entityMap|blocks|key|fjuv1|text|只需像这样修改正则表达式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|6dnhb|+/(http(s?):)([\/%7C.%7C\w%7C\s%7C-])*\.(?:jpg%7Cjpeg%7Cgif%7Cpng)/g+|code-block|syntax|javascript|c5ivu^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|I|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|J|9|@]|A|@]|B|$F|G]]|$2|H|4|-4|6|7|8|K|9|@]|A|@]|B|$]]]]

Just modify the regular expression like that:

<pre><code> /(http(s?):)([\/|.|\w|\s|-])*\.(?:jpg|jpeg|gif|png)/g 
</code></pre>

entityMap|blocks|key|d0rgu|text|您需要在正则表达式中对该/进行转义，因此这是正确的|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|m5lb|/(http(s?):)([\/%7C.%7C\w%7C\s%7C-])*\.(?:jpg%7Cjpeg%7Cgif%7Cpng)/g|code-block|syntax|javascript|1g6pl^0|C|1|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|M|9|@$A|N|B|O|C|D]]|E|@]|F|$]]|$2|G|4|H|6|I|8|P|9|@]|E|@]|F|$J|K]]|$2|L|4|-4|6|7|8|Q|9|@]|E|@]|F|$]]]]

You need to escape that <code>/</code> in regex, so this is correct 

<pre><code>/(http(s?):)([\/|.|\w|\s|-])*\.(?:jpg|jpeg|gif|png)/g
</code></pre>

entityMap|0|type|LINK|mutability|MUTABLE|data|url|https://regexr.com/4jm5v|blocks|key|5ojk2|text|通过折叠一些捕获组，以及从[]字符集中删除竖线字符，可以显著缩短现有表达式。字符集在不使用管道的情况下自动表现为"OR“语句。|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|d5rb4|/(https?:)([\/.\w\s-])*\.(?:(jpe?%7Cpn)g%7Cgif)/gi|code-block|syntax|javascript|30k9h|要剖析这个表达式并了解它的工作原理，可以在RegExr上查看它。|fqtco^0|D|2|0|0|L|6|0|0^^$0|$1|$2|3|4|5|6|$7|8]]]|9|@$A|B|C|D|2|E|F|U|G|@$H|V|I|W|J|K]]|L|@]|6|$]]|$A|M|C|N|2|O|F|X|G|@]|L|@]|6|$P|Q]]|$A|R|C|S|2|E|F|Y|G|@]|L|@$H|Z|I|10|A|11]]|6|$]]|$A|T|C|-4|2|E|F|12|G|@]|L|@]|6|$]]]]

You can shorten your existing expression significantly by collapsing some of your capture groups, as well as removing the pipe characters from your <code>[]</code> character set. Characters sets automatically behave like "OR" statements without the use of pipes.

<pre><code>/(https?:)([\/.\w\s-])*\.(?:(jpe?|pn)g|gif)/gi
</code></pre>

To dissect this expression and see why it works, take a look at it on <a href="https://regexr.com/4jm5v" rel="nofollow noreferrer">RegExr</a>.

entityMap|0|type|LINK|mutability|MUTABLE|data|url|https://regex101.com/r/ZXyOkQ/2|blocks|key|5ffse|text|正则表达式限制太多，因为它的域验证性能很差，所以您可以像这样放松它：|unstyled|depth|inlineStyleRanges|entityRanges|27hnf|https?:\/\/[%5E.]*?(\.[%5E.]%2B?)*\/.*?\.(jpg%7Cjpeg%7Cgif%7Cpng)|code-block|syntax|javascript|jppi|offset|length|6ajua^0|0|0|0|V|0|0^^$0|$1|$2|3|4|5|6|$7|8]]]|9|@$A|B|C|D|2|E|F|R|G|@]|H|@]|6|$]]|$A|I|C|J|2|K|F|S|G|@]|H|@]|6|$L|M]]|$A|N|C|8|2|E|F|T|G|@]|H|@$O|U|P|V|A|W]]|6|$]]|$A|Q|C|-4|2|E|F|X|G|@]|H|@]|6|$]]]]

That regex is too restrictive because it is poorly performing domain validation so you can loosen it up like this:

<pre><code>https?:\/\/[^.]*?(\.[^.]+?)*\/.*?\.(jpg|jpeg|gif|png)
</code></pre>

<a href="https://regex101.com/r/ZXyOkQ/2" rel="nofollow noreferrer">https://regex101.com/r/ZXyOkQ/2</a>

entityMap|blocks|key|753qo|text|您可以使用URL接口和test|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5qmoq|​|869rf|let+str+=+`https://images.pexels.com/photos/459225/pexels-photo-459225.jpeg?auto=compress&cs=tinysrgb&dpr=1&w=500`
let+str2+=+`https://images.pexels.com/photos/459225/pexels-photo-459225.exe?auto=compress&cs=tinysrgb&dpr=1&w=500`

let+isImage+=+(str)+=>{
++let+parsed+=+new+URL(str)
++return+/(?:jpe?g%7Cgif%7Cpng)$/i.test(parsed.pathname)
}+

console.log(isImage(str))
console.log(isImage(str2))|code-block|syntax|javascript|cg85s|2bv2c^0|5|3|B|4|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|P|9|@$A|Q|B|R|C|D]|$A|S|B|T|C|D]]|E|@]|F|$]]|$2|G|4|H|6|7|8|U|9|@]|E|@]|F|$]]|$2|I|4|J|6|K|8|V|9|@]|E|@]|F|$L|M]]|$2|N|4|H|6|7|8|W|9|@]|E|@]|F|$]]|$2|O|4|-4|6|7|8|X|9|@]|E|@]|F|$]]]]

You can use <code>URL</code> api and <code>test</code>

<div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>let str = `https://images.pexels.com/photos/459225/pexels-photo-459225.jpeg?auto=compress&amp;cs=tinysrgb&amp;dpr=1&amp;w=500`
let str2 = `https://images.pexels.com/photos/459225/pexels-photo-459225.exe?auto=compress&amp;cs=tinysrgb&amp;dpr=1&amp;w=500`

let isImage = (str) =&gt;{
 let parsed = new URL(str)
 return /(?:jpe?g|gif|png)$/i.test(parsed.pathname)
} 

console.log(isImage(str))
console.log(isImage(str2))</code></pre>
</div>
</div>

I have a string like this <code>https://images.pexels.com/photos/459225/pexels-photo-459225.jpeg?auto=compress&amp;cs=tinysrgb&amp;dpr=1&amp;w=500</code> and I need to define if this image-link or not. 

I found regular expression <code>/(http(s?):)([/|.|\w|\s|-])*\.(?:jpg|gif|png)/g</code> which define image if link has appropriate format in the end, but this regular expression doesn't work with such link <code>https://images.pexels.com/photos/459225/pexels-photo-459225.jpeg?auto=compress&amp;cs=tinysrgb&amp;dpr=1&amp;w=500</code>

I am not familiar with regular expressions, so can you help me to improve this regular expression and define such link. Thanks

Regular Expression to find Images links in string

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。

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 我有一个像这样的https://images.pexels.com/photos/459225/pexels-photo-459225.jpeg?auto=compress&cs=tinysrgb&dpr=1&w=500字符串，我需要定义这个图像是否链接。 我找到了定义图像的正则表达式/(http(s?):)([/|...

问用于在字符串中查找图像链接的正则表达式
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问用于在字符串中查找图像链接的正则表达式EN