Question

在下面的示例中，我正在发出网络请求以加载不同的电影类型，然后使用该请求来加载所有电影。接收器仅返回电影结果。我如何同时接收流派和电影？

struct Genre: Codable, Identifiable{
    let id: Int
    let name: String
    var movies: [Movie]?
}

struct Movie: Codable, Hashable, Identifiable {
    let title: String
    let id: Int
    let posterPath: String?
    let backdropPath : String?
    var tagline: String?
}

loadGenres() is AnyPublisher<[Genre], Error> 
fetchMoviesIn() is AnyPublisher<[Movie], Error>

class GenresViewModel: ObservableObject{
    @Published var genres = [Genre]()
    @Published var movies = [Movie]()
    var requests = Set<AnyCancellable>()
    
    init(){
        NetworkManager.shared.loadGenres()
            .flatMap{ genres in
                genres.publisher.flatMap{ genre in
                    NetworkManager.shared.fetchMoviesIn(genre)
                }
            }
            .collect()
            .retry(1)
            .receive(on: DispatchQueue.main)
            .sink(receiveCompletion: { completion in
                switch completion{
                case .finished:
                    print("Finished loading all movies in every genre")
                case .failure(let error):
                    print("Error: \(error)")
                }
            }, receiveValue: { [self] values in
                let allMovies = values.joined()
                self.movies = allMovies.map{$0}
            })
            .store(in: &self.requests)
    }
}

Answer 1

取决于您要收集的流派和电影的方式。

例如，您是否想要一种类型和该类型的电影列表？结果可能是(Genre, [Movies])的数组。

NetworkManager.shared.loadGenres()
   .flatMap { genres in
       genres.publisher.setFailureType(to: Error.self)
   }
   .flatMap { genre in
       NetworkManager.shared.fetchMoviesIn(genre)
          .map { movies in (genre, movies) } 
   }
   .collect()

或者，如果您想要一个(Genre, Movie)元组的数组，则这是一种类似的方法，但是需要另外一层.flatMap来获取单个电影

NetworkManager.shared.loadGenres()
   .flatMap { genres in
       genres.publisher.setFailureType(to: Error.self)
   }
   .flatMap { genre in
       NetworkManager.shared.fetchMoviesIn(genre)
          .flatMap { movies in
              movies.publisher.setFailureType(to: Error.self)
          }
          .map { movie in (genre, movie) }
   }
   .collect()

要回答您的评论问题，您想返回更新的Genre，可以返回而不是返回元组。请记住，由于Genre是一个结构，因此您需要创建对象的变量副本（genre闭包中可用的flatMap是常量），请更新副本，并返回：

NetworkManager.shared.loadGenres()
   .flatMap { genres in
       genres.publisher.setFailureType(to: Error.self)
   }
   .flatMap { genre in
       NetworkManager.shared.fetchMoviesIn(genre)
          .map { movies -> Genre in
             var genreCopy = genre
             genreCopy.movies = movies
             return genreCopy
          }
   }
   .collect()

链接快速合并发布者并接收每个结果

1 个答案: