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&param2=&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