blocks|key|1242663|text|其中的诀窍是让这个月第一次出现。这样，PHP就可以按最高值对数字进行比较。请看下面的示例：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1242664|$aDate+=+DateTime::createFromFormat('m-d',+'05-20');
$bDate+=+DateTime::createFromFormat('m-d',+'06-29');

if+($aDate->format('md')+>+$bDate->format('md'))+{
++++echo+"'aDate'+is+bigger+than+'bDate'";
}|code-block|syntax|javascript|1242665|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|$]]

The trick in this is to let the month come first. This way PHP can compare the numbers by highest value. Take a look at the following example:

<pre><code>$aDate = DateTime::createFromFormat('m-d', '05-20');
$bDate = DateTime::createFromFormat('m-d', '06-29');

if ($aDate-&gt;format('md') &gt; $bDate-&gt;format('md')) {
 echo "'aDate' is bigger than 'bDate'";
}
</code></pre>

blocks|key|1996097|text|使用类似|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1996098|$current_date+=+date("d-m");
$subscription+=+"03-06-2016";
$subscription_date++=+date("d-m",+strtotime($subscription));
if($current_date+==$subscription_date)
{
++++echo+"date+is+equal";
}else
{
++++echo+"date+is+not+equal";
}|code-block|syntax|javascript|1996099|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|$]]

use like 

<pre><code>$current_date = date("d-m");
$subscription = "03-06-2016";
$subscription_date = date("d-m", strtotime($subscription));
if($current_date ==$subscription_date)
{
 echo "date is equal";
}else
{
 echo "date is not equal";
}
</code></pre>

blocks|key|1242736|text|如果只需要检查j-n日期是否与当前日期相同，则不需要进行多个函数调用。因为您的输入不大于或小于，所以输入的格式并不重要。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1242737|代码：(演示)|1242738|$subscription+=+'29-11';
var_export(date("j-n")+===+$subscription);
//+at+the+moment,+the+result+is+true|code-block|syntax|javascript|1242739|j是一个月中没有任何前导零点的日子|unordered-list-item|1242740|n是今天的月份，没有任何前导零。|1242741|entityMap|0|LINK|mutability|MUTABLE|url|https://3v4l.org/rtZQ8^0|7|3|0|4|2|0|0|0|0|1|0|0|1|0^^$0|@$1|2|3|4|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|12|8|@]|D|@$9|13|A|14|1|15]]|E|$]]|$1|H|3|I|5|J|7|16|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|O|7|17|8|@$9|18|A|19|B|C]]|D|@]|E|$]]|$1|P|3|Q|5|O|7|1A|8|@$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|R|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]]|S|$T|$5|U|V|W|E|$X|Y]]]]

If you only need to check if the <code>j-n</code> date is the same as the current date, then you don't need to make more than one function call. Because you are not comparing greater than or less than, the format of your input is unimportant.
Code: (<a href="https://3v4l.org/rtZQ8" rel="nofollow noreferrer">Demo</a>)
<pre><code>$subscription = '29-11';
var_export(date(&quot;j-n&quot;) === $subscription);
// at the moment, the result is true
</code></pre>
<ul>
<li><code>j</code> is today's day of the month without any leading zeros and</li>
<li><code>n</code> is today's month without any leading zeros.</li>
</ul>

blocks|key|1242591|text|使用DATE_FORMAT()函数提取部分日期：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1242592|参考文献：日期格式|offset|length|1242593|SELECT+*+from+table_name+WHERE+DATE_FORMAT(subscription_date,+'%25d-%25m')+=+"05-05";|code-block|syntax|javascript|1242594|entityMap|0|LINK|mutability|MUTABLE|url|http://dev.mysql.com/doc/refman/5.7/en/date-and-time-functions.html#function_date-format^0|0|5|4|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@$D|U|E|V|1|W]]|A|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

Use DATE_FORMAT() function to extract part of date:

Ref: <a href="http://dev.mysql.com/doc/refman/5.7/en/date-and-time-functions.html#function_date-format" rel="nofollow">http://dev.mysql.com/doc/refman/5.7/en/date-and-time-functions.html#function_date-format</a>

