TypeScript handling type array or single value

Question

I have an interface like this. And I am wondering what is the best case to handle this case:

interface Ia {
  a: string
}
let a: Ia | Ia[]
console.log(a[0].a)

TS throws me an error:

Element implicitly has an 'any' type because expression of type '0' can't be used to index type 'Ia'.
  Property '0' does not exist on type 'Ia'.

I know I could use console.log((a as Ia[])[0].a) but it is not readable at all. Does anyone have better suggestions?

Also, the real use case type has many properties inside not the single one a: sting as in the example.

If you know that a is an array, why then you say it is Ia | Ia[] ? — Maciej Sikora
– Maciej Sikora, Commented Feb 7, 2020 at 11:50
@MaciejSikora I am getting the data structure and it might be a list or a single value. I will have a check for that, but my question is about options to handle the actual type assignation and to check if there are any alternatives to as — Lukas Ignatavičius
– Lukas Ignatavičius, Commented Feb 7, 2020 at 11:55

kaya3 · Accepted Answer · 2020-02-07 11:50:49Z

3

If a is not an array, then a[0] does not make sense. You need to test whether it is an array, and treat it differently depending on the result of that test.

interface Ia {
  a: string
}

let obj: Ia | Ia[] = { a: 'foo' };

function test(obj: Ia | Ia[]): void {
  console.log(Array.isArray(obj) ? obj[0].a : obj.a);
}

Or you can use this pattern, which takes advantage of control-flow type narrowing:

function test2(obj: Ia | Ia[]): void {
  if (Array.isArray(obj)) { obj = obj[0]; }
  // now obj: Ia is known not to be an array
  console.log(obj.a);
}

Playground Link

answered Feb 7, 2020 at 11:50

kaya3

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Comments

Lukas Ignatavičius · Accepted Answer · 2020-02-10 13:58:54Z

1

I have settled with:

interface Ia {
  a: string
}
let a: Ia | Ia[]
if (Array.isArray(a)){
  a = [a]
}
console.log(a[0].a)

Thank you @kaya3 and @maciej-sikora for suggestions

answered Feb 10, 2020 at 13:58

Lukas Ignatavičius

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Comments

Maciej Sikora · Accepted Answer · 2020-02-07 14:42:21Z

0

I propose create a getter for that, consider:

const getA = (x: Ia | Ia[]) => "a" in x ? x.a : x[0]?.a;

const a = getA({ a: 'I am a' });
const a2 = getA([{ a: 'I am also a ' }]);

Or uniform the type and use it always in the same way:

const uniform = (x: Ia | Ia[]): Ia => Array.isArray(a) ? a[0] : a;

Such function unify our type from the union to one member.

answered Feb 7, 2020 at 14:42

Maciej Sikora

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