CODEWITHSUNDEEP
typescript
ZiiMakc
ZiiMakc

Reputation: 36836

Transform an existing type into a new type with one property change

How to transform an existing type into a new type with one property change?

Typesctipt sandbox

Example:

type SomeComplexType = string // just for example

// cannot be changed
type Options<T> = {
    opt1: boolean | undefined;
    opt2: number;
    opt3?: SomeComplexType;
    opt4: T
}

// Can be changed

// How to change opt1 to accept only true and infer other option types?
type Keys = keyof Options<any>

let r: { [K in Keys]: K extends 'opt1' ? true : any }

// Good (should work)
r = { opt1: true, opt2: 2, opt3: '1', opt4: 1 }
r = { opt1: true, opt2: 2, opt3: '1', opt4: 'str' }

// Bad (should be error)
r = { opt1: false, opt2: 1, opt3: 'str', opt4: 1 } // opt1 should be true
r = { opt1: true, opt2: 'str', opt3: 'str', opt4: 1 } // opt2 should be number
r = { opt1: true, opt2: 'str', opt3: 1, opt4: 1 } // opt3 should be

Upvotes: 2

Views: 62

Answers (1)

jcalz
jcalz

Reputation: 328152

If you have an object type O and would like to make a new type where all the proptery keys and values are the same except that the property at key opt1 should be true, you can write it like this:

{ [K in keyof O]: K extends 'opt1' ? true : O[K] }

The syntax O[K] is an indexed access meaning "the type of the property of O with a key of type K".

Then your examples should work as desired, (assuming O is Options<any>):

// Good 
r = { opt1: true, opt2: 2, opt3: '1', opt4: 1 } // okay
r = { opt1: true, opt2: 2, opt3: '1', opt4: 'str' } // okay

// Bad 
r = { opt1: false, opt2: 1, opt3: 'str', opt4: 1 } // error!
r = { opt1: true, opt2: 'str', opt3: 'str', opt4: 1 } // error!
r = { opt1: true, opt2: 'str', opt3: 1, opt4: 1 } // error!

Playground link to code

Upvotes: 1

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