TypeScript 条件排除类型从接口中排除 [英] TypeScript conditional Exclude type exclude from interface

问题描述

根据文档，我可以使用预定义的排除类型从特定类型中排除某些属性:

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>;

但是，如果我有 Test 接口而不是 Test 类型，它似乎不起作用.如何在以下情况下获得类似的结果?

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??

推荐答案

要获得这种效果，您需要弄清楚哪些属性与 Function 匹配，然后有选择地排除它们.等效地，我们找到哪些属性与 Function 不匹配，然后有选择地包含它们.这不仅仅是排除联合的某些部分更复杂；它涉及映射类型以及条件类型.

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.

一种方法如下:

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;
// }

<小时>

检查 ExcludeMatchingProperties，我们可以注意到类型 { [K in keyof T]-?: T[K] 扩展了 V ?never : K }[keyof T] 将返回 T 中其属性不可分配给 V 的键.

---

Examining ExcludeMatchingProperties<T, V>, we can note that the type { [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.

如果 T 是 Test 并且 V 是 Function，这就变成了

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"]

，变成

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

，变成

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

或

"a" | "b" | never

或

"a" | "b"

一旦我们有了这些键，我们就从 T 中Pick 它们的属性(使用 Pick 实用程序类型):

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

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

成为想要的类型

{
  a: string,
  b: number
}

好的，希望有帮助；祝你好运！

Okay, hope that helps; good luck!

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