如何通过 TypeScript 中的映射类型删除属性

Question

这里是代码

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!

我想通过 RemoveProperties 删除“可见 / x / y”属性，但我只能用 never 替换它

Answer 1

您可以使用与Omit 类型相同的技巧：

// 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

您可以在mapped types 中使用as clauses 一次性过滤掉属性：

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

当as 子句中指定的类型解析为never 时，不会为该键生成任何属性。因此，as 子句可以用作过滤器 [.]

更多信息：Announcing TypeScript 4.1 - Key Remapping in Mapped Types

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