打字稿：polymorhpic此，超載簽名不是與功能實現兼容

Question

有人可以解釋爲什麼這不起作用：

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 { } 

這將使：

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

或

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

Typescript 2.5.2抱怨「過載簽名與函數實現不兼容」。

爲什麼需要這兩種形式？

考慮一個REST API返回序列化（JSON）型號：

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

然後一個通用服務（角），請求模型：

@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}`); 
    } 
} 

解決方案

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)); 
    } 
} 

這允許兩種形式同時保留abstract。

Answer 1

在我看來，有沒有必要有兩個：

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

第一種形式是足夠的，更具可讀性，然後代碼將是這樣的：

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({}, {}, {}); 

（code in playground）

但是，如果由於某種原因超出了我的範圍，您確實希望擁有這兩種形式，那麼您可以執行如下操作：

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, {}, {}, {}); 

（code in playground）

的代碼更compilated（我認爲沒有充分的理由）。
另外，我不得不從Model刪除abstract部分，否則有此錯誤：

The 'this' context of type 'typeof Model' is not assignable to method's 'this' of type 'ModelCtor'.
Cannot assign an abstract constructor type to a non-abstract constructor type.

對於這兩個行：

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

也許你可以用它玩多一點和修復那些錯誤沒有刪除abstract。