Variable name as string in JavaScript

Question

Is there a way to retrieve a variable's name in JavaScript without it being part of an object? Here's an example of what I'm trying to achieve:

var obj = {//Declared in a global scope
    init:function(){
        var var1 = /*value from input*/;
        var var2 = /*value from input*/;

        doCode({name:/*variable name as a string, "var1"*/,val:var1});
        doCode({name:/*variable name as a string, "var2"*/,val:var2});
        function checkNumber(args){
            if(typeof args.val != 'number')
                console.error("Error variable '"+arg.name+"' is not a number!");
            else
                console.log("Variable '"+arg.name+"' is a number!");
        }
    }
}
obj.init();

Except in very specific circumstances, like when the calling function only calls checkNumber once with one variable as its argument or the value is unique among some object you can access from the function, the answer is “no”. Why do you want to do this? — Ry-
– Ry- ♦, Commented Jun 28, 2013 at 13:50
@san.chez - depends on the scope, and what this really is, that's why I asked ? — adeneo
– adeneo, Commented Jun 28, 2013 at 13:53
@PhilippeTrépanier: Keep them all in one object, pass the name. Alternatively, pass the field and create a data-label attribute or something, then check arg.value and use arg.dataset.label. (Note that your typeof check probably doesn’t perform as intended if this is user input…) — Ry-
– Ry- ♦, Commented Jun 28, 2013 at 14:15

TulaGingerbread · Accepted Answer · 2013-06-28 14:22:16Z

1

This might work

var foo = 'bar'
var fooname
for (var i in window) {
  if (window[i] == foo) fooname = i
}
alert(fooname) // alerts 'foo'

upd Of course, that have some restrictions, for example you can't have two variables with same content, but it is the only way I still found.

edited Jun 28, 2013 at 14:22

answered Jun 28, 2013 at 14:17

TulaGingerbread

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1 Comment

PhilTrep Over a year ago

I wont use this in a global scope but with an object instead, thank you! +1 and correct.

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