blocks|key|1001387|text|$t1+=+strtotime(+'2006-04-14+11:30:00'+);
$t2+=+strtotime(+'2006-04-12+12:30:00'+);
$diff+=+$t1+-+$t2;
$hours+=+$diff+/+(+60+*+60+);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1001388|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>$t1 = strtotime( '2006-04-14 11:30:00' );
$t2 = strtotime( '2006-04-12 12:30:00' );
$diff = $t1 - $t2;
$hours = $diff / ( 60 * 60 );
</code></pre>

blocks|key|1214667|text|你的答案是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1214668|round((strtotime($day2)+-+strtotime($day1))/(60*60))|offset|length|style|CODE|1214669|entityMap^0|0|0|1G|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

your answer is:

<code>round((strtotime($day2) - strtotime($day1))/(60*60))</code>

blocks|key|1001503|text|获得两个日期(日期时间)之间的正确小时数的最简单方法是使用Unix时间戳中的差异，即使在夏令时更改的情况下也是如此。Unix时间戳是指自1970-01-01T00:00:00UTC以来经过的秒数，忽略闰秒(这是可以接受的，因为您可能不需要这种精度，而且很难将闰秒考虑在内)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1001504|将带有可选时区信息的datetime字符串转换为Unix时间戳的最灵活方法是构造一个DateTime对象(也可以使用DateTimeZone作为构造函数中的第二个参数)，然后调用它的getTimestamp方法。|offset|length|1001505|$str1+=+'2006-04-12+12:30:00';+
$str2+=+'2006-04-14+11:30:00';
$tz1+=+new+DateTimeZone('Pacific/Apia');
$tz2+=+$tz1;
$d1+=+new+DateTime($str1,+$tz1);+//+tz+is+optional,
$d2+=+new+DateTime($str2,+$tz2);+//+and+ignored+if+str+contains+tz+offset
$delta_h+=+($d2->getTimestamp()+-+$d1->getTimestamp())+/+3600;
if+($rounded_result)+{
+++$delta_h+=+round+($delta_h);
}+else+if+($truncated_result)+{
+++$delta_h+=+intval($delta_h);
}
echo+"Δh:+$delta_h\n";|code-block|syntax|javascript|1001506|entityMap|0|LINK|mutability|MUTABLE|url|http://php.net/manual/en/class.datetime.php|1|http://www.php.net/manual/en/class.datetimezone.php|2|http://www.php.net/manual/en/datetime.gettimestamp.php^0|0|16|8|0|1M|C|1|2J|C|2|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@$D|Y|E|Z|1|10]|$D|11|E|12|1|13]|$D|14|E|15|1|16]]|A|$]]|$1|F|3|G|5|H|7|17|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]|S|$5|N|O|P|A|$Q|T]]|U|$5|N|O|P|A|$Q|V]]]]

The easiest way to get the correct number of hours between two dates (datetimes), even across daylight saving time changes, is to use the difference in Unix timestamps. Unix timestamps are seconds elapsed since 1970-01-01T00:00:00 UTC, ignoring leap seconds (this is OK because you probably don't need this precision, and because it's quite difficult to take leap seconds into account).

The most flexible way to convert a datetime string with optional timezone information into a Unix timestamp is to construct a <a href="http://php.net/manual/en/class.datetime.php">DateTime</a> object (optionally with a <a href="http://www.php.net/manual/en/class.datetimezone.php">DateTimeZone</a> as a second argument in the constructor), and then call its <a href="http://www.php.net/manual/en/datetime.gettimestamp.php">getTimestamp</a> method.

<pre><code>$str1 = '2006-04-12 12:30:00'; 
$str2 = '2006-04-14 11:30:00';
$tz1 = new DateTimeZone('Pacific/Apia');
$tz2 = $tz1;
$d1 = new DateTime($str1, $tz1); // tz is optional,
$d2 = new DateTime($str2, $tz2); // and ignored if str contains tz offset
$delta_h = ($d2-&gt;getTimestamp() - $d1-&gt;getTimestamp()) / 3600;
if ($rounded_result) {
 $delta_h = round ($delta_h);
} else if ($truncated_result) {
 $delta_h = intval($delta_h);
}
echo "Δh: $delta_h\n";
</code></pre>

blocks|key|1001510|text|//Calculate+number+of+hours+between+pass+and+now
$dayinpass+=+"2013-06-23+05:09:12";
$today+=+time();
$dayinpass=+strtotime($dayinpass);
echo+round(abs($today-$dayinpass)/60/60);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1001511|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>//Calculate number of hours between pass and now
$dayinpass = "2013-06-23 05:09:12";
$today = time();
$dayinpass= strtotime($dayinpass);
echo round(abs($today-$dayinpass)/60/60);
</code></pre>

blocks|key|1001423|text|$day1+=+"2006-04-12+12:30:00"
$day1+=+strtotime($day1);
$day2+=+"2006-04-14+11:30:00"
$day2+=+strtotime($day2);

