blocks|key|4168045|text|你需要使用正则表达式来过滤掉你想要的标签。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4168046|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

You need to use regular expression to filter out the tag you want.

blocks|key|1851860|text|您可以使用str_replace|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1851861|$str++=++'<img+src="location-of-image.png"+style="width:100px;height:100px;"+title='sampletitle'+/>sampe+sample+sampple<br+/>when?+how?.';

$toBeReplaced+++=++'sampe+sample+sampple<br+/>when?+how?.';


$body+=+str_replace($toBeReplaced,+"",+$str);|code-block|syntax|javascript|1851862|entityMap|0|LINK|mutability|MUTABLE|url|http://php.net/manual/en/function.str-replace.php^0|5|B|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

You can use <a href="http://php.net/manual/en/function.str-replace.php" rel="nofollow">str_replace</a>

<pre><code>$str = '&lt;img src="location-of-image.png" style="width:100px;height:100px;" title='sampletitle' /&gt;&lt;p&gt;sampe sample sampple&lt;/p&gt;&lt;br /&gt;when? how?.';

$toBeReplaced = '&lt;p&gt;sampe sample sampple&lt;/p&gt;&lt;br /&gt;when? how?.';


$body = str_replace($toBeReplaced, "", $str);
</code></pre>

blocks|key|1522510|text|您可以使用preg_replace|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1522511|preg_replace('%5E(<img.*?/>).*$',+
+++++++++++++'$1',
+++++++++++++'<img+src="location-of-image.png"+style="width:100px;height:100px;"+title="sampletitle"+/>sampe+sample+sampple<br+/>when?+how?.');|code-block|syntax|javascript|1522512|entityMap|0|LINK|mutability|MUTABLE|url|http://php.net/manual/pt_BR/function.preg-replace.php^0|5|C|5|C|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

You can use <a href="http://php.net/manual/pt_BR/function.preg-replace.php" rel="nofollow"><code>preg_replace</code></a>:

<pre><code>preg_replace('^(&lt;img.*?/&gt;).*$', 
 '$1',
 '&lt;img src="location-of-image.png" style="width:100px;height:100px;" title="sampletitle" /&gt;&lt;p&gt;sampe sample sampple&lt;/p&gt;&lt;br /&gt;when? how?.');
</code></pre>

blocks|key|1527788|text|使用PHP+DOM：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1527789|$doc+=+new+DOMDocument();
$doc->loadXML('<img+src="location-of-image.png"+style="width:100px;height:100px;"++++title='sampletitle'+/>sampe+sample+sampple');

$doc+=+$doc->removeChild($doc->firstChild);|code-block|syntax|javascript|1527790|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 PHP DOM:

<pre><code>$doc = new DOMDocument();
$doc-&gt;loadXML('&lt;img src="location-of-image.png" style="width:100px;height:100px;" title='sampletitle' /&gt;&lt;p&gt;sampe sample sampple&lt;/p&gt;');

$doc = $doc-&gt;removeChild($doc-&gt;firstChild);
</code></pre>

blocks|key|1851920|text|好吧，我想我想通了。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1851921|这是我的答案，一个非常明显和简单的解决方案：|1851922|$str+=+'<img+src="location-of-image.png"+style="width:100px;height:100px"+title='smpl'/+sampe+sample+sampple<br+/>when?+how?.'

$str+=+strip_tags(+$str,+"<img>");+#For+safety+precaution.

$first+=+strpos($str,+'<img+');
$last+=+strpos($str,+'/>');
echo+"<".substr(+$str+,1,$last%2B1+);|code-block|syntax|javascript|1851923|感觉我得到了正确的答案。|1851924|感谢所有回复的人！:)|1851925|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Okay, I think I figure it out.

Here's my answer, a very obvious and easy solution:

<pre><code>$str = '&lt;img src="location-of-image.png" style="width:100px;height:100px" title='smpl'/ &lt;p&gt;sampe sample sampple&lt;/p&gt;&lt;br /&gt;when? how?.'

$str = strip_tags( $str, "&lt;img&gt;"); #For safety precaution.

$first = strpos($str, '&lt;img ');
$last = strpos($str, '/&gt;');
echo "&lt;".substr( $str ,1,$last+1 );
</code></pre>

Feels like i'm getting this answer right.

Thanks for everyone who answered! :)

I have am output that has something like this:

<pre><code>&lt;img src="location-of-image.png" style="width:100px;height:100px" title='smpl'/&gt;
&lt;p&gt;sampe sample sampple&lt;/p&gt;&lt;br /&gt;when? how?.
</code></pre>

My problem here is that I only want to get 

<pre><code>&lt;img src="location-of-image.png" style="width:100px;height:100px;" title='smpl'/&gt;
</code></pre>

and remove 

<pre><code>&lt;p&gt;sampe sample sampple&lt;/p&gt;&lt;br /&gt;when? how?.`
</code></pre>

So my expected output should be only like this:

<pre><code>&lt;img src="location-of-image.png" style="width:100px;height:100px;" title='smpl'/&gt;
</code></pre>

How would I do that in PHP? 
I know this is easier to be done in javascript/jquery, but I want it in PHP.

Getting the <img> tag in an HTML using 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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我有一个类似下面的输出：<img src="location-of-image.png" style="width:100px;height:100px" title='smpl'/>sampe sample sampple when? how?.我这里的问题是我只想<img src="loca...

问使用PHP获取HTML中的<img>标记
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问使用PHP获取HTML中的<img>标记EN