如果你正在处理web url，你可以使用:

function getExt(filepath){
     return filepath.split("?")[0].split("#")[0].split('.').pop();
}

getExt("../js/logic.v2.min.js") // js
getExt("http://example.net/site/page.php?id=16548") // php
getExt("http://example.net/site/page.html#welcome.to.me") // html
getExt("c:\\logs\\yesterday.log"); // log

演示:https://jsfiddle.net/squadjot/q5ard4fj/

我相信将来会有人缩小和/或优化我的代码。但是，到目前为止，我有200%的信心，我的代码在每一个独特的情况下工作(例如，只有文件名，相对，根相对，和绝对URL的，片段#标签，与查询?字符串，以及任何你可能决定扔给它的东西)，完美无缺，精确到极点。

为了证明，请访问:https://projects.jamesandersonjr.com/web/js_projects/get_file_extension_test.php

这是JSFiddle: https://jsfiddle.net/JamesAndersonJr/ffcdd5z3/

不要过于自信，或者自吹自擂，但我还没有看到任何代码块的这个任务(找到“正确的”文件扩展名，在电池不同的函数输入参数)，工作得很好。

注意:按照设计，如果给定的输入字符串不存在文件扩展名，它只返回一个空白字符串“”，而不是错误，也不是错误消息。

它有两个参数: 字符串:fileNameOrURL(自解释) 布尔值:showUnixDotFiles(是否显示以点“。”开头的文件)

注(2):如果你喜欢我的代码，一定要把它添加到你的js库中，和/或repo库中，因为我努力完善它，浪费它将是一种耻辱。所以，废话不多说，下面是:

function getFileExtension(fileNameOrURL, showUnixDotFiles)
    {
        /* First, let's declare some preliminary variables we'll need later on. */
        var fileName;
        var fileExt;
        
        /* Now we'll create a hidden anchor ('a') element (Note: No need to append this element to the document). */
        var hiddenLink = document.createElement('a');
        
        /* Just for fun, we'll add a CSS attribute of [ style.display = "none" ]. Remember: You can never be too sure! */
        hiddenLink.style.display = "none";
        
        /* Set the 'href' attribute of the hidden link we just created, to the 'fileNameOrURL' argument received by this function. */
        hiddenLink.setAttribute('href', fileNameOrURL);
        
        /* Now, let's take advantage of the browser's built-in parser, to remove elements from the original 'fileNameOrURL' argument received by this function, without actually modifying our newly created hidden 'anchor' element.*/ 
        fileNameOrURL = fileNameOrURL.replace(hiddenLink.protocol, ""); /* First, let's strip out the protocol, if there is one. */
        fileNameOrURL = fileNameOrURL.replace(hiddenLink.hostname, ""); /* Now, we'll strip out the host-name (i.e. domain-name) if there is one. */
        fileNameOrURL = fileNameOrURL.replace(":" + hiddenLink.port, ""); /* Now finally, we'll strip out the port number, if there is one (Kinda overkill though ;-)). */  
        
        /* Now, we're ready to finish processing the 'fileNameOrURL' variable by removing unnecessary parts, to isolate the file name. */
        
        /* Operations for working with [relative, root-relative, and absolute] URL's ONLY [BEGIN] */ 
        
        /* Break the possible URL at the [ '?' ] and take first part, to shave of the entire query string ( everything after the '?'), if it exist. */
        fileNameOrURL = fileNameOrURL.split('?')[0];

        /* Sometimes URL's don't have query's, but DO have a fragment [ # ](i.e 'reference anchor'), so we should also do the same for the fragment tag [ # ]. */
        fileNameOrURL = fileNameOrURL.split('#')[0];

        /* Now that we have just the URL 'ALONE', Let's remove everything to the last slash in URL, to isolate the file name. */
        fileNameOrURL = fileNameOrURL.substr(1 + fileNameOrURL.lastIndexOf("/"));

        /* Operations for working with [relative, root-relative, and absolute] URL's ONLY [END] */ 

        /* Now, 'fileNameOrURL' should just be 'fileName' */
        fileName = fileNameOrURL;
        
        /* Now, we check if we should show UNIX dot-files, or not. This should be either 'true' or 'false'. */  
        if ( showUnixDotFiles == false )
            {
                /* If not ('false'), we should check if the filename starts with a period (indicating it's a UNIX dot-file). */
                if ( fileName.startsWith(".") )
                    {
                        /* If so, we return a blank string to the function caller. Our job here, is done! */
                        return "";
                    };
            };
        
        /* Now, let's get everything after the period in the filename (i.e. the correct 'file extension'). */
        fileExt = fileName.substr(1 + fileName.lastIndexOf("."));

        /* Now that we've discovered the correct file extension, let's return it to the function caller. */
        return fileExt;
    };

享受吧!不客气!：

一行解决方案，也将考虑查询参数和url中的任何字符。

string.match(/(.*)\??/i).shift().replace(/\?.*/, '').split('.').pop()

// Example
// some.url.com/with.in/&ot.s/files/file.jpg?spec=1&.ext=jpg
// jpg

还有一个使用ES6解构的简单方法:

Const path = 'hello.world.txt' const[扩展，…nameParts] = path.split('.').reverse(); console.log(扩展:,扩展);

这个简单的解决方案

function extension(filename) {
  var r = /.+\.(.+)$/.exec(filename);
  return r ? r[1] : null;
}

测试

/* tests */ test('cat.gif', 'gif'); test('main.c', 'c'); test('file.with.multiple.dots.zip', 'zip'); test('.htaccess', null); test('noextension.', null); test('noextension', null); test('', null); // test utility function function test(input, expect) { var result = extension(input); if (result === expect) console.log(result, input); else console.error(result, input); } function extension(filename) { var r = /.+\.(.+)$/.exec(filename); return r ? r[1] : null; }

对于大多数应用程序，一个简单的脚本，如

return /[^.]+$/.exec(filename);

会工作得很好(由汤姆提供)。然而，这并非万无一失。如果提供了以下文件名，它将不起作用:

image.jpg?foo=bar

这可能有点过分，但我建议使用这样的url解析器来避免由于不可预测的文件名而导致的失败。

使用这个特定的函数，你可以得到这样的文件名:

var trueFileName = parse_url('image.jpg?foo=bar').file;

这将输出“image.jpg”，不带url变量。然后您就可以获取文件扩展名了。