从父类继承，扩展Typegoose并包含静态类型特定的方法

Question

我正在使用TypeScript和Node开发服务器，并使用Typegoose库将类映射到MongoDB文档。

我有以下两个类：

import { prop, getModelForClass, DocumentType, ReturnModelType, Typegoose } from '@typegoose/typegoose';

export default class Mode {
    // ...various attributes
    @prop() name?: string;
    private document?: DocumentType<Mode>;

    public get id(this: Mode): number {
        if (this.document)
            return this.document._id;
        else throw new Error('Looking for id on non-mapped Mode on database');
    }

    private static get model(): ReturnModelType<typeof Mode> {
        return getModelForClass(Mode);
    }

    private static attachDocument(document: DocumentType<Mode> | null): Mode | null {
        const instance: Mode | null = document as Mode | null;
        if (instance && document)
            instance.document = document;
        return instance;
    }
    // other methods...
}

import { prop, getModelForClass, DocumentType, ReturnModelType, Typegoose } from '@typegoose/typegoose';

export default class Player {
    // ...various attributes
    @prop() nickname: string;
    private document?: DocumentType<Player>;

    public get id(this: Player): number {
        if (this.document)
            return this.document._id;
        else throw new Error('Looking for id on non-mapped Player on database');
    }

    private static get model(): ReturnModelType<typeof Player> {
        return getModelForClass(Player);
    }

    private static query(document: DocumentType<Player> | null): Player | null {
        const instance: Player | null = document as Player | null;
        if (instance && document)
            instance.document = document;
        return instance;
    }
    // other methods...
}

很容易注意到有类似定义的方法：id()、model()和attachDocument()。我需要这些方法来将Typegoose的行为和查询的执行抽象到服务器的其余部分。有没有一种方法可以定义一个超类，使得这三个方法可以从Mode和Player中移除，并由它们从这个超类继承？

我在想这样的事情：

export default class Model<T> extends Typegoose {
    protected document: DocumentType<T>;

    public get id(this: T): number {
        if (this.document)
            return this.document._id;
        else throw new Error('Looking for id on non-mapped object on database');
    }

    private static get model(): ReturnModelType<typeof T> {
        return getModelForClass(T);
    }

    private static attachDocument(document: DocumentType<T> | null): T | null {
        const instance: T | null = document as T | null;
        if (instance && document)
            instance.document = document;
        return instance;
    }
}

export class Player extends Model<Player> { /* ... */ }
export class Mode extends Model<Mode> { /* ... */ }

然而，这似乎是不可行的，因为我不能将T作为参数传递给getModelForClass()方法。我已经看到它接受new () => T类型的参数，但是我还没有找到正确使用它的任何方法。

Answer 1

如果get model()不需要是private static方法，您可以尝试这样做：

class Model<T> extends Object {

    public get model(this: T): T {
        return this;
    }
}

class Player extends Model<Player> { }
class Topic extends Model<Topic> { }

const player = new Player();
const topic = new Topic();

console.log(player.model instanceof Player);
console.log(topic.model instanceof Topic)