Add array element during definition based on condition

Question

I define an array like this:

[{foo:0}, true === false && { foobar:1}, {bar:2}]

My expected result would be that the middle item is not added at all when the middle condition is not met:

[ { foo: 0 }, { bar: 2 } ]

in fact it adds false as an array item:

[ { foo: 0 }, false, { bar: 2 } ]

Is there a way to prevent adding the false while maintaining this lightweight syntax (I know I could always use push or the spread operator)

But if you want to keep your building terse you could just filter after: [{foo:0}, true === false && { foobar:1}, {bar:2}].filter(Boolean) — Denys Séguret
– Denys Séguret, Commented Mar 15, 2018 at 14:59
You could use a ternary like condition ? { foobar:1} : null and filter out null values at the end. — Zenoo
– Zenoo, Commented Mar 15, 2018 at 15:00

Nina Scholz · Accepted Answer · 2018-03-15 15:02:09Z

3

You could use concat with spread syntax and an empty array as neutral value.

var a = [].concat(...[
        { foo: 0 },
        true === false ? { foobar: 1 } : [],
        { bar: 2 }
    ]);

console.log(a);

With apply

var a = Array.prototype.concat.apply([], [
        { foo: 0 },
        true === false ? { foobar: 1 } : [],
        { bar: 2 }
    ]);

console.log(a);

answered Mar 15, 2018 at 15:02

Nina Scholz

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Comments

Oscar Paz · Accepted Answer · 2018-03-15 15:01:00Z

0

As Denys suggested, you could do this:

const arr = [{foo:0}, true === false && { foobar:1}, {bar:2}].filter(el => el !== false);
console.log(arr);

answered Mar 15, 2018 at 15:01

Oscar Paz

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2 Comments

Denys Séguret Over a year ago

el => el !== false? Please make it less verbose.

Oscar Paz Over a year ago

Less verbose? What do you suggest, !!el? Sorry, but the OP only wants to remove false values, nor 0, null or empty strings.

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