javascript jquery object-literal properties variables

How to use a variable for a key in a JavaScript object literal?


Why does the following work?

    { 'top' : 10 }, 10

Whereas this doesn’t work:

var thetop = 'top';
    { thetop : 10 }, 10

To make it even clearer: At the moment I’m not able to pass a CSS property to the animate function as a variable.



{ thetop : 10 } is a valid object literal. The code will create an object with a property named thetop that has a value of 10. Both the following are the same:

obj = { thetop : 10 };
obj = { "thetop" : 10 };

In ES5 and earlier, you cannot use a variable as a property name inside an object literal. Your only option is to do the following:

var thetop = "top";

// create the object literal
var aniArgs = {};

// Assign the variable property name with a value of 10
aniArgs[thetop] = 10; 

// Pass the resulting object to the animate method
    aniArgs, 10  

ES6 defines ComputedPropertyName as part of the grammar for object literals, which allows you to write the code like this:

var thetop = "top",
    obj = { [thetop]: 10 };

console.log(; // -> 10

You can use this new syntax in the latest versions of each mainstream browser.


  • I understand! Thank you! Shoudln’t this also work with eval? I mean it doesn’t work in my example at the moment, but I thin it should… 🙂

    – speendo

    Feb 16, 2010 at 16:38

  • 3

    @Marcel: eval() wouldn’t work inside an object literal for dynamic property names, you’d just get an error. Not only that, but eval() really shouldn’t be used for such things. There’s an excellent article on correct and incorrect usage of eval:

    – Andy E

    Feb 16, 2010 at 16:44

  • 2

    @AndyE consider updating “recent versions” and “current IE TP” with some more specific time like “Versions later than XXX” or “after 2014-mm” (I’d make change myself, but I don’t know what good values would be.

    Feb 26, 2015 at 1:59

  • 1

    for ES6, is there a way to leave out the property if the key is null/undefined? For example, in {[key] : "value"}, if key was null, it would give { null: "value"}, whereas I’d like the result to be {}

    Mar 10, 2018 at 14:52

  • wonderful solution, but why you know so much, even reading the whole specification could not get point:(

    – hugemeow

    Apr 13, 2018 at 22:41


With ECMAScript 2015 you are now able to do it directly in object declaration with the brackets notation: 

var obj = {
  [key]: value

Where key can be any sort of expression (e.g. a variable) returning a value.

So here your code would look like:

  [thetop]: 10
}, 10)

Where thetop will be evaluated before being used as key.



    ES5 quote that says it should not work

    Note: rules have changed for ES6:


    PropertyName :

    • IdentifierName
    • StringLiteral
    • NumericLiteral


    The production PropertyName : IdentifierName is evaluated as follows:

    1. Return the String value containing the same sequence of characters as the IdentifierName.

    The production PropertyName : StringLiteral is evaluated as follows:

    1. Return the SV [String value] of the StringLiteral.

    The production PropertyName : NumericLiteral is evaluated as follows:

    1. Let nbr be the result of forming the value of the NumericLiteral.
    2. Return ToString(nbr).

    This means that:

    • { theTop : 10 } is the exact same as { 'theTop' : 10 }

      The PropertyName theTop is an IdentifierName, so it gets converted to the 'theTop' string value, which is the string value of 'theTop'.

    • It is not possible to write object initializers (literals) with variable keys.

      The only three options are IdentifierName (expands to string literal), StringLiteral, and NumericLiteral (also expands to a string).


    • 7

      Downvoters: I was the first to quote any standard on this question. The ES6 quote on the top answer was edited after I answered, and that standard was not yet accepted at the time. If you happen to know why else I’m getting downvotes, please explain, I just want to learn. I might as well get the peer pressure badge…

      Jul 14, 2015 at 15:32

    • 1

      I guess the downvote was mostly because your answer doesn’t offer a solution, and is “not useful” in that regard. Voting towards peer pressure 🙂

      – Bergi

      Aug 2, 2015 at 21:22

    • 3

      @Bergi thanks for having the courage! 🙂 But I think I have answered the question directly: Q: “Why does the following work?”. A: because ES5 says so. “What to do about it?” is implicit. Did you downvote the top question for saying “It is impossible” without a standard quote before someone edited in the ES6 solution?

      Aug 3, 2015 at 5:17

    • Ah, right. I typically cite the specific questions in a blockquote to make it clear when I answer them directly. Quoting the standard can make a good answer even better, but currently your post doesn’t even answer the question imo. Stating what can make a key in ES5 doesn’t imply how they work. Surely thetop is an IdentifierName, so why does it not work? That question is still open.

      – Bergi

      Aug 3, 2015 at 13:10

    • 1

      @Bergi thanks again for explaining to me! I have updated it to make it more explicit. I hadn’t done it before because I though it was obvious, because we can write {a:1}.a, so a clearly not expand the variable value in the identifier case. But yes, explaining further is an improvement in this case.

      Aug 3, 2015 at 14:23