Javascript error - Can't run .split on undefined

Question

I imagine this is really simple but I just seem unable to achieve it. I have this snippet...

var choice = location.href.split("?")[1].split("=")[1];

It works fine until 'choice' is empty, then it throws the following error....

TypeError: location.href.split(...)[1] is undefined

I understand this is because you cant call split on a variable that is undefined, my solution i think is to store the variable and then split it if full or ignore it if undefined.

Can someone help me do this?

What do you get if you output just `location.href`? – boz Feb 05 '13 at 14:04 — boz, Feb 05 '13 at 14:04

score 3 · Accepted Answer · answered Feb 05 '13 at 14:12

3

Not super cool to do it like this, but if you expect only one parameter in query string you can do it like this:

var choice = (location.href.split("?")[1] || '').split("=")[1];

You will get choice as undedfined if it's not set in query string.

answered Feb 05 '13 at 14:12

dfsq

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colin-higgins · Answer 2 · 2013-02-05T14:07:31.877

There isn't enough info, but either the attribute href, of the 1 index of the first split is undefined, or the second item of the last split is undefined.

You would separate the split out into several statements.

var choice;
var firstSplit = location.href;

if (firstSplit) {
    var secondSplit = firstSplit.split("?");
}
if (secondSplit[1]) {
    choice = secondSplit[1].split("=");
}

Then you could use whatever returned from the last split.

score 0 · Answer 3 · edited May 23 '17 at 11:43

You would see this exception if you are attempting to split apart a URL that may or may not have a querystring. You should check that the querystring exists before attempting to manipulate it:

var choice;
var outerSplit = window.location.href.split('?');
var innerSplit;
if (outerSplit.length > 1) {
    innerSplit = outerSplit.split('=');
    if (innerSplit.length > 1) {
        choice = innerSplit[1];
    }
}

If you find yourself performing this type of querystring manipulation often, you might want to invest in a reusable function like the one from this question.

Javascript error - Can't run .split on undefined

3 Answers3