how to remove properties via mapped type in TypeScript

Question

Here is the code

class A {
    x = 0;
    y = 0;
    visible = false;
    render() {

    }
}

type RemoveProperties = {
    readonly [P in keyof T         


        

Answer 1

You can use the same trick as the Omit type uses:

// We take the keys of P and if T[P] is a Function we type P as P (the string literal type for the key), otherwise we type it as never. 
// Then we index by keyof T, never will be removed from the union of types, leaving just the property keys that were not typed as never
type JustMethodKeys<T> = ({[P in keyof T]: T[P] extends Function ? P : never })[keyof T];  
type JustMethods<T> = Pick<T, JustMethodKeys<T>>; 

Answer 2

TS 4.1

You can now use as clauses in mapped types to filter out properties in one go:

type Methods<T> = { [P in keyof T as T[P] extends Function ? P : never]: T[P] };
type A_Methods = Methods<A>;  // { render: () => void; }

When the type specified in an as clause resolves to never, no property is generated for that key. Thus, an as clause can be used as a filter[.]

Further info: Announcing TypeScript 4.1 Beta

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