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

我有以下两个课程：

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()。我需要这些方法来抽象 Typegoose 的行为以及对服务器其余部分的查询执行。有没有办法定义一个超类，以便它们可以从这个超类中删除和继承这三个Mode方法Player？

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> { /* ... */ }

但是，这似乎不可行，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 type 的参数new () => T，但我还没有找到任何正确使用它的方法。

4

1 回答 1

-1

如果get model()不需要是一种private static方法，你可以试试这个：


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)
于 2020-03-21T02:17:43.887 回答