mixins - TypeScript 中的 Mixin

Question

我正在玩 TypeScript，我有几个功能性 mixins，Eventable并且Settable，我想将它们混合到一个Model类中（假设它类似于 Backbone.js 模型）：

function asSettable() {
  this.get = function(key: string) {
    return this[key];
  };
  this.set = function(key: string, value) {
    this[key] = value;
    return this;
  };
}

function asEventable() {
  this.on = function(name: string, callback) {
    this._events = this._events || {};
    this._events[name] = callback;
  };
  this.trigger = function(name: string) {
    this._events[name].call(this);
  }
}

class Model {
  constructor (properties = {}) {
  };
}

asSettable.call(Model.prototype);
asEventable.call(Model.prototype);

上面的代码工作正常，但如果我尝试使用其中一种混合方法（如(new Model()).set('foo', 'bar').

我可以解决这个问题

interface为 mixins添加声明
declaring dummy ///get声明中set的方法ontriggerModel

虚拟声明是否有干净的方法？

score 12 · Accepted Answer

interfaces这是使用和方法处理mixins的一种static create()方法。接口支持多重继承，这样你就不必interfaces为你的 mixins 重新定义，并且该static create()方法会负责给你一个Model()as an的实例IModel（<any>需要强制转换来抑制编译器警告。）你需要复制所有您Model对which的成员定义IModel很糟糕，但这似乎是在当前版本的 TypeScript 中实现您想要的最简洁的方法。

编辑：我已经确定了一种稍微简单的方法来支持 mixins，甚至创建了一个帮助类来定义它们。可以在这里找到详细信息。

function asSettable() {
  this.get = function(key: string) {
    return this[key];
  };
  this.set = function(key: string, value) {
    this[key] = value;
    return this;
  };
}

function asEventable() {
  this.on = function(name: string, callback) {
    this._events = this._events || {};
    this._events[name] = callback;
  };
  this.trigger = function(name: string) {
    this._events[name].call(this);
  }
}

class Model {
  constructor (properties = {}) {
  };

  static create(): IModel {
      return <any>new Model();
  }
}

asSettable.call(Model.prototype);
asEventable.call(Model.prototype);

interface ISettable {
    get(key: string);
    set(key: string, value);
}

interface IEvents {
    on(name: string, callback);
    trigger(name: string);
}

interface IModel extends ISettable, IEvents {
}


var x = Model.create();
x.set('foo', 'bar');

score 3 · Accepted Answer

尽管仍然需要双重类型声明，但最简洁的方法是将 mixin 定义为模块：

module Mixin {
    export function on(test) {
        alert(test);
    }
};

class TestMixin implements Mixin {
    on: (test) => void;
};


var mixed = _.extend(new TestMixin(), Mixin); // Or manually copy properties
mixed.on("hi");

使用接口的另一种方法是使用类来破解它（尽管由于多重继承，您需要为 mixins 创建一个公共接口）：

var _:any;
var __mixes_in = _.extend; // Lookup underscore.js' extend-metod. Simply copies properties from a to b

class asSettable {
    getx(key:string) { // renamed because of token-clash in asEventAndSettable
        return this[key];
    }
    setx(key:string, value) {
        this[key] = value;
        return this;
    }
}

class asEventable {
    _events: any;
    on(name:string, callback) {
        this._events = this._events || {};
        this._events[name] = callback;
    }
    trigger(name:string) {
        this._events[name].call(this);
  }
}

class asEventAndSettable {
   // Substitute these for real type definitions
   on:any;
   trigger:any;
   getx: any;
   setx: any;
}

class Model extends asEventAndSettable {
    /// ...
}

var m = __mixes_in(new Model(), asEventable, asSettable);

// m now has all methods mixed in.

正如我对 Steven 的回答所评论的那样，mixins 确实应该是一个 TypeScript 功能。

score 2 · Accepted Answer

几年前在 Typescript 中构建了一种新方法，称为“mixin 类”。文档中没有很好地涵盖它，但是他们确实有一个很好的注释示例来很好地描述该模式。应用于您的情况，它可能看起来像：

type Constructor = new (...args: any[]) => {}

function Settable<TBase extends Constructor>(Base: TBase) {
  return class extends Base {
    _props: Record<string, any> = {};

    get(key: string) {
      return this._props[key];
    }

    set(key: string, value: any) {
      this._props[key] = value;
      return this;
    }
  }
}

function Eventable<TBase extends Constructor>(Base: TBase) {
  return class extends Base {
    _events: Record<string, () => void> = {};

    on(name: string, callback: () => void) {
      this._events[name] = callback;
    }

    trigger(name: string) {
      this._events[name].call(this);
    }
  }
}

class Model extends Settable(Eventable(Object)) {
  constructor(properties = {}) {
    super();
  }
}

这可以让您获得您想要的打字，例如，您可以在(new Model()).set('boo', 'bar')完全打字支持的情况下进行通话。没有虚拟声明。

score 1 · Accepted Answer

一种解决方案是不使用 typescript 类系统，而只使用类型和接口的系统，以及关键字“new”。

    //the function that create class
function Class(construct : Function, proto : Object, ...mixins : Function[]) : Function {
        //...
        return function(){};
}

module Test { 

     //the type of A
    export interface IA {
        a(str1 : string) : void;
    }

    //the class A 
    //<new () => IA>  === cast to an anonyme function constructor that create an object of type IA, 
    // the signature of the constructor is placed here, but refactoring should not work
    //Class(<IA> { === cast an anonyme object with the signature of IA (for refactoring, but the rename IDE method not work )
    export var A = <new () => IA> Class(

        //the constructor with the same signature that the cast just above
        function() { } ,

        <IA> {
            //!! the IDE does not check that the object implement all members of the interface, but create an error if an membre is not in the interface
            a : function(str : string){}
        }
    );


    //the type of B
    export interface IB {
        b() : void;
    }
    //the implementation of IB
    export class B implements IB { 
        b() { }
    }

    //the type of C
    export interface IC extends IA, IB{
        c() : void;
        mystring: string;
    }

     //the implementation of IC
    export var C = <new (mystring : string) => IC> Class(

        //public key word not work
        function(mystring : string) { 

            //problem with 'this', doesn't reference an object of type IC, why??
            //but google compiler replace self by this !!
            var self = (<IC> this);
            self.mystring = mystring;
        } ,

        <IC> {

            c : function (){},

            //override a , and call the inherited method
            a: function (str: string) {

                (<IA> A.prototype).a.call(null, 5);//problem with call and apply, signature of call and apply are static, but should be dynamic

                //so, the 'Class' function must create an method for that
                (<IA> this.$super(A)).a('');
            }

        },
        //mixins
        A, B
    );

}

var c = new Test.C('');
c.a('');
c.b();
c.c();
c.d();//ok error !

mixins - TypeScript 中的 Mixin

4 回答 4

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