<pre><code>SELECT * from table_name WHERE DATE_FORMAT(subscription_date, '%d-%m') = "05-05";
</code></pre>

blocks|key|1242616|text|使用DateTime()+PHP对象。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1242617|考虑到您有一个包含来自mysql查询结果的用户信息的数组：($userData['suscriptionDate'])|offset|length|style|CODE|1242618|$today+=+new+DateTime();

$userSuscription+=+new+DateTime($userData['suscriptionDate']);

if+(+$today->format('d')+==+$userSuscription->format('d')+&&+$today->format('m')+==+$userSuscription->format('m'))+{

++++echo+'Congratulations!!';
}|code-block|syntax|javascript|1242619|entityMap^0|0|U|S|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@$D|Q|E|R|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|S|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Use DateTime() PHP objects.

Considering you have an array with user info from mysql query result: (<code>$userData['suscriptionDate']</code>)

<pre><code>$today = new DateTime();

$userSuscription = new DateTime($userData['suscriptionDate']);

if ( $today-&gt;format('d') == $userSuscription-&gt;format('d') &amp;&amp; $today-&gt;format('m') == $userSuscription-&gt;format('m')) {

 echo 'Congratulations!!';
}
</code></pre>

blocks|key|1996110|text|这可能对你有帮助|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1996111|$sdate+=+$row['subscription_date'];
$date1+=+date("m-d");
$date2+=+date("m-d",strtotime($sdate))+;
if+($date1+==+$date2)+{

}|code-block|syntax|javascript|1996112|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|$]]

This may help you
<pre><code>$sdate = $row['subscription_date'];
$date1 = date(&quot;m-d&quot;);
$date2 = date(&quot;m-d&quot;,strtotime($sdate)) ;
if ($date1 == $date2) {

}
</code></pre>

blocks|key|1996136|text|我认为，比较更优雅的方法，尤其是当您有完整的日期和时间时，是diff类的Datetime函数：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1996137|$d1+=+new+Datetime();
$d2+=+new+Datetime('%2B3+months+%2B2+days+%2B3+hours');

$diff+=+$d1->diff($d2);
var_dump($diff->d);+//+2
var_dump($diff->m);+//+2
//+or+have+a+comparison+as+a+string
var_dump($diff->format('Difference+is+in+%25R%25a+days'));
//+output:+Difference+is+in+63+days|code-block|syntax|javascript|1996138|好好享受吧！链接到文档|1996139|entityMap|0|LINK|mutability|MUTABLE|url|https://www.php.net/manual/ru/dateinterval.format.php^0|U|4|10|8|0|0|6|5|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|D|@$9|11|A|12|1|13]]|E|$]]|$1|M|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

I think, more elegant way to compare, especially when you have a full date with time is <code>diff</code> function of <code>Datetime</code> class:

<pre><code>$d1 = new Datetime();
$d2 = new Datetime('+3 months +2 days +3 hours');

$diff = $d1-&gt;diff($d2);
var_dump($diff-&gt;d); // 2
var_dump($diff-&gt;m); // 2
// or have a comparison as a string
var_dump($diff-&gt;format('Difference is in %R%a days'));
// output: Difference is in 63 days
</code></pre>

Enjoy! <a href="https://www.php.net/manual/ru/dateinterval.format.php" rel="nofollow noreferrer">Link to doc</a>

I want to compare current date's day and month with subscription date's day and month only.
For example:
<pre><code>current date(d-m) = 3-6
</code></pre>
And I want compare it with any other d-m 
How should I do it in PHP
In my project condition is like birth date in which we don't compare year.

Check if date expression ("d-m" without leading zeros) is today

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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我希望将当前日期的日期和月份与订阅日期的日期和月份进行比较。例如：current date(d-m) = 3-6我想和其他的d-m做比较在我的项目条件下，我应该如何在PHP中这样做，就像我们不比较年份的出生日期一样。

问检查日期表达式(没有前导零的“d-m”)是否是今天。
EN

回答 7

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问检查日期表达式(没有前导零的“d-m”)是否是今天。EN

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