关闭包含在 UIHostingController 中的 SwiftUI 视图

我已将我的登录视图控制器重写为 SwiftUI View。SignInView包装在UIHostingController子类 (final class SignInViewController: UIHostingController<SignInView> {} ( 中，并在需要登录时以模式全屏显示。

一切正常，除了我不知道如何从SignInView中消除SignInViewController.我尝试添加：

@Environment(.isPresented) var isPresented

SignInView，并在登录成功时将其分配给false，但这似乎不会与 UIKit 互操作。如何关闭视图？

我发现了另一种似乎效果很好的方法，并且感觉比其他一些方法更干净。步骤：

将dismissAction属性添加到 SwiftUI 视图：

struct SettingsUIView: View {
    var dismissAction: (() -> Void)
    ...
}

当您想要关闭视图时，请调用dismissAction：

Button(action: dismissAction ) {
    Text("Done")
}

呈现视图时，请为其提供解除处理程序：

let settingsView = SettingsUIView(dismissAction: {self.dismiss( animated: true, completion: nil )})
let settingsViewController = UIHostingController(rootView: settingsView )
present( settingsViewController, animated: true )

更新：来自iOS 15 beta 1的发行说明：

是呈现的、演示模式和新的"消除操作"将关闭从 UIKit 呈现的托管控制器。(52556186(

<小时 />

我最终找到了一个比所提供的更简单的解决方案：


final class SettingsViewController: UIHostingController<SettingsView> {
    required init?(coder: NSCoder) {
        super.init(coder: coder, rootView: SettingsView())
        rootView.dismiss = dismiss
    }
    func dismiss() {
        dismiss(animated: true, completion: nil)
    }
}
struct SettingsView: View {
    var dismiss: (() -> Void)?
    
    var body: some View {
        NavigationView {
            Form {
                Section {
                    Button("Dimiss", action: dismiss!)
                }
            }
            .navigationBarTitle("Settings")
        }
    }
}

这里提供的所有答案都对我不起作用，可能是因为一些弱参考。这是我想出的解决方案：

创建视图和 UIHostingController：

let delegate = SheetDismisserProtocol()
let signInView = SignInView(delegate: delegate)
let host = UIHostingController(rootView: AnyView(signInView))
delegate.host = host
// Present the host modally

表消解器协议：

class SheetDismisserProtocol: ObservableObject {
    weak var host: UIHostingController<AnyView>? = nil
    func dismiss() {
        host?.dismiss(animated: true)
    }
}

必须驳回的观点：

struct SignInView: View {
    @ObservedObject var delegate: SheetDismisserProtocol
    var body: some View {
        Button(action: {
            self.delegate.dismiss()
        })
    }
}

另一种方法(在我看来相对容易(是在SwiftUI view中有一个可选的属性类型UIViewController，然后将其设置为视图控制器，该控制器将显示将包装您的SwiftUI视图的UIHostingController。

一个简单的设置视图：

struct SettingsView: View {
    
    var presentingVC: UIViewController?
    
    var body: some View {
        Button(action: {
            self.presentingVC?.presentedViewController?.dismiss(animated: true)
        }) {
            Text("Dismiss")
        }
    }
}

然后，当您使用 UIHostingController 从视图控制器呈现此视图时：

class ViewController: UIViewController {
    private func presentSettingsView() {
        var view = SettingsView()
        view.presentingVC = self
        let hostingVC = UIHostingController(rootView: view)
        present(hostingVC, animated: true, completion: nil)
    }
}

现在，正如您在SettingsView中Button的操作中看到的那样，我们将与ViewController交谈以关闭它所呈现的视图控制器，在我们的例子中，这将是包装SettingsView UIHostingController。

你可以只使用通知。

斯威夫特 5.1

在 SwiftUI 按钮处理程序中：

NotificationCenter.default.post(name: NSNotification.Name("dismissSwiftUI"), object: nil)

在 UIKit 视图控制器中：

NotificationCenter.default.addObserver(forName: NSNotification.Name("dismissSwiftUI"), object: nil, queue: nil) { (_) in
    hostingVC.dismiss(animated: true, completion: nil)
}

let rootView = SignInView();
let ctrl = UIHostingController(rootView: rootView);
ctrl.rootView.dismiss = {[weak ctrl] in
    ctrl?.dismiss(animated: true)
}
present(ctrl, animated:true, completion:nil);

注意：ctrl.rootView.dismiss而不是rootView.dismiss。

使用托管控制器演示器扩展环境值怎么样？它允许像presentationMode一样使用，从层次结构中的任何视图，并且易于重用和可扩展。定义新的环境值：

struct UIHostingControllerPresenter {
    init(_ hostingControllerPresenter: UIViewController) {
        self.hostingControllerPresenter = hostingControllerPresenter
    }
    private unowned var hostingControllerPresenter: UIViewController
    func dismiss() {
        if let presentedViewController = hostingControllerPresenter.presentedViewController, !presentedViewController.isBeingDismissed { // otherwise an ancestor dismisses hostingControllerPresenter - which we don't want.
            hostingControllerPresenter.dismiss(animated: true, completion: nil)
        }
    }
}
private enum UIHostingControllerPresenterEnvironmentKey: EnvironmentKey {
    static let defaultValue: UIHostingControllerPresenter? = nil
}
extension EnvironmentValues {
    /// An environment value that attempts to extend `presentationMode` for case where
    /// view is presented via `UIHostingController` so dismissal through
    /// `presentationMode` doesn't work.
    var uiHostingControllerPresenter: UIHostingControllerPresenter? {
        get { self[UIHostingControllerPresenterEnvironmentKey.self] }
        set { self[UIHostingControllerPresenterEnvironmentKey.self] = newValue }
    }
}

然后在需要时传递该值，例如：

let view = AnySwiftUIView().environment(.uiHostingControllerPresenter, UIHostingControllerPresenter(self))
let viewController = UIHostingController(rootView: view)
present(viewController, animated: true, completion: nil)
...

