Declare literal union type based on array of objects values

Question

Assuming I have an array of objects (simplified version):

const arrObj = [
  { id: "example1", other: "just" },
  { id: "example2", other: "an" },
  { id: "example3", other: "example" }
]

and I want to create a union type out of all ids

I tried:

const list = arrObj.map((item) => item.id) as const;
type ids = typeof list[number];

But this does not work as it seems I cannot create the array dynamically:

A 'const' assertions can only be applied to references to enum members, or string, number, boolean, array, or object literals.

The desired type output should be: "example1" | "example2" | "example3"

Type of the variable, access elements with number, then access id: typeof arrObj[number]["id"] — k.tten
– k.tten, Commented Aug 29, 2022 at 18:52
@kelly thanks for your comment. I want a literal type union. With that I only get "string". — supersize
– supersize, Commented Aug 29, 2022 at 18:56
Well then, append as const to your definition of arrObj. — k.tten
– k.tten, Commented Aug 29, 2022 at 18:57
No, but feel free to upvote and accept the one by Matthieu Riegler. — k.tten
– k.tten, Commented Aug 29, 2022 at 19:03

Matthieu Riegler · Accepted Answer · 2022-08-29 19:02:58Z

1

const arrObj = [
  { id: "example1", other: "just" },
  { id: "example2", other: "an" },
  { id: "example3", other: "example" }
] as const

type myIds = typeof arrObj[number]['id']

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answered Aug 29, 2022 at 19:02

Matthieu Riegler

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