TypeScript: Type of Key of Object?

Question

I'd like to define the following types:

type Item = {
    name: string,
    value: number | string
}


type ItemFilter<T extends Item> = {
    property: keyof Item;
    value: // ?????? I'd like to have here the type of Item[property]
}

How can I set value to have the type of item's property property?

I'd like to do the following:

const myItem: NumberItem = {
    name: 'phoneNumber',
    value: 1111111,
}

const valueFilter: ItemFilter<NumberItem> = {
    property: 'value',
    value: 11111, // SHOULD BE OK
}


const valueFilter: ItemFilter<NumberItem> = {
    property: 'value',
    value: 'STRING', // SHOULD FAIL
}

Answer 1

What I would do here is value constructor which would generate our proper filter value for our object. Consider the code:

const makeFilter = <T extends Record<string, any>> (item: T) => <K extends keyof T>(property: K, value: T[K]) => {
    return {
        property,
        value
    }
}

// TESTS
const myItem = {
    name: 'phoneNumber',
    value: 1111111,
}

const filter = makeFilter(myItem) // create function for specific object filtering

filter('value', 1231)  // ok
filter('value', 'a') // error as expected
filter('name', 1231)  // erorr as expected
filter('name', 'fine') // ok

Pay attention that we don't even need any Item type anymore, solution will work with every object.

Some explanation:

• makeFilter is a function which derives types by arguments
• it firstly get our initial object and return function which has arguments only for this specific object type
• filter = makeFilter(obj) creates our needed filter constructor for specific object
• using filter after has already proper narrowed types in the arguments
• property: K, value: T[K] means that value needs to be value of the object under the key - K

Why I have used function and not type directly.

Function allows us for type inference over arguments, so we don't need to repeat redundantly types. We just make filter for wanted object.

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