如何根据类方法的返回类型声明参数类型？

扬特拉布

例如：

    export const mock = <T>(provide, key: keyof T, value) => (){}
    class Service {
      do(): boolean {
        return true;
      }
    }
    // value have to be Boolean 
    mock<Service>(Service, "do", true)

value参数是T[key]函数的返回类型。我想知道是否有可能以某种方式声明值的类型。

约瑟夫船长

这是我的解决方案：

/**
 * If T is a Class, return union string type of class methods
 */
type ClassType<T> = T extends { new(): infer R } ? keyof R : never

/**
 * Check if property is a function/method
 */
type AnyFunc = (...args: any[]) => any

export const mock = <P,>(provide: P, key: ClassType<P>, value: P extends { new(): infer R } ? R[keyof R] extends AnyFunc ? ReturnType<R[keyof R]> : never : never) => {}

class Service {
  do(): boolean {
    return true;
  }
}



const reuslt = mock<typeof Service>(Service, 'do', false) // ok
const reuslt2 = mock<typeof Service>(Service, 'foo', false) // error, expected do
const reuslt3 = mock<typeof Service>(Service, 'do', 42) // error, expected boolean

请记住，TS会以不同方式对待Service类型typeof Service。

因为我不是解释大师，所以这里是您的示例：

const x: Service = {
  do: () => true
}

const y: typeof Service = class {
  do() {
    return true
  }
}

TypeScript类文档

更新句柄承诺

/**
 * If T is a Class, return union string type of class methods
 */
type ClassType<T> = T extends { new(): infer R } ? keyof R : never

/**
 * Check if property is a function/method
 */
type AnyFunc = (...args: any[]) => any

type PromiseReturnType<T> = T extends Promise<infer R> ? R : T

export const mock = <P,>(provide: P, key: ClassType<P>, value: P extends { new(): infer R } ? R[keyof R] extends AnyFunc ? PromiseReturnType<ReturnType<R[keyof R]>> : never : never) => { }

class Service {
  do() {
    return new Promise<number>(resolve => resolve(42))
  }
}


const reuslt = mock<typeof Service>(Service, 'do', false) // ok
const reuslt2 = mock<typeof Service>(Service, 'foo', false) // error, expected do
const reuslt3 = mock<typeof Service>(Service, 'do', Promise.resolve(42)) // error, expected number
const reuslt4 = mock<typeof Service>(Service, 'do', 42) // ok

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