Ternary operator return nothing

Question

is there a way to return nothing for ternary operator I mean

const a = [ 0 ? { name : "example" } : null ]

when i print a = [null]

or:

const a = [ 0 && {name:"example"}]

a will be [false]

i expected that a = [] for case 1

this might be a duplicate of https://stackoverflow.com/questions/44908159/how-to-define-an-array-with-conditional-elements — tbarbot, Jan 18 '22 at 10:44

score 5 · Accepted Answer · answered Jan 18 '22 at 10:43

5

You could spread an (empty) array.

console.log([...(0 ? [{ name : "example" }] : [])]);
console.log([...(1 ? [{ name : "example" }] : [])]);

answered Jan 18 '22 at 10:43

Nina Scholz

score 4 · Answer 2 · answered Jan 18 '22 at 10:42

4

No. It isn't possible. You are going to get a value back.

If you conditionally want a value, then either make the decision outside the array or filter the null values out afterwards.

e.g.

const a = 0 ? [ { name: "example" } ] : [];

or

const a = [ 0 ? { name : "example" } : null ].filter(Boolean);

answered Jan 18 '22 at 10:42

Quentin

yes i solved the problem with your first suggestion but i wondered is there antoher way – Berkay Bozkurt Jan 18 '22 at 11:11

score 1 · Answer 3 · answered Jan 18 '22 at 10:47

null is a perfectly valid array value in Javascript, as is undefined.

You can simply construct the array on demand...

const a = 0 ? [{ name : "example" }] : [];

or if this is part of a larger array construction, use splats...

const a = [
    'a',
    ...(0 ? [{name : "example"}] : []),
    'b'
];

console.log(a); //-> ['a', 'b']

3 Answers3