-d',+strtotime($Date.+'+%2B+1+days'));
echo+date('Y-m-d',+strtotime($Date.+'+%2B+2+days'));
?>|code-block|syntax|javascript|928257|并且它正确地输出：|928258|2010-09-18
2010-09-19|928259|entityMap^0|9|4|F|3|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

All you have to do is use <code>days</code> instead of <code>day</code> like this:

<pre><code>&lt;?php
$Date = "2010-09-17";
echo date('Y-m-d', strtotime($Date. ' + 1 days'));
echo date('Y-m-d', strtotime($Date. ' + 2 days'));
?&gt;
</code></pre>

And it outputs correctly:

<pre><code>2010-09-18
2010-09-19
</code></pre>

blocks|key|1142770|text|所有的代码都必须使用以下代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1142771|$nday+=+time()+%2B+(+24+*+60+*+60);++++
echo+'Now:+++++++'.+date('Y-m-d')+."\n";++++
echo+'Next+Day:+'.+date('Y-m-d',+$nday)+."\n";|code-block|syntax|javascript|1142772|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|$]]

All have to use bellow code:

<pre><code>$nday = time() + ( 24 * 60 * 60); 
echo 'Now: '. date('Y-m-d') ."\n"; 
echo 'Next Day: '. date('Y-m-d', $nday) ."\n";
</code></pre>

blocks|key|928318|text|您还可以使用以下格式|type|unstyled|depth|inlineStyleRanges|entityRanges|data|928319|strtotime("-3+days",+time());
strtotime("%2B1+day",+strtotime($date));|code-block|syntax|javascript|928320|您可以通过以下方式堆叠更改：|928321|strtotime("%2B1+day",+strtotime("%2B1+year",+strtotime($date)));|928322|注意这种方法与其他答案之间的区别:不需要连接%2B1+day和<timestamp>值，只需将时间戳作为strtotime的第二个参数传递即可。|offset|length|style|CODE|928323|entityMap^0|0|0|0|0|M|6|T|B|1E|9|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]|$M|Z|N|10|O|P]|$M|11|N|12|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|R|$]]

You can also use the following format

<pre><code>strtotime("-3 days", time());
strtotime("+1 day", strtotime($date));
</code></pre>

You can stack changes this way:

<pre><code>strtotime("+1 day", strtotime("+1 year", strtotime($date)));
</code></pre>

Note the difference between this approach and the one in other answers: instead of concatenating the values <code>+1 day</code> and <code>&lt;timestamp&gt;</code>, you can just pass in the timestamp as the second parameter of <code>strtotime</code>.

blocks|key|928368|text|下面是演示日期修改的一小段代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|928369|$date+=+date("Y-m-d");
//increment+2+days
$mod_date+=+strtotime($date."%2B+2+days");
echo+date("Y-m-d",$mod_date)+.+"\n";

//decrement+2+days
$mod_date+=+strtotime($date."-+2+days");
echo+date("Y-m-d",$mod_date)+.+"\n";

//increment+1+month
$mod_date+=+strtotime($date."%2B+1+months");
echo+date("Y-m-d",$mod_date)+.+"\n";

//increment+1+year
$mod_date+=+strtotime($date."%2B+1+years");
echo+date("Y-m-d",$mod_date)+.+"\n";|code-block|syntax|javascript|928370|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|$]]

Here is a small snippet to demonstrate the date modifications:

<pre><code>$date = date("Y-m-d");
//increment 2 days
$mod_date = strtotime($date."+ 2 days");
echo date("Y-m-d",$mod_date) . "\n";

//decrement 2 days
$mod_date = strtotime($date."- 2 days");
echo date("Y-m-d",$mod_date) . "\n";

//increment 1 month
$mod_date = strtotime($date."+ 1 months");
echo date("Y-m-d",$mod_date) . "\n";

//increment 1 year
$mod_date = strtotime($date."+ 1 years");
echo date("Y-m-d",$mod_date) . "\n";
</code></pre>

blocks|key|928425|text|使用天数的变量|type|unstyled|depth|inlineStyleRanges|entityRanges|data|928426|$myDate+=+"2014-01-16";
$nDays+=+16;
$newDate+=+strtotime($myDate+.+'%2B+'.$nDays.'days');
echo+new+Date('d/m/Y',+$soma);+//format+new+date+|code-block|syntax|javascript|928427|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|$]]

Using a variable for Number of days
<pre><code>$myDate = &quot;2014-01-16&quot;;
$nDays = 16;
$newDate = strtotime($myDate . '+ '.$nDays.'days');
echo new Date('d/m/Y', $soma); //format new date 
</code></pre>

blocks|key|928472|text|从PHP5.2开始，您可以对DateTime对象使用modify：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|928473|http://php.net/manual/en/datetime.modify.php|offset|length|928474|$Date1+=+'2010-09-17';
$date+=+new+DateTime($Date1);
$date->modify('%2B1+day');
$Date2+=+$date->format('Y-m-d');|code-block|syntax|javascript|928475|添加月份时要小心...(在较小的程度上，还有年份)|928476|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|18|0|0|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|U|8|@]|9|@$D|V|E|W|1|X]]|A|$]]|$1|F|3|G|5|H|7|Y|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|Z|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|C]]]]

