在打字稿中使用泛型强制执行类型并进行后续处理

Question

I have an abstract service which is taking a repository in its constructor and getting data with Sequelize. The abstract service looks like this:

export abstract class BaseService<T> {

  public constructor(private _repo: any) {

  }
  public findById(id: string) {
    return this._repo.findById(id).pipe(
      map(res => Response.ok(res)))
    )
  }
}

where T is the repository type:

export class UsersInfosRepository extends BaseRepository<UserInfosModel>{
  public findById(id: string): Observable<UserInfosModel | null> {
    return from(UserInfosModel.findById(id));
  }
}

UserInfosModel is a sequelize @Model class.

Now this works fine, but the only thing is that I wish to enforce the type to

_repo

with a generic T rather than use any. The problem is that the compiler won't recognise T as UsersInfosRepository and won't find the method findId. Has anyone got any suggestion on how to enforce the generic?

Cheers

Answer 1

So in the end I found the solution

I changed the abstract BaseRepository into an interface:

export interface IBaseRepository<T> {
  findById(id: string): Observable<T | null>;
}

and edited the BaseService to accept:

export abstract class BaseService<T extends IBaseRepository<U>, U> 

Where T is any Repository and U is any model.

export class UsersService extends BaseService<UsersInfosRepository, UserInfosModel>

hopefully this will be of help to anyone who like me struggled to find a solution including generics and sequelize.

Answer 2

You want to use <T extends Class> :

export abstract class BaseService<T extends UsersInfosRepository> {

  public constructor(private _repo: T) {

  }
  public findById(id: string) {
    return this._repo.findById(id).pipe(
      map(res => Response.ok(res)))
    )
  }
}

For more info: https://www.typescriptlang.org/docs/handbook/generics.html