Extending Interfaces Using Keys from an Array in Typescript

Question

Extending Interfaces Using Keys from an Array in Typescript

When dealing with a scenario where you want a pointfree omit, you can achieve this:

type PlainObject<T> = {[key: string]: T}

const omit = <K extends string>(
  name: K
) => <T, U extends PlainObject<T> & {
  [P in K]: T
}>(
  x: U,
): Partial<U> => {
  throw new Error ('Yet to implement !')
}

Is there a way to use K as an array of strings instead of just a string? Perhaps something like this:

const omit = <K extends string[]>(
  ...names: K
) => <T, U extends PlainObject<T> & {
  [P in K]: T // Type 'K' is not assignable to type 'string | number | symbol'
}>(
  x: U,
): Partial<U> => {
  throw new Error ('Yet to implement !')
}

The purpose here is to make the compiler flag any incorrect objects entered based on the keys passed:

omit('a', 'b')({b: 3, c: 4}) // => expect error

Thank you in advance
Seb

typescript typescript-typings

Answer 1

Answer №1

Absolutely, you can achieve this:

type ExcludeKey<T, K extends keyof T> = Pick<T, Exclude<keyof T, K>>;

declare function excludeKeys<K extends string>(...keys: K[]):
  <T extends Record<K, any>>(source: T) => ExcludeKey<T, K>;

Example of how to use it:

interface Person {
  name: string;
  age: number;
  job: symbol;
}

declare const person: Person;

excludeKeys('age', 'name')(person); // $ExpectType { job: symbol }
excludeKeys('foo', 'bar')(person);  // $ExpectError

Answer 2

Absolutely, you can achieve this:

type ExcludeKey<T, K extends keyof T> = Pick<T, Exclude<keyof T, K>>;

declare function excludeKeys<K extends string>(...keys: K[]):
  <T extends Record<K, any>>(source: T) => ExcludeKey<T, K>;

Example of how to use it:

interface Person {
  name: string;
  age: number;
  job: symbol;
}

declare const person: Person;

excludeKeys('age', 'name')(person); // $ExpectType { job: symbol }
excludeKeys('foo', 'bar')(person);  // $ExpectError

Extending Interfaces Using Keys from an Array in Typescript

Answer №1

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