TypeScript条件排除类型从接口排除

Question

According to documentation, I can use predefined Exclude type to exclude certain properties from certain type:

type Test = string | number | (() => void);

type T02 = Exclude<Test, Function>;

However if instead of Test type I'd have Test interface, it doesn't seems to work. How can I achieve similiar result in below case?

interface Test {
  a: string;
  b: number;
  c: () => void;
} 

// get somehow interface with excluded function properties??

Answer 1

To get this effect you need to figure out which properties match Function , and then selectively exclude them. Equivalently we find which properties do not match Function , and then selectively include them. This is more involved than just excluding some pieces of a union; it involves mapped types as well as conditional types .

One way to do this is as follows:

type ExcludeMatchingProperties<T, V> = Pick<
  T,
  { [K in keyof T]-?: T[K] extends V ? never : K }[keyof T]
>;

type T02 = ExcludeMatchingProperties<Test, Function>;
// type T02 = {
// a: string;
// b: number;
// }

---

Examining ExcludeMatchingProperties<T, V> , we can note that the type { [K in keyof T]-?: T[K] extends V ? never : K }[keyof T] { [K in keyof T]-?: T[K] extends V ? never : K }[keyof T] will return the keys in T whose properties are not assignable to V .

If T is Test and V is Function , this becomes something like

{ 
  a: string extends Function ? never : "a"; 
  b: number extends Function ? never : "b"; 
  c: ()=>void extends Function ? never : "c" 
}["a"|"b"|"c"]

, which becomes

{ a: "a"; b: "b"; c: never }["a"|"b"|"c"]

, which becomes

{ a: "a"; b: "b"; c: never }["a"] | 
{ a: "a"; b: "b"; c: never }["b"] | 
{ a: "a"; b: "b"; c: never }["c"]

, or

"a" | "b" | never

, or

"a" | "b"

Once we have those keys, we Pick their properties them from T (using the Pick<T, K> utility type ):

Pick<Test, "a" | "b">

which becomes the desired type

{
  a: string,
  b: number
}

Okay, hope that helps; good luck!

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