How to add a key value to javascript object,
i have an array like this :
var a = [20, 21, 22];
and i tried to convert it into a javascript object using this code :
var id = Object.assign({}, a);
the result is : {0 : 20, 1 : 21, 2 : 22}
but, how to change the result like an object below ? :
0 : {id : 20}, 1 : {id : 21}, 2 : {id : 22}
any help will be appreciated
Based on your comment "the format of the object supposed to be like this: 0 : {id : 20}, 1 : {id : 21}, 2 : {id : 22}", this looks like a job for .map():
var input = [ 20, 21, 22 ];
var output = input.map( function( value ) {
return { id: value };
});
console.log( output );Or, in modern JavaScript using an arrow function:
var input = [ 20, 21, 22 ];
var output = input.map( value => ({ id: value }) );
console.log( output );
You can't and shouldn't have an object like {id : 20, id: 21, id:22}. Each key in the object should be unique, otherwise it beats the purpose of being a 'key'.
If you, hypothetically, had an object like
var a = {id : 20, id: 21, id:22};
,then how are you planning to access those elements? What will a.id return and how will it help you?
If you want to get 0 : {id : 20}, 1 : {id : 21}, 2 : {id : 22}, you could do this this way:
let a = [20, 21, 22];
let obj = Array();
for(let i of a) {
obj.push({id: i});
}
0 : {id : 20}, 1 : {id : 21}, 2 : {id : 22}Create object like: var collections = [], a = [20, 21, 22];
Iterate array a([20, 21, 22]), and add individual item in collections like below which is array.
for (i = 0; i < a.length; i++) {
collections.push({id: a[i]});
}
Cheers...
{id : 20, id: 21, id:22} is equivalent to just 1 key/value pair. their is no way to have {id : 20, id: 21, id:22} as an object and keep values as 20, 21, 22
You can try making it into an array like
{id : [20, 21, 22]}
or
[{id:20}, {id:21}, {id:22}]
var a = [20, 21, 22];
var result = [];
for(let i=0; i<a.length; i++){
result.push({id: a[i]});
}
//console.log(result)
var id = Object.assign({}, result);
console.log(id);id{ 0 : 20, 1 : 21, 2 : 22}
push() make my browser unresponsed... but thanks for trying to help me btw :)