Get URL Parameters Using Javascript

Question

I am trying to get the ULR parameter, but my code does not run because at debugging it shows me an error which is: Cannot read property 'split' of undefined and I am not sure what I am doing wrong. For example: default.html?pid=405 I need to get in a JavaScript variable the 405 value. This is the code i am using:

<script type="text/javascript">
    function getUrlParameters(parameter, staticURL, decode) {
        var currLocation = (staticURL.length) ? staticURL : window.location.search,
            parArr = currLocation.split("?")[1].split("&"),
            returnBool = true;

        for (var i = 0; i < parArr.length; i++) {
            parr = parArr[i].value.split("=");
            if (parr[0] == parameter) {
                return (decode) ? decodeURIComponent(parr[1]) : parr[1];
                returnBool = true;
            } else {
                returnBool = false;
            }
        }

        if (!returnBool) return false;
    }
</script>

<script type="text/javascript">
    function runDepo()
    {
        var idParameter = getUrlParameters("pid", "", true);
    }

Can somebody help me? Thanks in advance.

Possible duplicate of [How can I get query string values in JavaScript?](http://stackoverflow.com/questions/901115/how-can-i-get-query-string-values-in-javascript) — Aeolingamenfel, Apr 29 '16 at 20:14
If `parArr` is actually an array then you would just use `parArr[i].split("=")` — Jonathan Michalik, Apr 29 '16 at 20:16
This is not a duplicate because I need to create that function and call the variable in a another function, besides my issue is that the script is showing me an error which is NOT RECOGNIZE the split!! THIS IS NOT A DUPLICATE! — Lanshore, Apr 29 '16 at 20:17
I also tried ´parArr[i].split("=")´ but it still showing me the same error — Lanshore, Apr 29 '16 at 20:18
check what is the value that `window.location.search` returns. Because if neither window.location.search nor staticURL have a `?`, it will crash when accessing the first position of the array — Bruno, Apr 29 '16 at 20:19
@Lanshore No need to yell. Notice that they said "possible duplicate." And it's not the `split` isn't recognized, it's that you're trying to use `split` on `undefined`. It's the same thing as typing `undefined.split('=')`. — Mike Cluck, Apr 29 '16 at 20:23

Shady Alset · Answer 1 · 2016-04-29T20:47:16.697

2

Just remove .value from parr = parArr[i].value.split("=");

function getUrlParameters(parameter, staticURL, decode) {
        var currLocation = (staticURL.length) ? staticURL : window.location.search,
            parArr = currLocation.split("?")[1].split("&"),
            returnBool = true;

        for (var i = 0; i < parArr.length; i++) {
            parr = parArr[i].split("=");
            if (parr[0] == parameter) {
                return (decode) ? decodeURIComponent(parr[1]) : parr[1];
                returnBool = true;
            } else {
                returnBool = false;
            }
        };

        if (!returnBool) return false;
    };
    
    document.getElementById('Result').value = getUrlParameters("pid", "default.html?pid=405", true);

    alert(getUrlParameters("pid", "default.html?pid=405", true));

PID : <input type="text" id="Result" placeholder="Result">

edited Apr 29 '16 at 20:47

answered Apr 29 '16 at 20:19

Shady Alset

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I tried this, but should I have to code this `alert(getUrlParameters("pid", "default.html?pid=405", true));` is there any way not to have to write it? – Lanshore Apr 29 '16 at 20:35
@Lanshore no it's just for testing functionality, but where do you want displaying result ? – Shady Alset Apr 29 '16 at 20:42
Post this as answer, I cannot answer myself, I solved it by my own. – Lanshore Apr 29 '16 at 21:18
`function getUrlVars() { var vars = {}; var parts = window.location.href.replace(/[?&]+([^=&]+)=([^&]*)/gi, function (m, key, value) { vars[key] = value; }); return vars; } var first = getUrlVars()["pid"];` – Lanshore Apr 29 '16 at 21:18
This is a better way to get what I needed. – Lanshore Apr 29 '16 at 21:19

Get URL Parameters Using Javascript

1 Answers1