并享受使用

@Environment(.uiHostingControllerPresenter) private var uiHostingControllerPresenter
...
uiHostingControllerPresenter?.dismiss()

你以其他方式去的地方

@Environment(.presentationMode) private var presentationMode
...
presentationMode.wrappedValue.dismiss() // .isPresented = false

我遇到了同样的问题，多亏了这篇文章，我可以写一个混合解决方案，以提高这篇文章解决方案的可用性：

final class RootViewController<Content: View>: UIHostingController<AnyView> {
    init(rootView: Content) {
        let dismisser = ControllerDismisser()
        let view = rootView
            .environmentObject(dismisser)
        super.init(rootView: AnyView(view))
        dismisser.host = self
    }
    @objc required dynamic init?(coder aDecoder: NSCoder) {
        super.init(coder: aDecoder)
    }
}
final class ControllerDismisser: ObservableObject {
    var host: UIHostingController<AnyView>?
    func dismiss() {
        host?.dismiss(animated: true)
    }
}

这样，我可以将此控制器初始化为普通的UIHosting控制器

let screen = RootViewController(rootView: MyView())

注意：我使用.environmentObject将对象传递给需要它的视图。这样就无需将其放入初始值设定项中，也无需通过所有视图层次结构传递它

这是 Xcode 12 中的一个错误(很可能也是 Xcode 的早期版本(。它已在 Xcode 13.0 beta 5 中得到解决，并希望在 Xcode 13.0 的稳定版本中继续解决。也就是说，如果您能够使用 Xcode 13 构建并面向 iOS 15(或更高版本(，则首选 EnvironmentValues.dismiss 属性而不是已弃用的 EnvironmentValues.presentationMode 属性，如下所示：

struct MyView: View {
    
    @Environment(.dismiss) var dismiss
    
    var body: some View {
        Button("Dismiss") { dismiss() }
    }
}

如果您无法使用 Xcode 13 构建并面向 iOS 15，请选择此线程中提出的解决方法之一。

我不确定isPresented是否会在未来的版本中连接到View的UIHostingController。您应该提交有关它的反馈。

同时，请参阅此答案，了解如何从View访问UIViewController。

然后，你可以做self.viewController?.dismiss(...).

我遇到了类似的问题，提出了UIDocumentPickerViewController的实例。

在这种情况下，UIDocumentPickerViewController 以模态方式呈现 ( sheet (，它与您的略有不同 - 但该方法也可能适合您。

我可以通过遵守 UIViewControllerRepresentable 协议并添加一个回调来关闭Coordinator内的视图控制器来使其工作。

代码示例：

SwiftUI Beta 5

struct ContentProviderButton: View {
    @State private var isPresented = false
    var body: some View {
        Button(action: {
            self.isPresented = true
        }) {
            Image(systemName: "folder").scaledToFit()
        }.sheet(isPresented: $isPresented) { () -> DocumentPickerViewController in
            DocumentPickerViewController.init(onDismiss: {
                self.isPresented = false
            })
        }
    }
}
/// Wrapper around the `UIDocumentPickerViewController`.
struct DocumentPickerViewController {
    private let supportedTypes: [String] = ["public.image"]
    // Callback to be executed when users close the document picker.
    private let onDismiss: () -> Void
    init(onDismiss: @escaping () -> Void) {
        self.onDismiss = onDismiss
    }
}
// MARK: - UIViewControllerRepresentable
extension DocumentPickerViewController: UIViewControllerRepresentable {
    typealias UIViewControllerType = UIDocumentPickerViewController
    func makeUIViewController(context: Context) -> DocumentPickerViewController.UIViewControllerType {
        let documentPickerController = UIDocumentPickerViewController(documentTypes: supportedTypes, in: .import)
        documentPickerController.allowsMultipleSelection = true
        documentPickerController.delegate = context.coordinator
        return documentPickerController
    }
    func updateUIViewController(_ uiViewController: DocumentPickerViewController.UIViewControllerType, context: Context) {}
    // MARK: Coordinator
    func makeCoordinator() -> Coordinator {
        Coordinator(self)
    }
    class Coordinator: NSObject, UIDocumentPickerDelegate {
        var parent: DocumentPickerViewController
        init(_ documentPickerController: DocumentPickerViewController) {
            parent = documentPickerController
        }
        func documentPicker(_ controller: UIDocumentPickerViewController, didPickDocumentsAt urls: [URL]) {
            // TODO: handle user selection
        }
        func documentPickerWasCancelled(_ controller: UIDocumentPickerViewController) {
            parent.onDismiss()
        }
    }
}

我相信

你可以直接使用环境变量来消除：

@Environment(.presentationMode) var presentationMode
var body: some View {
    Button("Dismiss") {
        presentationMode.wrappedValue.dismiss()
    }
}

iOS 15 及以上

版本

struct MyView: View {
@Environment(.dismiss) var dismiss
    var body: some View {
        NavigationView {
            Text("Hello World")
                .toolbar {
                    ToolbarItem(placement: .navigationBarTrailing) {
                        Button("Dismiss") {
                            dismiss()
                        }
                    }
                }
       }
   }
}

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