Check out this code snippet:
class A {
x = 0;
y = 0;
visible = false;
render() {
}
}
type RemoveProperties<T> = {
readonly [P in keyof T]: T[P] extends Function ? T[P] : never//;
};
var a = new A() as RemoveProperties<A>
a.visible // never
a.render() // ok!
I'm trying to use the RemoveProperties function to eliminate the "visible / x / y" properties, but it seems that I can only replace them with 'never'.
In TypeScript 4.1, a new feature allows you to utilize as clauses within mapped types for efficient property filtering:
type Methods<T> = { [P in keyof T as T[P] extends Function ? P : never]: T[P] };
type A_Methods = Methods<A>; // { render: () => void; }
By using an
asclause with a type that resolves tonever, you can exclude properties from the final result. This makesasa powerful tool for filtering.
For more details, check out: Announcing TypeScript 4.1 - Key Remapping in Mapped Types
To experiment with this feature, visit the TypeScript Playground: Playground
If you want to achieve a similar effect like the Omit type does, you can follow this trick:
// This code snippet takes the keys of P and checks if T[P] is a Function. If it is, we type P as P (the string literal type for the key), otherwise we type it as never.
// Then we index by keyof T, removing any instances of never from the union of types, which results in only the property keys that were not typed as never being retained.
type JustMethodKeys<T> = ({[P in keyof T]: T[P] extends Function ? P : never })[keyof T];
type JustMethods<T> = Pick<T, JustMethodKeys<T>>;
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