获取函数/类构造函数的参数类型

Question

我正在尝试做我不确定在TypeScript中可能做的事情：从函数推断参数类型/返回类型。

例如

function foo(a: string, b: number) {
  return `${a}, ${b}`;
}

type typeA = <insert magic here> foo; // Somehow, typeA should be string;
type typeB = <insert magic here> foo; // Somehow, typeB should be number;

我的用例是尝试创建一个包含构造函数和参数的配置对象：

例如：

interface IConfigObject<T> {
    // Need a way to compute type U based off of T.
    TypeConstructor: new(a: U): T;
    constructorOptions: U;
}

// In an ideal world, could infer all of this from TypeConstructor

class fizz {
    constructor(a: number) {}
}

const configObj : IConfigObj = {
    TypeConstructor: fizz;
    constructorOptions: 13; // This should be fine
}

const configObj2 : IConfigObj = {
    TypeConstructor: fizz;
    constructorOptions: 'buzz'; // Should be a type error, since fizz takes in a number
}

有人可以帮我吗？谢谢！

Answer 1

使用TypeScript 2.8，您可以使用新的extends关键字：

type FirstArgument<T> = T extends (arg1: infer U, ...args: any[]) => any ? U : any;
type SecondArgument<T> = T extends (arg1: any, arg2: infer U, ...args: any[]) => any ? U : any;

let arg1: FirstArgument<typeof foo>; // string;
let arg2: SecondArgument<typeof foo>; // number;
let ret: ReturnType<typeof foo>; // string;

Answer 2

对于提取构造函数参数类型的用例，我将给出更直接的答案。

type GetConstructorArgs<T> = T extends new (...args: infer U) => any ? U : never

class Foo {
    constructor(foo: string, bar: number){
        //
    }
}

type FooConstructorArgs = GetConstructorArgs<typeof Foo> // [string, number]

Answer 3

添加了打字稿2.8 conditional types with type inference

添加了TypeScript 3.0 rest-elements-in-tuple-types，因此您现在就可以获取Array类型的所有参数。

type ArgumentsType<T extends (...args: any[]) => any> = T extends (...args: infer A) => any ? A : never;

type Func = (a: number, b: string) => boolean;
type Args = ArgumentsType<Func> // type Args = [number, string];
type Ret = ReturnType<Func> // type Ret = boolean;

您可以这样使用它：

const func = (...args: Args): Ret => { // type the rest parameters and return type
  const [a, b] = args; // spread the arguments into their names
  console.log(a, b); // use the arguments like normal
  return true;
};

// Above is equivalent to:
const func: Func = (a, b) => {
  console.log(a, b);
  return true;
}

Answer 4

Typescript现在具有ConstructorParameters内置，类似于Parameters内置。确保您传递的是类类型，而不是实例：

ConstructorParameters<typeof SomeClass>

Answer 5

这种方法怎么样：

interface IConfigObject<T, U> {
    TypeConstructor: new(a: U) => T;
    constructorOptions: U;
}

class fizz {
    constructor(a: number) {}
}

function createConfig<U, T>(cls: { new (arg: U): T }, arg: U): IConfigObject<T, U> {
    return {
        TypeConstructor: cls,
        constructorOptions: arg
    }
}

const configObj = createConfig(fizz, 3); // ok
const configObj2 = createConfig(fizz, "str"); // error

（code in playground）

---

编辑

您可以有一个索引类型变量：

const configs: { [name: string]: IConfigObject<any, any> } = {
    config1: createConfig(fizz, 3),
    config2: createConfig(fizz, "str"), // error
    config3: createConfig(buzz, "str")
}