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返回自身的复杂函数的Typescript类型声明

[英]Typescript type declaration for a complex function that returns itself

我有以下javascript函数,该函数返回包含对象参数中给定的其他方法的函数。
我认为代码比解释会更容易理解:

var scopeFunction = (object) => {

  // create main function if it doesn't exist
  if (!object.main) object.main = function(){
    alert("No main function for this scopeFunction");
  };

  // add all object keys to the main function that will be return
  _.each(object, function(d,i){ object.main[i] = d; });

  // return main function
  return object.main;

};

我想在打字稿中正确定义此代码,这就是我所做的,但是当我尝试访问返回的函数对象的键时,原子打字稿(这是我对其进行测试的地方)会引发错误。 这是我当前的代码外观:

// TYPES
namespace ScopeFunction {

  export interface Function<Obj extends MakerObject, Key extends keyof Obj>  {
    (): Obj["main"];
    [key: Key]: Obj[Key];
  }
  export interface MainFunction {
    (...args:any[]) : any;
  }
  export interface MakerObject {
    [key: string]: any;
    main?: MainFunction;
  }

  export type Maker = <Obj extends MakerObject, Key extends keyof Obj>(object:Obj) => ScopeFunction.Function<Obj, Key>;

}

// FUNC
var scopeFunction:ScopeFunction.Maker = (object) => {

  // create main function if it doesn't exist
  if (!object.main) object.main = function(){
    alert("No main function for this scopeFunction");
  };

  // add all object keys to the main function that will be return
  _.each(object, function(d,i){ object.main[i] = d; });

  // return main function
  return object.main;

};

// TEST
var test = scopeFunction({
  a: 1,
  b: "3",
  main: () => { console.log("testLog"); return 0; }
})

var test1 = test(); // WORKS OK
var test2 = test.main(); // ALERT: Property 'main' doesn't exist on type 'Function<{ a: number; b: string; main: () => number }, "main" | "a" | "b">'
var test3 = test.a; // ALERT: Property 'a' doesn't exist on type 'Function<{ a: number; b: string; main: () => number }, "main" | "a" | "b">'

知道我的问题出在哪里吗?

您的代码有几个问题:

  1. 定义不编译[key: Key]: Obj[Key]无效,并且indexer参数必须为numberstring (仅那些类型有效)。 您需要改用映射类型。
  2. (): Obj["main"]不会是与Obj["main"]相同类型的调用签名,而是将返回main属性的任何函数。
  3. main函数的类型过于通用,因此不会保留任何参数类型。

一种符合您期望的解决方案可能是:

namespace ScopeFunction {

    export type Function<Obj extends MakerObject<(...args: any[]) => any>> = Obj['main'] & {
        [P in keyof Obj]: Obj[P];
    }
    export interface MakerObject<TMain extends (...args: any[]) => any> {
        main?: TMain;
    }

    export type Maker = <TObj extends MakerObject<(...args: any[]) => any>>(object: TObj) => ScopeFunction.Function<TObj>;

}

// FUNC
var scopeFunction: ScopeFunction.Maker = (object) => {

    // create main function if it doesn't exist
    if (!object.main) object.main = function () {
        alert("No main function for this scopeFunction");
    };

    // return main function
    return Object.assign(object.main, object);
};


// TEST
var test = scopeFunction({
    a: 1,
    b: "3",
    main: (param: number) => { console.log("testLog"); return param; }
})

var test1 = test(10);
var test2 = test.main(10); 
var test3 = test.a;

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