URLSearchParams.get()
The get()
method of the URLSearchParams
interface returns the first value associated to the given search parameter.
Note: This feature is available in Web Workers
Syntax
URLSearchParams.get(name)
Parameters
- name
-
The name of the parameter to return.
Return value
A USVString
if the given search parameter is found; otherwise, null
.
Examples
If the URL of your page is https://example.com/?name=Jonathan&age=18
you could parse out the 'name' and 'age' parameters using:
let params = new URLSearchParams(document.location.search.substring(1)); let name = params.get("name"); // is the string "Jonathan" let age = parseInt(params.get("age"), 10); // is the number 18
Requesting a parameter that isn't present in the query string will return null
:
let address = params.get("address"); // null
Specifications
Specification |
---|
URL Standard (URL) # dom-urlsearchparams-get |
Browser compatibility
Desktop | Mobile | |||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|
Chrome | Edge | Firefox | Internet Explorer | Opera | Safari | WebView Android | Chrome Android | Firefox for Android | Opera Android | Safari on IOS | Samsung Internet | |
get |
49 |
17 |
29 |
No |
36 |
10.1 |
49 |
49 |
29 |
36 |
10.3 |
5.0 |
© 2005–2021 MDN contributors.
Licensed under the Creative Commons Attribution-ShareAlike License v2.5 or later.
https://developer.mozilla.org/en-US/docs/Web/API/URLSearchParams/get