我正在阅读有关TypeScript 2.8 - Conditional Types的内容,我在Typescript文档中看到了很多示例,其中列出了已解决的版本的类型作为旁边的注释:
type TypeName<T> =
T extends string ? "string" :
T extends number ? "number" :
T extends boolean ? "boolean" :
T extends undefined ? "undefined" :
T extends Function ? "function" :
"object";
type T0 = TypeName<string>; // "string"
type T1 = TypeName<"a">; // "string"
type T2 = TypeName<true>; // "boolean"
type T3 = TypeName<() => void>; // "function"
type T4 = TypeName<string[]>; // "object"
例如TypeName<true>
实际上是boolean
。这对于更复杂的场景变得更加有用,在这些场景中,您可以看到实际构建的类型:
type FunctionPropertyNames<T> = { [K in keyof T]: T[K] extends Function ? K : never }[keyof T];
type FunctionProperties<T> = Pick<T, FunctionPropertyNames<T>>;
type NonFunctionPropertyNames<T> = { [K in keyof T]: T[K] extends Function ? never : K }[keyof T];
type NonFunctionProperties<T> = Pick<T, NonFunctionPropertyNames<T>>;
interface Part {
id: number;
name: string;
subparts: Part[];
updatePart(newName: string): void;
}
type T40 = FunctionPropertyNames<Part>; // "updatePart"
type T41 = NonFunctionPropertyNames<Part>; // "id" | "name" | "subparts"
type T42 = FunctionProperties<Part>; // { updatePart(newName: string): void }
type T43 = NonFunctionProperties<Part>; // { id: number, name: string, subparts: Part[] }
当出现关于类型的混淆时,这当然可以帮助我,所以我想知道是否有任何方法让编译器提取相同的类型信息?