Question

我有此方法返回用户可见

  internal func getUser() -> Observable<User> {
    let result = withRealm("getting user") { realm -> Observable<User> in
      let realm = try Realm()
      let user = realm.objects(User.self).first
      return Observable.from(optional: user)
    }
    return result ?? .empty()
  }

现在，我想拥有另一种方法，如果用户登录，该方法将返回我，该怎么做？

  private var _isLoggedIn: Observable<User> {

    return getUser().count > 0 //error

  }

Answer 1

我建议改用枚举以摆脱可选项，以便于管理：

enum LoginState {
  case loggedIn(user: User)
  case loggedOut
}

internal func getLoginState() -> Observable<LoginState> {
  let user = withRealm("getting user") { realm -> User? in
    let realm = try Realm()
    return realm.objects(User.self).first
  }
  if let user = user {
    return Observable.just(.loggedIn(user))
  } else {
    return Observable.just(.loggedOut)
  }
}

private var _isLoggedIn: Observable<Bool> {
  return getLoginState().map {
    switch $0 {
    case .loggedIn: return true
    case .loggedOut: return false
    }
  }
}

如何检查rxswift中的Observable计数？

1 个答案: