Using Variable for Property Name of Object - Javascript

Question

saw a few answers related to this, but none answer this version of the subject in question.

Consider the following: (linkto: jsfiddle)

$(function(){

arrKeys = [];
objArr = [];

nameArr = ['name1','name2','name3','name4'];
descArr = ['desc1','desc2','desc3','desc4'];
allValues = {name:  nameArr, desc: descArr};

arrKeys[0] = 'name';
arrKeys[1] = 'desc';

    first = arrKeys.shift(); // returns 'name'

    $(allValues[first]).each(function (key,value) { 

        console.log(first); //returns 'name'
        objArr[key] = {first:value}; //the problem

    });

    console.log(objArr);


});

With console.log(objArr) producing the following array of objects like so:

[Object, Object, Object, Object] 0: Object first: "name1" 1: Object first: "name2" 2: Object first: "name3" 3: Object first: "name4" length: 4

The issue is that I'd like the property "first" to be the value of the var first (which is "name".. So instead, the result would be:

[Object, Object, Object, Object] 0: Object name: "name1" 1: Object name: "name2" 2: Object name: "name3" 3: Object name: "name4" length: 4

(linkto: jsfiddle)

score 10 · Accepted Answer · edited Mar 02 '19 at 16:47

10

To set variables as key names you have to use bracket notation;

console.log(first); // returns 'name'
var obj = {};
obj[first] = value;
objArr[key] = obj; // no longer a problem

Sorry it's more verbose :(

Edit;

In ES6 you can now use computed-property-names;

const key = 'name';
const value = 'james';

const obj = {
  [key]: value
};

edited Mar 02 '19 at 16:47

Ele

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answered Jun 22 '13 at 18:53

Jivings

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score 2 · Answer 2 · answered Jun 22 '13 at 18:56

2

var x = {}
x[first] = value
objArr[key] = x

Javascript object literal syntax treats unquoted keys as if they were quoted strings instead of variable names. You have to use bracket syntax to set properties with names computed at runtime.

(Why are you wrapping your code in a jQuery call? And you really should be using var declarations.)

answered Jun 22 '13 at 18:56

user2357112

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I'm not using those global vars, just when throwing that example together for this specific question. The jquery is nonessential. Thank you for your response. – Fezzik Jun 23 '13 at 16:05

Using Variable for Property Name of Object - Javascript

2 Answers2

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