blocks|key|1433196|text|第一组后的/%5E.*\/(.*)\.?(.*)$/g是您的文件名，第二组是扩展名。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1433197|var+myString+=+"filePath/long/path/myfile.even.with.dotes.TXT";
var+myRegexp+=+/%5E.*\/(.*)\.(.*)$/g;
var+match+=+myRegexp.exec(myString);
alert(match[1]);++//+myfile.even.with.dotes
alert(match[2]);++//+TXT|code-block|syntax|javascript|1433198|即使你的文件名包含一个以上的点或者根本不包含点(没有扩展名)，这也是有效的。|1433199|编辑：|BOLD|1433200|这是针对linux的，对于windows使用这个/%5E.*\\(.*)\.?(.*)$/g+(在linux中目录分隔符是/在windows中是\+)|1433201|entityMap^0|5|K|0|0|0|0|3|0|O|K|1N|1|1Y|1|0^^$0|@$1|2|3|4|5|6|7|T|8|@$9|U|A|V|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|W|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|X|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|Y|8|@$9|Z|A|10|B|O]]|D|@]|E|$]]|$1|P|3|Q|5|6|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|C]|$9|16|A|17|B|C]]|D|@]|E|$]]|$1|R|3|-4|5|6|7|18|8|@]|D|@]|E|$]]]|S|$]]

<code>/^.*\/(.*)\.?(.*)$/g</code> after this first group is your file name and second group is extention.

<pre><code>var myString = "filePath/long/path/myfile.even.with.dotes.TXT";
var myRegexp = /^.*\/(.*)\.(.*)$/g;
var match = myRegexp.exec(myString);
alert(match[1]); // myfile.even.with.dotes
alert(match[2]); // TXT
</code></pre>

This works even if your filename contains more then one dotes or doesn't contain dots at all (has no extention). 
EDIT: 
This is for linux, for windows use this <code>/^.*\\(.*)\.?(.*)$/g</code> (in linux directory separator is <code>/</code> in windows is <code>\</code> )

blocks|key|4400622|text|假设所有文件都有一个扩展名，您可以使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4400623|var+regexAll+=+/[%5E\\]*\.(\w%2B)$/;|code-block|syntax|javascript|4400624|然后你可以这样做|4400625|var+total+=+path.match(regexAll);
var+filename+=+total[0];
var+extension+=+total[1];|4400626|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Assuming that all files do have an extension, you could use

<pre><code>var regexAll = /[^\\]*\.(\w+)$/;
</code></pre>

Then you can do

<pre><code>var total = path.match(regexAll);
var filename = total[0];
var extension = total[1];
</code></pre>

blocks|key|1429820|text|您可以在正则表达式中使用组来实现以下目的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1429821|var+regex+=+/%5E([%5E\\]*)\.(\w%2B)$/;
var+matches+=+filename.match(regex);

if+(matches)+{
++++var+filename+=+matches[1];
++++var+extension+=+matches[2];
}|code-block|syntax|javascript|1429822|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|$]]

You can use groups in your regular expression for this:

<pre><code>var regex = /^([^\\]*)\.(\w+)$/;
var matches = filename.match(regex);

if (matches) {
 var filename = matches[1];
 var extension = matches[2];
}
</code></pre>

blocks|key|361518|text|我认为这是一种更好的方法，因为只匹配有效的目录、文件名和扩展名。并且还对路径、文件名和文件扩展名进行分组。并且也仅适用于空路径的文件名。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|361519|%5E([\w\/]*?)([\w\.]*)\.(\w)$|code-block|syntax|javascript|361520|测试用例|361521|the/p0090Aath/fav.min.icon.png
the/p0090Aath/fav.min.icon.html
the/p009_0Aath/fav.m45in.icon.css
fav.m45in.icon.css
favicon.ico|361522|输出|361523|[the/p0090Aath/][fav.min.icon][png]
[the/p0090Aath/][fav.min.icon][html]
[the/p009_0Aath/][fav.m45in.icon][css]
[][fav.m45in.icon][css]
[][favicon][ico]|361524|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

I think this is a better approach as matches only valid directory, file names and extension. and also groups the path, filename and file extension. And also works with empty paths only filename.

<pre><code>^([\w\/]*?)([\w\.]*)\.(\w)$
</code></pre>

Test cases

<pre><code>the/p0090Aath/fav.min.icon.png
the/p0090Aath/fav.min.icon.html
the/p009_0Aath/fav.m45in.icon.css
fav.m45in.icon.css
favicon.ico
</code></pre>

Output

<pre><code>[the/p0090Aath/][fav.min.icon][png]
[the/p0090Aath/][fav.min.icon][html]
[the/p009_0Aath/][fav.m45in.icon][css]
[][fav.m45in.icon][css]
[][favicon][ico]
</code></pre>

blocks|key|361561|text|这将识别甚至/home/someUser/.aaa/.bb.c|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|361562|function+splitPathFileExtension(path){
++++var+parsed+=+path.match(/%5E(.*\/)(.*)\.(.*)$/);
++++return+[parsed[1],+parsed[2],+parsed[3]];
}|code-block|syntax|javascript|361563|entityMap^0|6|P|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

