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How to get the URL without any parameters in JavaScript?

开发者 https://www.devze.com 2023-03-11 21:26 出处:网络
If I use: alert(window.location.href); I get everything including query strings. Is there a way to just get the main url part, for exa开发者_如何学编程mple:

If I use:

alert(window.location.href);

I get everything including query strings. Is there a way to just get the main url part, for exa开发者_如何学编程mple:

http://mysite.com/somedir/somefile/

instead of

http://mysite.com/somedir/somefile/?foo=bar&loo=goo


This is possible, but you'll have to build it manually from the location object:

location.protocol + '//' + location.host + location.pathname


Every answer is rather convoluted. Here:

var url = window.location.href.split('?')[0];

Even if a ? isn't present, it'll still return the first argument, which will be your full URL, minus query string.

It's also protocol-agnostic, meaning you could even use it for things like ftp, itunes.etc.


I'm LATE to the party, but I had to solve this recently, figured I'd share the wealth.

const url = window.location.origin + window.location.pathname
//http://example.com/somedir/somefile/

window.location.origin will give you the base url, in our test case: http://example.com

window.location.pathname will give you the route path (after the base url), in our test case /somedir/somefile

SOLUTION 2

You can simply do the following to get rid of the query parameters.

const url = window.location.href.split('?')[0]


Use indexOf

var url = "http://mysite.com/somedir/somefile/?aa";

if (url.indexOf("?")>-1){
url = url.substr(0,url.indexOf("?"));
}


You can concat origin and pathname, if theres present a port such as example.com:80, that will be included as well.

location.origin + location.pathname


Just one more alternative using URL

var theUrl = new URL(window.location.href);
theUrl.search = ""; //Remove any params
theUrl //as URL object
theUrl.href //as a string


Use the URL() constructor, then extract and concatenate the origin and pathname. This will automatically strip the search (aka query) parameters from the url, leaving the scheme, domain, port and pathname only.

const url = new URL('http://localhost:8080/index.html?search=foo&other=bar');
console.log(url.origin + url.pathname);

As a note, this type of transformation is usually referred to as normalization, specifically in this case URI Normalization. There may already exist libraries that accomplish this more robustly with more options in your environment.


var url = "tp://mysite.com/somedir/somefile/?foo=bar&loo=goo"    

url.substring(0,url.indexOf("?"));


You can use a regular expression: window.location.href.match(/^[^\#\?]+/)[0]


If you look at the documentation you can take just the properties you're interested in from the window object i.e.

protocol + '//' + hostname + pathname
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