用ObservableObject修饰的协议

Question

我创建了一个符合JSONService的协议ObservableObject。

protocol JSONService {
    var screenModel: ScreenModel? { get set }
    func load(_ resourceName: String) async throws 
}

接下来，我创建了符合LocalService的ObservableObject，如下所示：

class LocalService: JSONService, ObservableObject {
    
    @Published var screenModel: ScreenModel?
    
    func load(_ resourceName: String) async throws {
        
        // some code
        
    }
    
}

现在，当我在视图(SwiftUI)中创建一个属性时，我会得到一个错误：

struct ContentView: View {
    
    @StateObject private var jsonService: any JSONService

Type 'any JSONService' cannot conform to 'ObservableObject'

如何在协议中使用@StateObject？

Answer 1

您只需将一致性添加到protocol。

protocol JSONService: ObservableObject, AnyObject {
    var screenModel: ScreenModel? { get set }
    func load(_ resourceName: String) async throws
}

然后，可以将泛型添加到View中。

struct GenericOOView<JS: JSONService>: View {
    @StateObject private var jsonService: JS
    
    init(jsonService: JS){
        _jsonService = StateObject(wrappedValue: jsonService)
    }

    var body: some View {
        Text("Hello, World!")
    }
}