URL最后一部分正则表达式

Question

I need to get the last part of my URL with html on the end. So if I have this url http://step/build/index.html I need to get only index.html . I need to do this with javascript

let address = http://step/build/index.html;
let result = address.match(/html/i);

I tried this, but it doesn't work for me, maybe I make some mistakes. How do I get the last segment of URL using regular expressions Could someone help me and explain it in details? Thank you.

Answer 1

You could split it on slashes and then fetch the last item:

 let address = "http://step/build/index.html"; let result = address.split("/").pop(); console.log(result) 

Answer 2

You can extract the .html filename part using this /[^/]+\\.html/i RegEx.

See the code below.

 const regex = /[^/]+\\.html/i; let address = "http://step/build/index.html"; let result = address.match(regex); console.log(result); 

The same RegEx will also match the filename incase the URL has additional parameters.

 const regex = /[^/]+\\.html/i; let address = "http://step/build/index.html?name=value"; let result = address.match(regex); console.log(result); 

Answer 3

You could use split which returns an array to split on a forward slash and then use pop which removes the last element from the array and returns that:

 let address = "http://step/build/index.html".split('/').pop(); console.log(address); 

If you have querystring parameters which could for example start with ? or # , you might use split again and get the first item from the array:

 let address2 = "\\"http://step/build/index.html?id=1&cat=2" .split('/') .pop() .split(/[?#]/)[0]; console.log(address2); 

Answer 4

Here's a non-regex approach. Should be more reliable/appropriate at the job, depending on whether you'll need other URL-specific parts:

 // Note the ?foo=bar part, that URL.pathname will ignore below let address = 'http://step/build/index.html?foo=bar'; let url = new URL(address); // Last part of the path console.log(url.pathname.split('/').pop()); // Query string console.log(url.search); // Whole data console.log(url);