From PHP 5.2 on you can use modify with a DateTime object:

<a href="http://php.net/manual/en/datetime.modify.php">http://php.net/manual/en/datetime.modify.php</a>

<pre><code>$Date1 = '2010-09-17';
$date = new DateTime($Date1);
$date-&gt;modify('+1 day');
$Date2 = $date-&gt;format('Y-m-d');
</code></pre>

Be careful when adding months... (and to a lesser extent, years)

blocks|key|1143003|text|这里有一个简单的方法来解决这个问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1143004|<?php
+++$date+=+"2015-11-17";
+++echo+date('Y-m-d',+strtotime($date.+'+%2B+5+days'));
?>|code-block|syntax|javascript|1143005|输出将为：|1143006|2015-11-22|1143007|从这里找到了解决方案-+How+to+Add+Days+to+Date+in+PHP|offset|length|1143008|entityMap|0|LINK|mutability|MUTABLE|url|http://www.codexworld.com/how-to-guides/add-days-to-date-in-php/^0|0|0|0|0|C|U|0|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]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|10|8|@]|9|@$M|11|N|12|1|13]]|A|$]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

Here has an easy way to solve this.

<pre><code>&lt;?php
 $date = "2015-11-17";
 echo date('Y-m-d', strtotime($date. ' + 5 days'));
?&gt;
</code></pre>

Output will be:

<pre><code>2015-11-22
</code></pre>

Solution has found from here - <a href="http://www.codexworld.com/how-to-guides/add-days-to-date-in-php/" rel="nofollow">How to Add Days to Date in PHP</a>

blocks|key|1143022|text|以下是您查询的最简单的解决方案|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1143023|$date=date_create("2013-03-15");+//+or+your+date+string
date_add($date,date_interval_create_from_date_string("40+days"));//+add+number+of+days+
echo+date_format($date,"Y-m-d");+//set+date+format+of+the+result|code-block|syntax|javascript|1143024|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|$]]

Here is the simplest solution to your query

<pre><code>$date=date_create("2013-03-15"); // or your date string
date_add($date,date_interval_create_from_date_string("40 days"));// add number of days 
echo date_format($date,"Y-m-d"); //set date format of the result
</code></pre>

blocks|key|928545|text|要将特定天数添加到日期，请使用以下函数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|928546|function+add_days_to_date($date1,$number_of_days){
++++/*
++++//$date1+is+a+string+representing+a+date+such+as+'2021-04-17+14:34:05'
++++//$date1+=date('Y-m-d+H:i:s');+
++++//+function+date+without+a+secrod+argument+returns+the+current+datetime+as+a+string+in+the+specified+format
++++*/
++++$str+='+%2B+'.+$number_of_days.+'+days';
++++$date2=+date('Y-m-d+H:i:s',+strtotime($date1.+$str));
++++return+$date2;+//$date2+is+a+string
}//[end+function]|code-block|syntax|javascript|928547|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|$]]

To add a certain number of days to a date, use the following function.
<pre><code>function add_days_to_date($date1,$number_of_days){
 /*
 //$date1 is a string representing a date such as '2021-04-17 14:34:05'
 //$date1 =date('Y-m-d H:i:s'); 
 // function date without a secrod argument returns the current datetime as a string in the specified format
 */
 $str =' + '. $number_of_days. ' days';
 $date2= date('Y-m-d H:i:s', strtotime($date1. $str));
 return $date2; //$date2 is a string
}//[end function]
</code></pre>

I have a date returned as part of a mySQL query in the form <code>2010-09-17</code>

I would like to set the variables $Date2 to $Date5 as follows:

<code>$Date2 = $Date + 1</code>

<code>$Date3 = $Date + 2</code>

etc..

so that it returns <code>2010-09-18</code>, <code>2010-09-19</code> etc...

I have tried 

<pre><code>date('Y-m-d', strtotime($Date. ' + 1 day'))
</code></pre>

but this gives me the date BEFORE <code>$Date</code>.

What is the correct way to get my Dates in the format form 'Y-m-d' so that they may be used in another query?

Adding days to $Date in PHP

我有一个日期以2010-09-17的形式作为mySQL查询的一部分返回我想将变量$Date2设置为$Date5，如下所示：$Date2 = $Date + 1$Date3 = $Date + 2等等。这样它就会返回2010-09-18、2010-09-19等。我试过了date('Y-m-d', strtotime($D...

问在PHP语言中向$Date添加天数
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