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Home/ Questions/Q 834933
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T04:44:49+00:00 2026-05-15T04:44:49+00:00

Ok I’ve been learning some of the more advanced aspects of Javascript and now

  • 0

Ok I’ve been learning some of the more advanced aspects of Javascript and now trying to use this I’m stuck.

Here is my code:

function Data(){}

function init(state){
  var item;
  item=new Data();
  item.fieldrid=17;
  item.description='foo';
  state.push(item);
};


function findInState(state,fieldrid) {
    for (var item in state) {
        alert(item.fieldrid); //prints undefined
        if (item.fieldrid == fieldrid) {
            return item;
        }
    }
    return null;
}

var s=[];
init(s);
alert(s[0].fieldrid); //prints 17 (expected)
alert(findInState(s,17).fieldrid); //exception here. function returns null.

A running example is here at jsbin

Why does this not work? I would expect the alert in findInState to yield 17 but instead it yields undefined.

What am I doing wrong?

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  1. Editorial Team
    Editorial Team
    2026-05-15T04:44:50+00:00Added an answer on May 15, 2026 at 4:44 am
    for (var item in state)
    

    You should not loop over an array using for..in.
    It works as expected when using for (var i = 0; i < state.length; i++).

    See https://developer.mozilla.org/en/Core_JavaScript_1.5_Reference/Statements/for…in:

    Although it may be tempting to use this as a way to iterate over an Array, this is a bad idea.

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