Getting references to local variables created during eval() in JavaScript

Question

In the scenario below, how can I get references to the variables declared during eval() if I do not know their names?

function test() {
  eval("var myVariable = 5");
  var locals = magic() // TODO What should we do here?
  alert(locals["myVariable"]); // returns myVariable
}

Just a note: JavaScript being evaluated comes from a trusted source.

Do you have any ability to affect what that code you're passing to eval() looks like? In other words, can you change that code at all in order to make this work, or are you stuck with code coming from another source, and you cannot change it? — Pointy
– Pointy, Commented Jun 16, 2010 at 14:34
Well if you can make the code explicitly declare an object with a predictable name, and then put other (unpredictable) properties inside that object (see @tomalak's answer), then you can use that. But eval() does not create a new scope, and as far as I know there's no way to get a reference to your local "closure" scope in a way that'd let you treat it like an object. — Pointy
– Pointy, Commented Jun 16, 2010 at 14:41

Philippe Leybaert · Accepted Answer · 2010-06-16 14:30:26Z

2

eval() runs in the same scope as the caller, so this will work:

function test() {
  eval("var myVariable = 5");
  var locals = {};
  locals.myVariable = myVariable; // TODO What should we do here?
  alert(locals["myVariable"]); // returns myVariable
}

But you can't determine what variables were declared in the eval() call (if that's what you want)

answered Jun 16, 2010 at 14:30

Philippe Leybaert

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Tomalak · Accepted Answer · 2010-06-16 14:32:46Z

1

function test() {
  eval("var locals = {myVariable: 5};");
  alert(locals["myVariable"]);
}

works for me. eval() does not create a new scope.

answered Jun 16, 2010 at 14:32

Tomalak

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1 Comment

dpq Over a year ago

I've clarified the question to point out that I do not know the names of those variables beforehand.

Kevin Le - Khnle · Accepted Answer · 2010-06-16 14:33:02Z

1

Simple as :

eval("var myVariable = 5");
//no magic is needed
alert(myVariable); // returns myVariable

answered Jun 16, 2010 at 14:33

Kevin Le - Khnle

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1 Comment

dpq Over a year ago

What if I don't know the variables' names? I.e. I need to find out their names first.

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