$diffHours+=+round(($day2+-+$day1)+/+3600);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1001424|我猜strtotime()函数接受这种日期格式。|unstyled|1001425|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>$day1 = "2006-04-12 12:30:00"
$day1 = strtotime($day1);
$day2 = "2006-04-14 11:30:00"
$day2 = strtotime($day2);

$diffHours = round(($day2 - $day1) / 3600);
</code></pre>

I guess strtotime() function accept this date format.

blocks|key|1215723|text|<?
+++++$day1+=+"2014-01-26+11:30:00";
+++++$day1+=+strtotime($day1);
+++++$day2+=+"2014-01-26+12:30:00";
+++++$day2+=+strtotime($day2);

+++$diffHours+=+round(($day2+-+$day1)+/+3600);

+++echo+$diffHours;

?>|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1215724|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>&lt;?
 $day1 = "2014-01-26 11:30:00";
 $day1 = strtotime($day1);
 $day2 = "2014-01-26 12:30:00";
 $day2 = strtotime($day2);

 $diffHours = round(($day2 - $day1) / 3600);

 echo $diffHours;

?&gt;
</code></pre>

blocks|key|1001447|text|此函数可帮助您计算$doj1和$doj这两个给定日期之间的精确年份和月份。它返回示例4.3，表示4年零3个月。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1001448|<?php
++++function+cal_exp($doj1)
++++{
++++++++$doj1=strtotime($doj1);
++++++++$doj=date("m/d/Y",$doj1);+//till+date+or+any+given+date

++++++++$now=date("m/d/Y");
++++++++//$b=strtotime($b1);
++++++++//echo+$c=$b1-$a2;
++++++++//echo+date("Y-m-d+H:i:s",$c);
++++++++$year=date("Y");
++++++++//$chk_leap=is_leapyear($year);

++++++++//$year_diff=365.25;

++++++++$x=explode("/",$doj);
++++++++$y1=explode("/",$now);

++++++++$yy=$x[2];
++++++++$mm=$x[0];
++++++++$dd=$x[1];

++++++++$yy1=$y1[2];
++++++++$mm1=$y1[0];
++++++++$dd1=$y1[1];
++++++++$mn=0;
++++++++$mn1=0;
++++++++$ye=0;
++++++++if($mm1>$mm)
++++++++{
++++++++++++$mn=$mm1-$mm;
++++++++++++if($dd1<$dd)
++++++++++++{
++++++++++++++++$mn=$mn-1;
++++++++++++}
++++++++++++$ye=$yy1-$yy;
++++++++}
++++++++else+if($mm1<$mm)
++++++++{
++++++++++++$mn=12-$mm;
++++++++++++//$mn=$mn;

++++++++++++if($mm!=1)
++++++++++++{
++++++++++++++++$mn1=$mm1-1;
++++++++++++}

++++++++++++$mn%2B=$mn1;
++++++++++++if($dd1>$dd)
++++++++++++{
++++++++++++++++$mn%2B=1;
++++++++++++}

++++++++++++$yy=$yy%2B1;
++++++++++++$ye=$yy1-$yy;
++++++++}
++++++++else
++++++++{
++++++++++++$ye=$yy1-$yy;
++++++++++++$ye=$ye-1;

++++++++++++$mn=12-1;

++++++++++++if($dd1>$dd)
++++++++++++{
++++++++++++++++$ye%2B=1;
++++++++++++++++$mn=0;
++++++++++++}
++++++++}

++++++++$to=$ye."+year+and+".$mn."+months";
++++++++return+$ye.".".$mn;

++++++++/*return+daysDiff($x[2],$x[0],$x[1]);
+++++++++$days=dateDiff("/",$now,$doj)/$year_diff;
++++++++$days_exp=explode(".",$days);
++++++++return+$years_exp=$days;+//number+of+years+exp*/
++++}
?>|code-block|syntax|javascript|1001449|entityMap^0|9|5|F|4|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

This function helps you to calculate exact years and months between two given dates, <code>$doj1</code> and <code>$doj</code>. It returns example 4.3 means 4 years and 3 month.

