Type array so that find does not return undefined

[half portal]

Is it possible to type an array where the items match a string literal union so that .find will never return undefined. For example:

const COLUMNS = ['id', 'name', 'date'] as const
type COLUMN = (typeof COLUMNS)[number]

const columns = [ { name: 'id', databaseName: 'FOOBAR_ID' }, { name: 'name', databaseName: 'FOOBAR_NAME', } { name: 'date', databaseName: 'FOOBAR_DATE" } ]

// This is of type { name: string, databaseName: string } | undefined
// and ideally it wouldn't be undefined since name is type COLUMN
const getColumn = (name: COLUMN) => columns.find((column) => column.name === name)

[bronze raven]

you can use the ! operator

[half portal]

yea I thought about that, but isn't it considered "hacky" ? lol

[bronze raven]

no, casts are no hacky when used right

if you know stuff better than the compiler, then it's the only way to tell it about it

[half portal]

true, I was just hoping there was a "proper" way to do it

[bronze raven]

https://www.typescriptlang.org/docs/handbook/2/everyday-types.html#non-null-assertion-operator-postfix-

[bronze raven]

this is the proper way, at least in this case, only cast when you really need to it, like here

otherwise yes, that would be unsafe

[half portal]

Is there a way to actually type columns so that if COLUMNS changes, it warns if its missing or has extra entries?

[bronze raven]

nope, not when using arrays at least

would be possible, but using objects

(with Record<COLUMN, > and some transform)