Get escaped URL parameter

Question

I'm looking for a jQuery plugin that can get URL parameters, and support this search string without outputting the JavaScript error: "malformed URI sequence". If there isn't a jQuery plugin that supports this, I need to know how to modify it to support this.

?search=%E6%F8%E5

The value of the URL parameter, when decoded, should be:

æøå

(the characters are Norwegian).

I don't have access to the server, so I can't modify anything on it.

Answer 1

$.urlParam = function(name){
  var results = new RegExp('[\\?&]' + name + '=([^&#]*)').exec(top.window.location.href); 
  return (results !== null) ? results[1] : 0;
}

$.urlParam("key");

Answer 2

function getURLParameters(paramName) 
{
        var sURL = window.document.URL.toString();  
    if (sURL.indexOf("?") > 0)
    {
       var arrParams = sURL.split("?");         
       var arrURLParams = arrParams[1].split("&");      
       var arrParamNames = new Array(arrURLParams.length);
       var arrParamValues = new Array(arrURLParams.length);     
       var i = 0;
       for (i=0;i<arrURLParams.length;i++)
       {
        var sParam =  arrURLParams[i].split("=");
        arrParamNames[i] = sParam[0];
        if (sParam[1] != "")
            arrParamValues[i] = unescape(sParam[1]);
        else
            arrParamValues[i] = "No Value";
       }

       for (i=0;i<arrURLParams.length;i++)
       {
                if(arrParamNames[i] == paramName){
            //alert("Param:"+arrParamValues[i]);
                return arrParamValues[i];
             }
       }
       return "No Parameters Found";
    }

}

Answer 3

There's a lot of buggy code here and regex solutions are very slow. I found a solution that works up to 20x faster than the regex counterpart and is elegantly simple:

    /*
    *   @param      string      parameter to return the value of.
    *   @return     string      value of chosen parameter, if found.
    */
    function get_param(return_this)
    {
        return_this = return_this.replace(/\?/ig, "").replace(/=/ig, ""); // Globally replace illegal chars.

        var url = window.location.href;                                   // Get the URL.
        var parameters = url.substring(url.indexOf("?") + 1).split("&");  // Split by "param=value".
        var params = [];                                                  // Array to store individual values.

        for(var i = 0; i < parameters.length; i++)
            if(parameters[i].search(return_this + "=") != -1)
                return parameters[i].substring(parameters[i].indexOf("=") + 1).split("+");

        return "Parameter not found";
    }

console.log(get_param("parameterName"));

Regex is not the be-all and end-all solution, for this type of problem simple string manipulation can work a huge amount more efficiently. Code source.

Answer 4

Just in case you guys have the url like localhost/index.xsp?a=1#something and you need to get the param not the hash.

var vars = [], hash, anchor;
var q = document.URL.split('?')[1];
if(q != undefined){
    q = q.split('&');
    for(var i = 0; i < q.length; i++){
        hash = q[i].split('=');
        anchor = hash[1].split('#');
        vars.push(anchor[0]);
        vars[hash[0]] = anchor[0];
    }
}

Answer 5

Slight modification to the answer by @pauloppenheim , as it will not properly handle parameter names which can be a part of other parameter names.

Eg: If you have "appenv" & "env" parameters, redeaing the value for "env" can pick-up "appenv" value.

Fix:

var urlParamVal = function (name) {
    var result = RegExp("(&|\\?)" + name + "=(.+?)(&|$)").exec(location.search);
    return result ? decodeURIComponent(result[2]) : "";
};

Answer 6

For example , a function which returns value of any parameters variable.

function GetURLParameter(sParam)
{
    var sPageURL = window.location.search.substring(1);
    var sURLVariables = sPageURL.split('&');
    for (var i = 0; i < sURLVariables.length; i++)
    {
        var sParameterName = sURLVariables[i].split('=');
        if (sParameterName[0] == sParam)
        {
            return sParameterName[1];
        }
    }
}​

And this is how you can use this function assuming the URL is,

"http://example.com/?technology=jquery&blog=jquerybyexample".

var tech = GetURLParameter('technology');
var blog = GetURLParameter('blog');

So in above code variable "tech" will have "jQuery" as value and "blog" variable's will be "jquerybyexample".