<pre><code>&lt;?php
 function cal_exp($doj1)
 {
 $doj1=strtotime($doj1);
 $doj=date("m/d/Y",$doj1); //till date or any given date

 $now=date("m/d/Y");
 //$b=strtotime($b1);
 //echo $c=$b1-$a2;
 //echo date("Y-m-d H:i:s",$c);
 $year=date("Y");
 //$chk_leap=is_leapyear($year);

 //$year_diff=365.25;

 $x=explode("/",$doj);
 $y1=explode("/",$now);

 $yy=$x[2];
 $mm=$x[0];
 $dd=$x[1];

 $yy1=$y1[2];
 $mm1=$y1[0];
 $dd1=$y1[1];
 $mn=0;
 $mn1=0;
 $ye=0;
 if($mm1&gt;$mm)
 {
 $mn=$mm1-$mm;
 if($dd1&lt;$dd)
 {
 $mn=$mn-1;
 }
 $ye=$yy1-$yy;
 }
 else if($mm1&lt;$mm)
 {
 $mn=12-$mm;
 //$mn=$mn;

 if($mm!=1)
 {
 $mn1=$mm1-1;
 }

 $mn+=$mn1;
 if($dd1&gt;$dd)
 {
 $mn+=1;
 }

 $yy=$yy+1;
 $ye=$yy1-$yy;
 }
 else
 {
 $ye=$yy1-$yy;
 $ye=$ye-1;

 $mn=12-1;

 if($dd1&gt;$dd)
 {
 $ye+=1;
 $mn=0;
 }
 }

 $to=$ye." year and ".$mn." months";
 return $ye.".".$mn;

 /*return daysDiff($x[2],$x[0],$x[1]);
 $days=dateDiff("/",$now,$doj)/$year_diff;
 $days_exp=explode(".",$days);
 return $years_exp=$days; //number of years exp*/
 }
?&gt;
</code></pre>

blocks|key|1215857|text|要传递unix时间戳，请使用以下表示法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1215858|$now++++++++=+time();
$now++++++++=+new+DateTime("@$now");|code-block|syntax|javascript|1215859|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 pass a unix timestamp use this notation

<pre><code>$now = time();
$now = new DateTime("@$now");
</code></pre>

blocks|key|1215921|text|首先，您应该从一个日期范围创建一个interval对象。仅通过这句话中使用的措辞，就可以很容易地识别出所需的基本抽象。有一个间隔作为一个概念，还有更多的方法来实现它，包括已经提到的方法--来自一系列日期。因此，间隔如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1215922|$interval+=
++++new+FromRange(
++++++++new+FromISO8601('2017-02-14T14:27:39%2B00:00'),
++++++++new+FromISO8601('2017-03-14T14:27:39%2B00:00')
++++);|code-block|syntax|javascript|1215923|FromISO8601具有相同的语义:它是由from+iso8601-formatted+string创建的datetime对象，因此得名。|offset|length|style|CODE|1215924|当您有一个间隔时，您可以随心所欲地对其进行格式化。如果你需要一个完整的小时数，你可以有|1215925|(new+TotalFullHours($interval))->value();|1215926|如果你想要一个有上限的总时数，你可以这样做：|1215927|(new+TotalCeiledHours($interval))->value();|1215928|有关此方法的更多信息和一些示例，请查看this+entry。|1215929|entityMap|0|LINK|mutability|MUTABLE|url|https://meringue.readthedocs.io/en/latest/quick_start.html^0|0|0|0|B|M|T|0|0|0|0|0|J|A|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|16|8|@$I|17|J|18|K|L]|$I|19|J|1A|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|1B|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|1C|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|1D|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|1E|8|@]|9|@]|A|$E|F]]|$1|U|3|V|5|6|7|1F|8|@]|9|@$I|1G|J|1H|1|1I]]|A|$]]|$1|W|3|-4|5|6|7|1J|8|@]|9|@]|A|$]]]|X|$Y|$5|Z|10|11|A|$12|13]]]]

First, you should create an interval object from a range of dates. By the wording used in this sentence alone, one can easily identify basic abstractions needed. There is an interval as a concept, and a couple of more ways to implement it, include the one already mentioned -- from a range of dates. Thus, an interval looks like that:

<pre><code>$interval =
 new FromRange(
 new FromISO8601('2017-02-14T14:27:39+00:00'),
 new FromISO8601('2017-03-14T14:27:39+00:00')
 );
</code></pre>

<code>FromISO8601</code> has the same semantics: it's a datetime object created <code>from iso8601-formatted string</code>, hence the name.

