JavaScript set object key by variable [duplicate]
I am building some objects in JavaScript and pushing those objects into an array, I am storing the key I want to use in a variable then creating my objects like so:
var key = "happyCount";
myArray.push( { key : someValueArray } );
but when I try to examine my array of objects for every object the key is "key"
instead of the value of the variable key. Is there any way to set the value of the key from a variable?
Fiddle for better explanation: http://jsfiddle.net/Fr6eY/3/
Solution 1:
You need to make the object first, then use []
to set it.
var key = "happyCount";
var obj = {};
obj[key] = someValueArray;
myArray.push(obj);
UPDATE 2021:
Computed property names feature was introduced in ECMAScript 2015 (ES6) that allows you to dynamically compute the names of the object properties in JavaScript object literal notation.
const yourKeyVariable = "happyCount";
const someValueArray= [...];
const obj = {
[yourKeyVariable]: someValueArray,
}
Solution 2:
In ES6, you can do like this.
var key = "name";
var person = {[key]:"John"}; // same as var person = {"name" : "John"}
console.log(person); // should print Object { name="John"}
var key = "name";
var person = {[key]:"John"};
console.log(person); // should print Object { name="John"}
Its called Computed Property Names, its implemented using bracket notation( square brackets) []
Example: { [variableName] : someValue }
Starting with ECMAScript 2015, the object initializer syntax also supports computed property names. That allows you to put an expression in brackets [], that will be computed and used as the property name.
For ES5, try something like this
var yourObject = {};
yourObject[yourKey] = "yourValue";
console.log(yourObject );
example:
var person = {};
var key = "name";
person[key] /* this is same as person.name */ = "John";
console.log(person); // should print Object { name="John"}
var person = {};
var key = "name";
person[key] /* this is same as person.name */ = "John";
console.log(person); // should print Object { name="John"}