链接 swift 合并发布者并接收每个结果

Question

在下面的示例中，我发出一个网络请求来加载不同的电影类型，然后使用它来加载所有电影。接收器只返回电影结果。我怎样才能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

取决于你想如何收集流派和电影。

例如，您是否想要一个流派和该流派的电影列表？结果可能是(Genre, [Movies]).

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，如果你想要一个(Genre, Movie)元组数组，那么它是一种类似的方法，但有一个额外的级别.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，您可以返回它而不是返回元组。请记住，由于Genre是一个结构体，您需要创建对象的变量副本（闭包中的genre可用变量flatMap是一个常量），更新副本，然后返回：

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()