When you have an interval, you can format it however you like. If you need a number of full hours, you can have

<pre><code>(new TotalFullHours($interval))-&gt;value();
</code></pre>

If you want a ceiled total hours, here you go:

<pre><code>(new TotalCeiledHours($interval))-&gt;value();
</code></pre>

For more about this approach and some examples, check out <a href="https://meringue.readthedocs.io/en/latest/quick_start.html" rel="nofollow noreferrer">this entry</a>.

blocks|key|1216066|text|除了@fyrye's+very+helpful+answer之外，这是对前面提到的bug+(this+one)的一种不错的解决方法，即DatePeriod在进入夏季时减去1小时，但在离开夏季时不加1小时(因此欧洲/柏林的3月份有正确的743小时，但10月份有744小时，而不是745小时)：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1216067|计算一个月的小时数(或任何时间跨度)，考虑两个方向的DST转换|1216068|function+getMonthHours(string+$year,+string+$month,+\DateTimeZone+$timezone):+int
{
++++//+or+whatever+start+and+end+\DateTimeInterface+objects+you+like
++++$start+=+new+\DateTimeImmutable($year+.+'-'+.+$month+.+'-01+00:00:00',+$timezone);
++++$end+=+new+\DateTimeImmutable((new+\DateTimeImmutable($year+.+'-'+.+$month+.+'-01+23:59:59',+$timezone))->format('Y-m-t+H:i:s'),+$timezone);
++++
++++//+count+the+hours+just+utilizing+\DatePeriod,+\DateInterval+and+iterator_count,+hell+yeah!
++++$hours+=+iterator_count(new+\DatePeriod($start,+new+\DateInterval('PT1H'),+$end));
++++
++++//+find+transitions+and+check,+if+there+is+one+that+leads+to+a+positive+offset
++++//+that+isn't+added+by+\DatePeriod
++++//+this+is+the+workaround+for+https://bugs.php.net/bug.php?id=75685
++++$transitions+=+$timezone->getTransitions((int)$start->format('U'),+(int)$end->format('U'));
++++if+(2+===+count($transitions)+&&+$transitions[0]['offset']+-+$transitions[1]['offset']+>+0)+{
++++++++$hours+%2B=+(round(($transitions[0]['offset']+-+$transitions[1]['offset'])/3600));
++++}
++++
++++return+$hours;
}

$myTimezoneWithDST+=+new+\DateTimeZone('Europe/Berlin');
var_dump(getMonthHours('2020',+'01',+$myTimezoneWithDST));+//+744
var_dump(getMonthHours('2020',+'03',+$myTimezoneWithDST));+//+743
var_dump(getMonthHours('2020',+'10',+$myTimezoneWithDST));+//+745,+finally!

$myTimezoneWithoutDST+=+new+\DateTimeZone('UTC');
var_dump(getMonthHours('2020',+'01',+$myTimezoneWithoutDST));+//+744
var_dump(getMonthHours('2020',+'03',+$myTimezoneWithoutDST));+//+744
var_dump(getMonthHours('2020',+'10',+$myTimezoneWithoutDST));+//+744|code-block|syntax|javascript|1216069|附注:如果你检查一个(更长的)时间跨度，这会导致超过这两个转换，我的解决方案不会接触到计数的小时数，以减少有趣的副作用的可能性。在这种情况下，必须实现更复杂的解决方案。可以遍历所有找到的转换，将当前转换与上一个转换进行比较，并检查它是否是带有DST+true->false的转换。|1216070|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/35801738/2564552|1|https://bugs.php.net/bug.php?id=75685^0|2|S|0|1A|8|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]|$A|10|B|11|1|12]]|C|$]]|$1|D|3|E|5|6|7|13|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|14|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|6|7|15|8|@]|9|@]|C|$]]|$1|M|3|-4|5|6|7|16|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]|U|$5|P|Q|R|C|$S|V]]]]

