打字稿:Polymorhpic this，过载签名与函数实现不兼容

Question

有人可以解释为什么这不起作用：

abstract class Model {
    static deserialize<T extends Model>(this: ( new() => T ), object: any): T;
    static deserialize<T extends Model>(ctor: ( new() => T ), object: any): T {
        return new ctor();
    }
    static deserializeArray<T extends Model>(this: ( new() => T ), ...objects: any[]): T[];
    static deserializeArray<T extends Model>(ctor: ( new() => T ), ...objects: any[]): T[] {
        return objects.map(object => Model.deserialize(ctor, object));
    }
}
class MyModel extends Model { }

将允许：

let myModel = MyModel.deserialize({});
let myModels = MyModel.deserializeArray({}, {}, {});

或

let myModel = Model.deserialize(MyModel, {});
let myModels = Model.deserializeArray(MyModel, {}, {}, {});

打字稿2.5.2抱怨"超载签名与函数实现不兼容"。

为什么两种表格都需要？

考虑返回序列化（JSON）模型的REST API：

class MyModelController {
    get(id: number) {
        let myModel = ... some db/service call ...
        return myModel.serialize();
    }
}

，然后是通用服务（Angular）请求模型：

@Injectable()
abstract class HttpService {
    constructor(private http: Http) { }
    errorHandler(response) {
        ...
    }
    get<T extends Model>(ModelType: ( new() => T ), endpoint: string): Observable<T> {
        return this.http.get(endpoint)
            // we can't call ModelType.deserialize() here...
            .map(response => Model.deserialize(ModelType, response.json()))
            .catch(response => this.errorHandler(response));
    }
}
@Injectable()
class MyModelService extends HttpService {
    get(id: number) {
       return super.get(MyModel, `/api/models/${id}`);
    }
}

解决方案

abstract class Model {
    static deserialize<T extends Model>(this: ( new() => T), object: {}): T;
    static deserialize<T extends Model>(this: Function & { prototype: Model }, ctor: ( new() => T ), object: {});
    static deserialize<T extends Model>(this: ( new() => T), first: ( new() => T ) | {}, second?: any) {
        return typeof first === "function" ? new first() : new this(); 
    }
    static deserializeArray<T extends Model>(this: ( new() => T), ...objects: {}[]): T[];
    static deserializeArray<T extends Model>(this: Function & { prototype: Model }, ctor: ( new() => T ), ...objects: {}[]): T[];
    static deserializeArray<T extends Model>(this: ( new() => T), first: ( new() => T ) | {}[], second?: {}[]): T[] {
        const ctor = typeof first === "function" ? first : this;
        const objects = typeof first === "function" ? second : first;
        return objects.map(object => Model.deserialize(ctor, object));
    }
}

这允许在保留abstract的同时两种表单。

Answer 1

我认为，没有必要两者：

let myModel1 = MyModel.deserialize({});
let myModel2 = Model.deserialize(MyModel, {});

第一个形式足够且更可读，然后代码看起来像这样：

type ModelCtor<T extends Model> = {
    new(): T;
    deserialize<T extends Model>(object: any): T;
    deserializeArray<T extends Model>(...objects: any[]): T[];
};
abstract class Model {
    static deserialize<T extends Model>(this: ModelCtor<T>, object: any): T {
        return new this();
    }
    static deserializeArray<T extends Model>(this: ModelCtor<T>, ...objects: any[]): T[] {
        return objects.map(object => this.deserialize(object));
    }
}
class MyModel extends Model { }
let myModel = MyModel.deserialize({});
let myModels = MyModel.deserializeArray({}, {}, {});

（操场上的代码）

但是，如果出于某种原因，您确实想拥有两种表格，那么您可以做类似的事情：

type ModelCtor<T extends Model> = {
    new(): T;
    deserialize<T extends Model>(object: any): T;
    deserialize<T extends Model>(ctor: ModelCtor<T>, object: any): T;
    deserializeArray<T extends Model>(...objects: any[]): T[];
    deserializeArray<T extends Model>(ctor: ModelCtor<T>, ...objects: any[]): T[];
};
class Model {
    static deserialize<T extends Model>(this: ModelCtor<T>, object: any): T;
    static deserialize<T extends Model, S extends Model>(this: ModelCtor<T>, ctor: ModelCtor<S>, object: any): S;
    static deserialize<T extends Model, S extends Model>(this: ModelCtor<T>, first: ModelCtor<S> | any, second?: any): S {
        return second === undefined ? new this() : new first();
    }
    static deserializeArray<T extends Model>(this: ModelCtor<T>, ...objects: any[]): T[];
    static deserializeArray<T extends Model, S extends Model>(this: ModelCtor<T>, ctor: ModelCtor<S>, ...objects: any[]): S[];
    static deserializeArray<T extends Model, S extends Model>(this: ModelCtor<T>, ...objects: any[]): S[] {
        const ctor: ModelCtor<S> = typeof objects[0] === "function" ? objects[0] : this as any;
        return objects.map(object => ctor.deserialize(ctor, object));
    }
}
class MyModel extends Model { }
let myModel1 = MyModel.deserialize({});
let myModels1 = MyModel.deserializeArray({}, {}, {});
let myModel2 = Model.deserialize(MyModel, {});
let myModels2 = Model.deserializeArray(MyModel, {}, {}, {});

（操场上的代码）

代码更加汇编（我认为这没有充分的理由）。
另外，我必须从Model中删除abstract零件，否则有此错误：

类型"类型"模型的"此"上下文无法分配给方法的" type" of Type'Modelctor'。
无法将抽象的构造函数分配给非抽象构造函数。

对于这两种行：

let myModel2 = Model.deserialize(MyModel, {});
let myModels2 = Model.deserializeArray(MyModel, {}, {}, {});

也许您可以随身携带一点并修复这些错误而无需删除abstract。