Typescript - keyof without a function

Question

I am facing the following issue and I am not sure how to approach it. I want to assign a name of property to a string property of another object. Not sure however how to use key of in this case.

I managed to do it using "validation" function, just to have compile errors, but I would like to skip declaring a function that does nothing. Any ideas?

Titian Cernicova-Dragomir · Accepted Answer · 2018-06-01 10:46:23Z

1

Type assertions are there to let you do things the compiler thinks are invalid. Just declare the type of const explicitly, :

interface IGame { id: number}
const incalidColumn: keyof IGame = 'invalild' // error
const validColumn: keyof IGame = 'id'

answered Jun 1, 2018 at 10:46

Titian Cernicova-Dragomir

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2 Comments

Daniel Stoyanoff Over a year ago

The problem is that I need to pass it to an object property that is already defined as string and I would like to avoid creating const and then assigning { key: keyOfGame("id") }

Titian Cernicova-Dragomir Over a year ago

@DanielStoyanoff your question was with variables and I answered it in that context. Unfortunately there is no other way to do it without an extra function if you are assigning to a field that is declared as type string

Estus Flask · Accepted Answer · 2018-06-01 11:07:32Z

1

It's possible to use same approach as in this answer in order to segregate type checking from JS output:

type CheckGameProp<T extends keyof IGame> = T;

const invalidColumn = "invalidColumn";
{ type _checkGameProp = CheckGameProp<typeof invalidColumn>; }

const validColumn = "id";
{ type _checkGameIdProp = CheckGameProp<typeof validColumn>; }

_checkGameIdProp dummy type can be scoped to avoid problems with type name choice.

Depending on how invalidColumn and validColumn are used, it may be better to check them where they are used instead.

answered Jun 1, 2018 at 11:07

Estus Flask

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3 Comments

Daniel Stoyanoff Over a year ago

Almost the same as creating a function, which I eventually did in a separate file -> keyOf<T>("prop")

Estus Flask Over a year ago

But types don't pollute compiled code, if this was a concern for you. And as it was mentioned, it may be reasonable to move type checks to a place where constants are used. This isn't shown in the question.

Daniel Stoyanoff Over a year ago

Runtime type checking is not required. The ideas was to do it for compile time.

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