In addition to <a href="https://stackoverflow.com/a/35801738/2564552">@fyrye's very helpful answer</a> this is an okayish workaround for the mentioned bug (<a href="https://bugs.php.net/bug.php?id=75685" rel="nofollow noreferrer">this one</a>), that DatePeriod substracts one hour when entering summertime, but doesn't add one hour when leaving summertime (and thus Europe/Berlin's March has its correct 743 hours but October has 744 instead of 745 hours):
<h2>Counting the hours of a month (or any timespan), considering DST-transitions in both directions</h2>
<pre><code>function getMonthHours(string $year, string $month, \DateTimeZone $timezone): int
{
 // or whatever start and end \DateTimeInterface objects you like
 $start = new \DateTimeImmutable($year . '-' . $month . '-01 00:00:00', $timezone);
 $end = new \DateTimeImmutable((new \DateTimeImmutable($year . '-' . $month . '-01 23:59:59', $timezone))-&gt;format('Y-m-t H:i:s'), $timezone);
 
 // count the hours just utilizing \DatePeriod, \DateInterval and iterator_count, hell yeah!
 $hours = iterator_count(new \DatePeriod($start, new \DateInterval('PT1H'), $end));
 
 // find transitions and check, if there is one that leads to a positive offset
 // that isn't added by \DatePeriod
 // this is the workaround for https://bugs.php.net/bug.php?id=75685
 $transitions = $timezone-&gt;getTransitions((int)$start-&gt;format('U'), (int)$end-&gt;format('U'));
 if (2 === count($transitions) &amp;&amp; $transitions[0]['offset'] - $transitions[1]['offset'] &gt; 0) {
 $hours += (round(($transitions[0]['offset'] - $transitions[1]['offset'])/3600));
 }
 
 return $hours;
}

$myTimezoneWithDST = new \DateTimeZone('Europe/Berlin');
var_dump(getMonthHours('2020', '01', $myTimezoneWithDST)); // 744
var_dump(getMonthHours('2020', '03', $myTimezoneWithDST)); // 743
var_dump(getMonthHours('2020', '10', $myTimezoneWithDST)); // 745, finally!

$myTimezoneWithoutDST = new \DateTimeZone('UTC');
var_dump(getMonthHours('2020', '01', $myTimezoneWithoutDST)); // 744
var_dump(getMonthHours('2020', '03', $myTimezoneWithoutDST)); // 744
var_dump(getMonthHours('2020', '10', $myTimezoneWithoutDST)); // 744
</code></pre>
P.S. If you check a (longer) timespan, which leads to more than those two transitions, my workaround won't touch the counted hours to reduce the potential of funny side effects. In such cases, a more complicated solution must be implemented. One could iterate over all found transitions and compare the current with the last and check if it is one with DST true-&gt;false.

blocks|key|1001782|text|$diff_min+=+(+strtotime(+$day2+)+-+strtotime(+$day1+)+)+/+60+/+60;
$total_time++=+$diff_min;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1001783|你可以试试这个。|unstyled|1001784|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>$diff_min = ( strtotime( $day2 ) - strtotime( $day1 ) ) / 60 / 60;
$total_time = $diff_min;
</code></pre>
You can try this one.

blocks|key|1001529|text|这在我的项目中是有效的。我想，这会对你有帮助的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1001530|如果日期是过去的，则invert将为1。|1001531|如果日期在将来，则invert将为0。|1001532|$defaultDate+=+date('Y-m-d');+++
$datetime1+++=+new+DateTime('2013-03-10');++
$datetime2+++=+new+DateTime($defaultDate);++
$interval++++=+$datetime1->diff($datetime2);++
$days++++++++=+$interval->format('%25a');
$invert++++++=+$interval->invert;|code-block|syntax|javascript|1001533|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|$]]

This is working in my project. I think, This will be helpful for you.

If Date is in past then invert will 1. 
If Date is in future then invert will 0.

<pre><code>$defaultDate = date('Y-m-d'); 
$datetime1 = new DateTime('2013-03-10'); 
$datetime2 = new DateTime($defaultDate); 
$interval = $datetime1-&gt;diff($datetime2); 
$days = $interval-&gt;format('%a');
$invert = $interval-&gt;invert;
</code></pre>

How do I calculate the difference between two dates in hours?

For example:

<pre><code>day1=2006-04-12 12:30:00
day2=2006-04-14 11:30:00
</code></pre>

In this case the result should be 47 hours.

Calculate number of hours between 2 dates in PHP

如何计算两个日期之间的差额(小时)？例如：day1=2006-04-12 12:30:00day2=2006-04-14 11:30:00在这种情况下，结果应该是47小时。

问在PHP中计算两个日期之间的小时数
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