在 watchOS 中使用 environmentObject [英] Using environmentObject in watchOS

问题描述

我试图在 watchOS6 应用程序中使用 environmentObject 将我的数据模型绑定到我的视图.

I am trying to use environmentObject in a watchOS6 app to bind my data model to my view.

我在 Xcode 11 中创建了一个简单、独立的 Watch 应用.

I have created a simple, stand-alone Watch app in Xcode 11.

我创建了一个新的 DataModel 类

I created a new DataModel class

import Combine
import Foundation
import SwiftUI

final class DataModel: BindableObject {

    let didChange = PassthroughSubject<DataModel,Never>()

    var aString: String = "" {
        didSet {
            didChange.send(self)
        }
    }

}

在我的 ContentView 结构中，我使用 @EnvironmentObject -

In my ContentView struct I bind this class using @EnvironmentObject -

struct ContentView : View {

    @EnvironmentObject private var dataModel: DataModel

    var body: some View {
        Text($dataModel.aString.value)
    }
}

最后，我尝试将 DataModel 的实例注入到 HostingController 类中的环境中 -

Finally, I attempt to inject an instance of the DataModel into the environment in the HostingController class -

class HostingController : WKHostingController<ContentView> {
    override var body: ContentView {
        return ContentView().environmentObject(DataModel())
    }
}

但是，我收到一个错误:

But, I get an error:

Cannot convert return expression of type '_ModifiedContent<ContentView, _EnvironmentKeyWritingModifier<DataModel?>>' to return type 'ContentView'

错误是因为 WKHostingController 是一个泛型，需要一个具体的类型 - WKHostingController 在这种情况下.

The error is because the WKHostingController is a generic that needs a concrete type - WKHostingController<ContentView> in this case.

类似的方法与 iOS 应用中的 UIHostingController 完美配合，因为 UIHostingController 不是通用类.

A similar approach works perfectly with UIHostingController in an iOS app because UIHostingController isn't a generic class.

是否有其他方法可以将环境注入 watchOS 视图?

Is there some other way to inject the environment to a watchOS view?

推荐答案

你可以使用类型擦除，AnyView 在 SwiftUI View 的情况下.

You can use type erasure, AnyView in the case of SwiftUI View.

我会重构 WKHostingController 以返回 AnyView.

I would refactor WKHostingController to return AnyView.

这对我来说似乎编译得很好.

This seems to compile fine on my end.

class HostingController : WKHostingController<AnyView> {
    override var body: AnyView {
        return AnyView(ContentView().environmentObject(DataModel()))
    }
}

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