How to get URL parameter using jQuery or plain JavaScript?
Solution from 2022
We have: http://example.com?sent=yes
let searchParams = new URLSearchParams(window.location.search)
Does sent exist?
searchParams.has('sent') // true
Is it equal to "yes"?
let param = searchParams.get('sent')
and then just compare it.
Best solution here.
var getUrlParameter = function getUrlParameter(sParam) {
var sPageURL = window.location.search.substring(1),
sURLVariables = sPageURL.split('&'),
sParameterName,
i;
for (i = 0; i < sURLVariables.length; i++) {
sParameterName = sURLVariables[i].split('=');
if (sParameterName[0] === sParam) {
return sParameterName[1] === undefined ? true : decodeURIComponent(sParameterName[1]);
}
}
return false;
};
And this is how you can use this function assuming the URL is,http://dummy.com/?technology=jquery&blog=jquerybyexample
.
var tech = getUrlParameter('technology');
var blog = getUrlParameter('blog');
jQuery code snippet to get the dynamic variables stored in the url as parameters and store them as JavaScript variables ready for use with your scripts:
$.urlParam = function(name){
var results = new RegExp('[\?&]' + name + '=([^&#]*)').exec(window.location.href);
if (results==null) {
return null;
}
return decodeURI(results[1]) || 0;
}
example.com?param1=name¶m2=&id=6
$.urlParam('param1'); // name
$.urlParam('id'); // 6
$.urlParam('param2'); // null
example params with spaces
http://www.jquery4u.com?city=Gold Coast
console.log($.urlParam('city'));
//output: Gold%20Coast
console.log(decodeURIComponent($.urlParam('city')));
//output: Gold Coast