How to get the URL without any parameters in JavaScript?

Question

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 example:

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

instead of

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

possible duplicate of Remove querystring from URL

Alex Angas
– Alex Angas

2014-01-20 05:02:07 +00:00
Commented Jan 20, 2014 at 5:02 — Alex Angas
– Alex Angas, Commented Jan 20, 2014 at 5:02

Dan Passaro · Accepted Answer · 2025-09-17 02:43:44Z

426

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

location.origin + location.pathname

edited Sep 17 at 2:43

Dan Passaro

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answered Jun 6, 2011 at 20:12

lonesomeday

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13 Comments

JonnyReeves Over a year ago

Note that you don't need to supply the location.protocol as all modern browsers (including IE6) will automatically inherit the current protocol. ie: '//' + location.host + location.pathname

lonesomeday Over a year ago

@JonnyReeves If you're using it in the current document, that's true: good point. There are occasions when it might be necessary, e.g. if outputting the URL as plain text.

izogfif Over a year ago

It's missing port, so this will return incorrect result for page http://www.example.com:8080/asdf.html?foo=bar

Coderer Over a year ago

You can save a few characters with location.origin, which I believe also addresses @izogfif's concern.

Air Over a year ago

@izogfif (really old post I know) You said this solution is missing port, but I don't believe it is. location.hostname returns the host name without the port, location.host should return the complete host value (port included). Just so others are aware of the difference between the two.

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Oddman · Accepted Answer · 2016-01-01 23:35:38Z

276

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.

edited Jan 1, 2016 at 23:35

answered Feb 22, 2015 at 19:31

Oddman

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8 Comments

Bahattin Ungormus Over a year ago

in case you need to remove the anchors window.location.href.split(/[?#]/)[0];

mehmet Over a year ago

indeed every answer is rather convoluted

Oddman Over a year ago

@mehmet all ears if you have a simpler approach :)

mehmet Over a year ago

@Oddman, I absolutely don't remember what I was thinking then, but I guess that might be an affirmative statement :))

Oddman Over a year ago

@Vishal--JAVA--CQ "better" is subjective, but it's simpler.

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Robo Rick · Accepted Answer · 2020-09-23 14:41:20Z

44

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]

edited Sep 23, 2020 at 14:41

answered Sep 23, 2020 at 14:32

Robo Rick

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1 Comment

Arash Over a year ago

Solution 2 is not complete and will include the hashes (#) if come before the QSPs.

zr0gravity7 · Accepted Answer · 2023-07-01 16:20:55Z

17

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://example.com/somedir/?foo=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.

edited Jul 1, 2023 at 16:20

answered Jan 17, 2022 at 10:14

zr0gravity7

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Comments

Niklas · Accepted Answer · 2011-06-06 20:13:24Z

15

Use indexOf

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

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

answered Jun 6, 2011 at 20:13

Niklas

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1 Comment

gouessej Over a year ago

It doesn't remove #.

AamirR · Accepted Answer · 2019-03-27 12:00:26Z

12

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

answered Mar 27, 2019 at 12:00

AamirR

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1 Comment

gouessej Over a year ago

This solution doesn't work with Microsoft Internet Explorer <= 10: developer.mozilla.org/en-US/docs/Web/API/Location/origin

digitalWestie · Accepted Answer · 2020-09-22 22:27:57Z

12

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

answered Sep 22, 2020 at 22:27

digitalWestie

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Comments

Mic · Accepted Answer · 2011-06-06 20:25:43Z

5

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

answered Jun 6, 2011 at 20:25

Mic

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Comments

Headshota · Accepted Answer · 2011-06-06 20:13:16Z

4

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

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

answered Jun 6, 2011 at 20:13

Headshota

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1 Comment

actimel Over a year ago

This presumes, that parameter is defined. Much better is url.split('?')[0]

Anupam Samanta · Accepted Answer · 2023-10-30 16:32:50Z

1

You can do this

window.location.origin + window.location.pathname

answered Oct 30, 2023 at 16:32

Anupam Samanta

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Comments

planetjones · Accepted Answer · 2011-06-06 20:14:19Z

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

answered Jun 6, 2011 at 20:14

planetjones

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1 Comment

gouessej Over a year ago

You lose the port number with your suggestion, lonesomeday's solution is better in my humble opinion.

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