Is it possible to use a variable value as part of another variables name?

Question

var b = '1';

How can I create a variable name using the contents of variable b? I am doing this in a each loop, so b will change.

I am wanting the final product to be something like:

var trip1 = '';

How would I accomplish something like that?

var trip + b = ''; var trip.b = '';

If you're always modifying the name by adding a numeric qualifier, then perhaps what you really want is an Array. — Pointy
– Pointy, Commented Dec 8, 2012 at 14:40
@ATLChris : aggreeing to what Pointy said , use an array like var b0=array[0] , var b1=array[1] , ... and so on — Hussain Akhtar Wahid 'Ghouri'
– Hussain Akhtar Wahid 'Ghouri', Commented Dec 8, 2012 at 14:42

T.J. Crowder · Accepted Answer · 2012-12-08 14:36:10Z

No, but you can do it with properties of an object:

var obj = {}; obj.b = '1' obj['trip' + obj.b] = ''; // or var obj = {}; var b = '1' obj['trip' + b] = '';

In JavaScript, you can access properties in two ways:

In the latter case, the string doesn't have to be a string literal, it can be the result of any expression, hence the examples above.

score 1 · Accepted Answer · 2012-12-08 14:41:48Z

I am doing this in a each loop

Since you are iterating over them, the best way to do this is to use an array.

var trip = [1,2,3,4,5]; for (var i = 0; i < trip.length; i++){ alert(trip[i]); }

To address your question, you could store your values as properties of an object:

var obj = {}; var b = '1'; obj['trip' + b] = "";

If you are doing this in the global scope, you could also add the variable as a property of the global object:

var b = '1'; window['trip' + b] = ""; trip1; // == ""

gherkins · Accepted Answer · 2012-12-08 14:41:53Z

you can access global variables via the window object, using a dynmamic key like this:

var b = '1'; trip1 = 'hello'; console.log(window['trip'+b]);

Tim Joyce · Accepted Answer · 2012-12-08 14:54:58Z

It sounds like you are looking for the eval() method.

var b = '1'; (eval('var trip'+b+' = 3')); // Is the same as writing var trip1 = 3; console.log(trip1); //Output of variable trip1 is 3

Although, do a quick google search for why eval(); is bad before you employ this technique.

4 Answers 4