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是否可以用仍然强制类型安全的方式编写泛型?

interface types design-patterns generics typescript

0

MusicDev · 技术社区 · 6 年前

假设我有多个具有类似方法的类。在本例中,一组具有saveModel方法的服务:

public async saveModel(newModel: IModel): Promise<IModel> {
    return await newModel.save();
}

我创建的通用方法如下所示:

public async saveModel<P extends Document>(newModel: P): Promise<P> {
    return await newModel.save();
}

我的所有服务(使用该方法的服务)都扩展了一个包含该泛型方法的类,所以现在在这些服务中,我可以调用 answerService.saveModel(newAnswer) 这样就省去了答案。所有的东西都很漂亮,直到我意识到我也可以在里面放一个IQuestion类型的东西。如果我愿意,还有一个IDinosaur类型的对象,只要它扩展了Mongoose文档接口。

我有没有办法强制每个服务中使用某些接口?如何确保answerService只保存IAnswer类型的对象?answerService是否需要实现一个具有以下saveModel签名的接口:

interface IAnswerService {
    saveModel(newModel: IAnswer): Promise<IAnswer>;
}

这是整件事的基本情况:

实体服务.ts

import { Document } from 'mongoose';
export class EntityService implements IEntityService {
  public async saveModel<P extends Document>(newModel: P): Promise<P> {
    try {
      const savedModel = await newModel.save();
      return savedModel;
    } catch (err) {
      console.log(err);
      return null;
    }
  }
}

应答服务

import { Answer, IAnswer } from '@models/answer.model';
import { EntityService } from '@classes/EntityService';

class AnswerService extends EntityService {
  private static instance: AnswerService;

  private constructor() {
    super();
  }

  static getInstance(): AnswerService {
    if (!AnswerService.instance) {
      AnswerService.instance = new AnswerService();
    }

    return AnswerService.instance;
  }
}

const answerService = AnswerService.getInstance();
export default answerService;

答案.model.ts

import mongoose, { Schema, Document, Model } from 'mongoose';

export interface IAnswer extends Document {
  answerText: string;
  posterID: string;
}

const AnswerSchema: Schema = new Schema({
  answerText: {type: String, required: true},
  posterID: {type: String, required: true}
}, {
  minimize: false
});

export const Answer: Model<IAnswer> = mongoose.model<IAnswer>('Answer', AnswerSchema);

所有其他接口都继承自IAnswer所做的相同事情(模型、模式、文档)。

AnswerService的用法通常如下所示:

import answerService from 'answer.service';
import { Answer } from 'answer.model';

const answer = new Answer({
    answerText: 'Stuff',
    posterID: '123456789'
})

answerService.saveModel(answer);

1 回复 | 直到 6 年前

1

jcalz 6 年前

据我所知,你想要 EntityService 处理的特定子类型 Document . 如果是,那你就不想 saveModel() 通用的 P extends Document ;你想要整个 实体服务 类的泛型 P . 以及 实体服务 应该指定泛型类型参数。例如:

// P moved to EntityService from saveModel    
export class EntityService<P extends Document> {
  public async saveModel(newModel: P): Promise<P> {
    try {
      const savedModel = await newModel.save();
      return savedModel;
    } catch (err) {
      console.log(err);
      return null!;
    }
  }
}

// subclass sets specific value for P   
class AnswerService extends EntityService<IAnswer> {
  private static instance: AnswerService;   
  private constructor() {
    super();
  }    
}

以后,如果你有 answer 类型 IAnswer 以及 document 只知道是类型 文件 ,然后是 AnswerService 只允许你 saveModel(answer) 而不是 saveModel(document) :

AnswerService.getInstance().saveModel(answer); // okay
AnswerService.getInstance().saveModel(document); // error, Document is not IAnswer

好吧,希望能帮上忙;祝你好运!

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