Question

我得到了method expression is not of function type。使用typescript 1.7.5。这是我的例子：

interface IEntity {
    new(parameters:{id:number}): this;
}

class Repository<T extends IEntity> {
    constructor(private model:T) {}

    create():any {
        return new this.model({id: 1});
    }
}

我怎样才能让它发挥作用？

应该是这样的：

class Model implements IEntity {}

let r = new Repository(Model);
let object = r.create();

Answer 1

这应该有用。

interface IEntity {
    id:number;
    show(): number;
}

class Repository<T extends IEntity> {
    constructor(private model: {new(id:number): T; }){} //Notice the empty constructor

    create():T {
        return new this.model(2);
    }
}

class Model implements IEntity {
    id:number;

    constructor (id:number){
        this.id = id;
    }
    show():number {
        return this.id;
    }
}

let r = new Repository(Model);
let object = r.create();
console.log(object.show());

这是@DennisJaamann和@John White答案的组合。

Answer 2

我认为语法应该如下：

class Repository<T extends IEntity> {
    constructor(){} //Notice the empty constructor

    create():T {
        return new T({id: 1});
    }
}

class Model implements IEntity {}

let r = new Repository<Model>(); //Specify the specific type of IEntity <T>
let object = r.create();

@see http://www.typescriptlang.org/Handbook#generics-generic-classes

干杯

如何使用工厂实例化通用对象

2 个答案: