Typescript 返回基类中工厂方法的类型 [英] Typescript return type of factory method in base class

问题描述

来自 Python 背景，我在尝试使 Typescript 对以下继承模型满意时遇到了一些问题.

Coming from a Python background, I encountered a few problems while trying to make Typescript happy with the following inheritance model.

给定以下基类:

// model.ts
export class Model {

    serialize() {}
    static deserialize(dbRow) {}

    static async filter(options) {
        const dbRows = await getData(options);

        return dbRows.map(row => this.deserialize(row));
    }
}

以及以下扩展它的类:

// user.ts
export class User extends Model {
    serialize() {......}

    static deserialize(dbRow) {......return new User(......)}

    someOtherMethod() {.....}
}

我如何定义 Model.filter 以便 const users = await User.filter({.....}) 将自动推断用户类型为 User[]?现在，推断的类型是 Model[].

How could I define Model.filter so that const users = await User.filter({.....}) will automatically infer the type of users as User[]? Right now, the inferred type is Model[].

我知道我可以做到 const users = await User.filter({....}) 但我想知道是否有任何方法建议将 Model.filter 的返回类型作为类构造函数，以便正确推断子类的实例.>

I know I can do const users = <User[]> await User.filter({....}) but I wonder if there is any way of suggesting the return type of Model.filter as the class constructor so that instances of child classes will be inferred correctly.

推荐答案

您可以使用以下解决方法(在 这个线程):

You can use the following workaround (mentioned several times in this thread):

type ModelContructor<T> = new () => T;

class Model {
    // ...
    static async filter<T extends Model>(this: ModelContructor<T> & typeof Model, options): Promise<T[]> {
        const dbRows = await getData(options);

        return dbRows.map(row => this.deserialize(row));
    }
}

这里我们使用伪造的 this 参数，它强制 T 在 User.filter 时被解析为 User调用.

Here we use fake this parameter, which enforces T to be resolved as User when User.filter called.

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