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typescript - 打字稿：可以用任意类型实例化

问题描述

您好，我正在尝试使用 ddd，但是在使用 ts 对数据库进行操作后，我在建模映射器以返回我的域的实例时遇到问题：

这是我的域名：

export interface ILocationProps {
  stret_address: string;
  cep_code: CepCode;
}
export class LocationSeats extends AggregateRoot<ILocationProps> {
  get StretAddress(): string {
    return this.props.stret_address;
  }
  get CepCode(): CepCode {
    return this.props.cep_code;
  }
  private constructor(props: ILocationProps, id?: UniqueEntityID) {
    super(props, id);
  }
  public static create(
    props: ILocationProps,
    id?: UniqueEntityID,
  ): Result<LocationSeats> {
    const guardedProps = [
      { argument: props.cep_code, argumentName: 'cep_code' },
      { argument: props.stret_address, argumentName: 'stret_address' },
    ];

    const guardResult = Guard.againstNullOrUndefinedBulk(guardedProps);

    if (!guardResult.succeeded) {
      return Result.fail<LocationSeats>(guardResult.message);
    }

    const location = new LocationSeats(props, id);

    const idWasProvided = !!id;

    if (!idWasProvided) {
      location.when(new LocationCreatedEvent(location));
    }

    return Result.ok<LocationSeats>(location);
  }
}

这是我的映射器：

export default interface IMapperr<T> {
  toPersistence(t: any): Result<T>;
  toDomain(raw: any): Result<T>;
}
@singleton()
export class LocationMapper<Location = LocationSeats> implements IMapper<Location> {
  constructor() {}
  public toPersistence(t: any) {
    throw new Error('Method not implemented.');
  }
  private validate() {}
  public toDomain(raw: any): Result<Location> {
    const cepCodeorError = CepCode.create(raw.cep_code);
    const locationOrError = LocationSeats.create(
      {
        cep_code: cepCodeorError.getValue(),
        stret_address: raw.street_address,
      },
      new UniqueEntityID(raw.id),
    );
    return locationOrError;
  }
}

这是我的结果类：

export class Result<T> {
  public isSuccess: boolean;

  public isFailure: boolean;

  public error?: T | string;

  private _value?: T;

  constructor(isSuccess: boolean, error?: T | string, value?: T) {
    if (isSuccess && error) {
      throw new Error(
        'InvalidOperation: A result cannot be successful and contain an error',
      );
    }
    if (!isSuccess && !error) {
      throw new Error(
        'InvalidOperation: A failing result needs to contain an error message',
      );
    }

    this.isSuccess = isSuccess;
    this.isFailure = !isSuccess;
    this.error = error;
    this._value = value;

    Object.freeze(this);
  }

  getValue(): T {
    if (!this.isSuccess) {
      console.log(this.error);

      throw new Error(
        "Can't get the value of an error result. Use 'errorValue' instead.",
      );
    }
    if (!this._value)
      throw new Error(
        "Can't get the value of an error result. Use 'errorValue' instead.",
      );
    return this._value;
  }

  errorValue(): T {
    return this.error as T;
  }

  static ok<U>(value?: U): Result<U> {
    return new Result<U>(true, undefined, value);
  }

  static fail<U>(error: any): Result<U> {
    return new Result<U>(false, error);
  }

  static combine(results: Result<any>[]): Result<any> {
    for (const result of results) {
      if (result.isFailure) return result;
    }
    return Result.ok();
  }
}

但我收到了这个错误：

在我的 toDomain 函数上：

 public toDomain(raw: any): Result<Location> {
    const cepCodeorError = CepCode.create(raw.cep_code);
    const locationOrError = LocationSeats.create(
      {
        cep_code: cepCodeorError.getValue(),
        stret_address: raw.street_address,
      },
      new UniqueEntityID(raw.id),
    );
    return locationOrError;
  }

错误：

Type 'Result<LocationSeats>' is not assignable to type 'Result<Location>'.
  Type 'LocationSeats' is not assignable to type 'Location'.
    'Location' could be instantiated with an arbitrary type which could be unrelated to 'LocationSeats'.ts(2322)

我不知道为什么我的创建函数返回一个类型：结果

标签： typescript

解决方案

问题是您已标记toDomain为返回 aLocation这是一个没有约束的模板参数，但随后您将其硬编码为 return LocationSeats.create。Typescript 不知道是否LocationSeats.create与您可能选择传递的任何泛型类型兼容Location。

设计对正在构建的类型通用的工厂函数的问题经常让刚接触 Typescript 的人感到困惑。原因通常是，与其他一些语言不同，泛型类型参数不能用作运行时对象，您可以在其上调用静态方法，例如您的create方法。它们在运行时被完全擦除。

在 Typescript 中，要创建通用工厂，您必须在运行时实际将构造函数引用传递给您的工厂。

像这样的东西可能接近你想要的（精简一下以显示我想要突出显示的部分）：

class LocationSeats extends AggregateRoot<ILocationProps> {
  private constructor (props: ILocationProps, id?: UniqueEntityID) {
    super(props, id)
  }

  static create (props: ILocationProps, id?: UniqueEntityID): Result<LocationSeats> {
    return Result.ok(new LocationSeats(props, id))
  }
}

// We need an explicit factory interface for creating these objects
interface LocationCreator<T> {
  create (props: ILocationProps, id?: UniqueEntityID): Result<T>
}

class LocationMapper<Location> {
  // Need to accept the Location creator at runtime (Location is inferred)
  constructor (private creator: LocationCreator<Location>) {}

  toDomain (): Result<Location> {
    return this.creator.create({
        cep_code: '',
        stret_address: '',
      })
  }
}

// Need to pass the LocationSeats constructor in at runtime
const mapper = new LocationMapper(LocationSeats)

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