This will recognize even <code>/home/someUser/.aaa/.bb.c</code>:

<pre><code>function splitPathFileExtension(path){
 var parsed = path.match(/^(.*\/)(.*)\.(.*)$/);
 return [parsed[1], parsed[2], parsed[3]];
}
</code></pre>

blocks|key|4400803|text|我知道这是一个古老的问题，但这里有另一个解决方案，可以处理名称中的多个点，也可以在根本没有扩展名(或只有‘.’的扩展名)的情况下：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4400804|/%5E(.*?)(\.[%5E.]*)?$/|offset|length|style|CODE|4400805|一次取一块：|4400806|%5E|4400807|锚定到字符串的开头(以避免部分匹配)|4400808|(.*?)|4400809|匹配任意字符.，0或更多次*，惰性?+(如果后面的可选扩展可以匹配，不要只抓取它们)，并将它们放在第一个捕获组(+)中。|4400810|(\.|4400811|使用(为扩展启动第二个捕获组。该组以文字.字符开始(我们使用\对其进行转义，这样.就不会被解释为“匹配任何字符”)。|4400812|[%5E.]*|4400813|定义字符集[]。通过指定this+is+an+inverted+character+set+%5E匹配不在集合中的字符。匹配0个或更多非.字符以获取文件扩展名*的其余部分。我们这样指定它，这样它就不会匹配像foo.bar.baz这样的早期文件名，错误地给出了一个带有多个点的扩展名，而不是.bar.baz，而不仅仅是.baz。.不需要在[]中进行转义，因为所有内容(除了%5E)都是字符集的文字。|4400814|)?|4400815|结束第二个捕获组)并指出整个组是可选的?，因为它可能没有扩展名。|4400816|$|4400817|锚定到字符串的末尾(同样，避免部分匹配)|4400818|如果你使用的是ES6，你甚至可以使用析构来获取一行结果：|4400819|[,filename,+extension]+=+/%5E(.*?)(\.[%5E.]*)?$/.exec('foo.bar.baz');，文件名为'foo.bar'，扩展名为'.baz'。|4400820|'foo'给了'foo'+and+''|4400821|'foo.'提供了'foo'和'.'|4400822|'.js'提供了''和'.js'|4400823|entityMap^0|0|0|J|0|0|0|1|0|0|0|5|0|6|1|D|1|H|1|1J|1|1L|1|0|0|3|0|2|1|K|1|U|1|14|1|0|0|5|0|5|2|1A|1|1U|1|25|1|2T|B|3Y|8|4C|4|4H|1|4M|2|53|1|0|0|2|0|8|1|J|1|0|0|1|0|0|0|0|1T|1Y|9|2C|6|0|0|5|7|C|0|0|6|9|5|F|3|0|0|5|8|2|B|5|0^^$0|@$1|2|3|4|5|6|7|1J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1K|8|@$D|1L|E|1M|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|1N|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|1O|8|@$D|1P|E|1Q|F|G]]|9|@]|A|$]]|$1|L|3|M|5|6|7|1R|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|1S|8|@$D|1T|E|1U|F|G]]|9|@]|A|$]]|$1|P|3|Q|5|6|7|1V|8|@$D|1W|E|1X|F|G]|$D|1Y|E|1Z|F|G]|$D|20|E|21|F|G]|$D|22|E|23|F|G]|$D|24|E|25|F|G]]|9|@]|A|$]]|$1|R|3|S|5|6|7|26|8|@$D|27|E|28|F|G]]|9|@]|A|$]]|$1|T|3|U|5|6|7|29|8|@$D|2A|E|2B|F|G]|$D|2C|E|2D|F|G]|$D|2E|E|2F|F|G]|$D|2G|E|2H|F|G]]|9|@]|A|$]]|$1|V|3|W|5|6|7|2I|8|@$D|2J|E|2K|F|G]]|9|@]|A|$]]|$1|X|3|Y|5|6|7|2L|8|@$D|2M|E|2N|F|G]|$D|2O|E|2P|F|G]|$D|2Q|E|2R|F|G]|$D|2S|E|2T|F|G]|$D|2U|E|2V|F|G]|$D|2W|E|2X|F|G]|$D|2Y|E|2Z|F|G]|$D|30|E|31|F|G]|$D|32|E|33|F|G]|$D|34|E|35|F|G]]|9|@]|A|$]]|$1|Z|3|10|5|6|7|36|8|@$D|37|E|38|F|G]]|9|@]|A|$]]|$1|11|3|12|5|6|7|39|8|@$D|3A|E|3B|F|G]|$D|3C|E|3D|F|G]]|9|@]|A|$]]|$1|13|3|14|5|6|7|3E|8|@$D|3F|E|3G|F|G]]|9|@]|A|$]]|$1|15|3|16|5|6|7|3H|8|@]|9|@]|A|$]]|$1|17|3|18|5|6|7|3I|8|@]|9|@]|A|$]]|$1|19|3|1A|5|6|7|3J|8|@$D|3K|E|3L|F|G]|$D|3M|E|3N|F|G]|$D|3O|E|3P|F|G]]|9|@]|A|$]]|$1|1B|3|1C|5|6|7|3Q|8|@$D|3R|E|3S|F|G]|$D|3T|E|3U|F|G]]|9|@]|A|$]]|$1|1D|3|1E|5|6|7|3V|8|@$D|3W|E|3X|F|G]|$D|3Y|E|3Z|F|G]|$D|40|E|41|F|G]]|9|@]|A|$]]|$1|1F|3|1G|5|6|7|42|8|@$D|43|E|44|F|G]|$D|45|E|46|F|G]|$D|47|E|48|F|G]]|9|@]|A|$]]|$1|1H|3|-4|5|6|7|49|8|@]|9|@]|A|$]]]|1I|$]]

