链接swift组合发布者并接收每个结果

Question

在下面的例子中，我提出了一个网络请求来加载不同的电影类型，然后用它来加载所有的电影。接收器只返回电影结果。我怎么能同时收到这两种类型的电影？

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

取决于您希望如何收集流派和电影。

例如，你想要一个类型和该类型的电影列表吗？结果可能是CCD_ 1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，如果你想要一个(Genre, Movie)元组的数组，那么这是一个类似的方法，但需要额外的.flatMap级别来获得单个电影

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，您需要返回更新的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，您需要创建对象的变量副本(flatMap闭包中可用的genre是一个常量(，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()