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.
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);
/([^/?&#]+)(?![^/]*/)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)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);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);
.html?