结果与组合无法获得结果

Posted 2023-02-24

【中文标题】结果与组合无法获得结果

【英文标题】：Result with combine can't get results

【发布时间】：2021-04-04 06:02:37

【问题描述】：

我写的一切都是正确的，代码本身正在运行，但它给了我一个错误调用“fetchPokemon()”的结果未使用，这可能是什么问题？

听到我的代码：ModelView 类

 import Foundation
 import Combine
 class NetworkManager: ObservableObject 

let baseuRL = "https://pokeapi.co/api/v2/pokemon"

@Published var pokemon: [Pokemon] = []
var error: Error?
var cancellables: Set<AnyCancellable> = []

func fetchPokemon() -> Future<[Pokemon], Error> 
    return Future<[Pokemon], Error>  promice in
        guard let url = URL(string: "\(self.baseuRL)") else 
            return promice(.failure(ApiError.unknowed))
        
        URLSession.shared.dataTaskPublisher(for: url)
            .tryMap  (data, response) -> Data in
                guard let http = response as? HTTPURLResponse,
                      http.statusCode == 200 else 
                    throw ApiError.responseError
                
                return data
            
            .decode(type: PokemonList.self, decoder: JSONDecoder())
            .receive(on: RunLoop.main)
            .sink(receiveCompletion:  completion in
                switch completion 
                case .finished:
                    break
                case .failure(let error):
                    print(error)
                
            , receiveValue: 
                promice(.success($0.results))
            )
            .store(in: &self.cancellables)
    



 struct ContentView: View 
@StateObject var net = NetworkManager()
var body: some View 
    List(net.pokemon, id: \.self)  pokemon in
        Text(pokemon.name)
    .onAppear 
        net.fetchPokemon()
    

【参考方案1】：

您的 fetchPokemon 函数返回一个 Future，但您没有对它做任何事情 - 这就是您收到未使用错误的原因。

此外，在该函数中，您正在返回您的承诺，但没有对结果做任何事情。因此，您需要处理 Future 并对这些结果进行处理。

它可能看起来像下面这样：

class NetworkManager: ObservableObject 
    
    let baseuRL = "https://pokeapi.co/api/v2/pokemon"
    
    @Published var pokemon: [Pokemon] = []
    var error: Error?
    var cancellables: Set<AnyCancellable> = []
    
    //New function here:
    func runFetch() 
        fetchPokemon().sink  (completion) in
            //handle completion, error
         receiveValue:  (pokemon) in
            self.pokemon = pokemon //do something with the results from your promise
        .store(in: &cancellables)
    
    
    private func fetchPokemon() -> Future<[Pokemon], Error> 
        return Future<[Pokemon], Error>  promice in
            guard let url = URL(string: "\(self.baseuRL)") else 
                return promice(.failure(ApiError.unknowed))
            
            URLSession.shared.dataTaskPublisher(for: url)
                .tryMap  (data, response) -> Data in
                    guard let http = response as? HTTPURLResponse,
                          http.statusCode == 200 else 
                        throw ApiError.responseError
                    
                    return data
                
                .decode(type: PokemonList.self, decoder: JSONDecoder())
                .receive(on: RunLoop.main)
                .sink(receiveCompletion:  completion in
                    switch completion 
                    case .finished:
                        break
                    case .failure(let error):
                        print(error)
                    
                , receiveValue: 
                    promice(.success($0.results))
                )
                .store(in: &self.cancellables)
        
    



struct ContentView: View 
    @StateObject var net = NetworkManager()
    var body: some View 
        List(net.pokemon, id: \.self)  pokemon in
            Text(pokemon.name)
        .onAppear 
            net.runFetch() //call runFetch instead of fetchPokemon
        
    

由于您没有包含 PokemonList 的代码，我对其内容做了一个假设：

struct PokemonList: Codable 
    var results: [Pokemon]

如果类型不同，您必须将receiveValue 中发生的内容更改为runFetch。

【讨论】：

谢谢，一切正常，但我有点不明白为什么不可能在一个函数中做到这一点，事实证明你需要在一个函数中创建一个会话并在秒

您可以一次性完成所有操作。您可以将接收器和存储连接到您最初拥有的末尾，而不是返回 Future。不过，它看起来有点乱。