I know this is an old question, but here's another solution that can handle multiple dots in the name and also when there's no extension at all (or an extension of just '.'): 
<code>/^(.*?)(\.[^.]*)?$/</code> 

Taking it a piece at a time: 
<code>^</code> 
Anchor to the start of the string (to avoid partial matches) 

<code>(.*?)</code> 
Match any character <code>.</code>, 0 or more times <code>*</code>, lazily <code>?</code> (don't just grab them all if the later optional extension can match), and put them in the first capture group <code>(</code> <code>)</code>. 

<code>(\.</code> 
Start a 2nd capture group for the extension using <code>(</code>. This group starts with the literal <code>.</code> character (which we escape with <code>\</code> so that <code>.</code> isn't interpreted as "match any character"). 

<code>[^.]*</code> 
Define a character set <code>[]</code>. Match characters not in the set by specifying this is an inverted character set <code>^</code>. Match 0 or more non-<code>.</code> chars to get the rest of the file extension <code>*</code>. We specify it this way so that it doesn't match early on filenames like <code>foo.bar.baz</code>, incorrectly giving an extension with more than one dot in it of <code>.bar.baz</code> instead of just <code>.baz</code>.
<code>.</code> doesn't need escaped inside <code>[]</code>, since everything (except<code>^</code>) is a literal in a character set. 

<code>)?</code> 
End the 2nd capture group <code>)</code> and indicate that the whole group is optional <code>?</code>, since it may not have an extension. 

<code>$</code> 
Anchor to the end of the string (again, to avoid partial matches)

If you're using ES6 you can even use destructing to grab the results in 1 line: 
<code>[,filename, extension] = /^(.*?)(\.[^.]*)?$/.exec('foo.bar.baz');
</code>
which gives the filename as <code>'foo.bar'</code> and the extension as <code>'.baz'</code>. 
<code>'foo'</code> gives <code>'foo' and ''</code> 
<code>'foo.'</code> gives <code>'foo'</code> and <code>'.'</code> 
<code>'.js'</code> gives <code>''</code> and <code>'.js'</code>

blocks|key|1429909|text|(?!\w%2B).(\w%2B)(\s)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1429910|查找一个或多个单词\w%2B，取反(?!+)以使单词不显示在结果中，指定分隔符.，查找第一个单词(\w%2B)并忽略可能在空格(\s)之后的单词|1429911|entityMap^0|0|H|0|9|3|F|5|11|1|1A|5|1N|4|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|I|$]]

<code>(?!\w+).(\w+)(\s)</code>

Find one or more word (s) <code>\w+</code>, negate <code>(?! )</code> so that the word (s) are not shown on the result, specify the delimiter <code>.</code>, find the first word <code>(\w+)</code> and ignore the words that are after a possible blank space <code>(\s)</code>

I'm sure this must be easy enough, but I'm struggling...

<pre><code>var regexFileName = /[^\\]*$/; // match filename
var regexFileExtension = /(\w+)$/; // match file extension

function displayUpload() {
 var path = $el.val(); //This is a file input
 var filename = path.match(regexFileName); // returns file name
 var extension = filename[0].match(regexFileExtension); // returns extension

 console.log("The filename is " + filename[0]);
 console.log("The extension is " + extension[0]);
}
</code></pre>

The function above works fine, but I'm sure it must be possible to achieve with a single regex, by referencing different parts of the array returned with the .match() method. I've tried combining these regex but without success.

Also, I'm not using a string to test it on in the example, as console.log() escapes the backslashes in a filepath and it was starting to confuse me :)

Match filename and file extension from single Regex

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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我相信这一定很简单，但我在努力...var regexFileName = /[^\\]*$/; // match filenamevar regexFileExtension = /(\w+)$/; // match file extensionfunction displayUpload() {    var pa...

问匹配单个Regex中的文件名和文件扩展名
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