关闭包含在 UIHostingController 中的 SwiftUI 视图

Question

我已将我的登录视图控制器重写为 SwiftUI View。 SignInView 包含在 UIHostingController 子类 (final class SignInViewController: UIHostingController<SignInView> {}) 中，并在需要登录时以全屏模式呈现。

一切正常，除了我不知道如何从 SignInView 中删除 SignInViewController。我尝试添加：

@Environment(\.isPresented) var isPresented

in SignInView 并在登录成功时将其分配给 false，但这似乎无法与 UIKit 互操作。我怎样才能关闭视图？

Answer 1

我不确定 isPresented 是否会在未来的版本中连接到 View 的 UIHostingController。你应该 submit feedback 了解一下。

同时，请参阅了解如何从您的 View 访问 UIViewController。

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

Answer 2

我在呈现 UIDocumentPickerViewController.

的实例时遇到了类似的问题

在这种情况下，UIDocumentPickerViewController 以模态方式呈现 (sheet)，这与您的稍有不同 -- 但该方法也可能适用于您。

我可以通过遵守 UIViewControllerRepresentable 协议并添加一个回调来关闭 Coordinator.

中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()
        }
    }
}

Answer 3

更新： 来自 iOS 15 beta 1 的发行说明：

isPresented, PresentationMode, and the new DismissAction action dismiss a hosting controller presented from UIKit. (52556186)

我最终找到了一个比所提供的解决方案简单得多的解决方案：


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")
        }
    }
}

Answer 4

您可以只使用通知。

Swift 5.1

在 SwiftUI 按钮处理程序中：

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

在UI套件视图控制器中：

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

Answer 5

这里提供的所有答案都不适合我，可能是因为某些参考资料不充分。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

SheetDismisserProtocol:

class SheetDismisserProtocol: ObservableObject {
    weak var host: UIHostingController<AnyView>? = nil

    func dismiss() {
        host?.dismiss(animated: true)
    }
}

不得不驳回的观点：

struct SignInView: View {
    @ObservedObject var delegate: SheetDismisserProtocol

    var body: some View {
        Button(action: {
            self.delegate.dismiss()
        })
    }
}

Answer 6

我发现另一种方法似乎很有效，而且感觉比其他一些方法更简洁。步骤：

添加一个 dismissAction 属性到 SwiftUI 视图：

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

当您想要关闭视图时调用 dismissAction：

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

当您呈现视图时，为其提供解雇处理程序：

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

present( settingsViewController, animated: true )

Answer 7

我遇到了同样的问题，感谢这个 post，我可以编写一个混合解决方案，以提高这个 post 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，我就可以将这个控制器初始化为普通的 UIHostingController

let screen = RootViewController(rootView: MyView())

注意：我使用 .environmentObject 将对象传递给我需要它的视图。这样就不需要把它放在初始化器中，或者通过所有的视图层次结构传递它

Answer 8

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的 UIViewController 然后将其设置为viewController 将显示 UIHostingController，它将包装您的 SwiftUI 视图。

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 对话以关闭它正在呈现的视图控制器，这在我们的case 将是包装 SettingsView.

的 UIHostingController

Answer 9

如何使用托管控制器演示器扩展环境值？它允许从层次结构的任何视图中像 presentationMode 一样使用，并且很容易重用和扩展。定义您的新环境值：

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 }
    }
}

然后在需要时传递值，例如：

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

享受使用

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

你去哪里

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

Answer 10

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

注意：ctrl.rootView.dismiss不是rootView.dismiss

Answer 11

这是 Xcode 12 中的错误（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 的稳定版本中继续得到解决。也就是说，如果您能够使用 Xcode 13 和目标 iOS 15（或更高）进行构建，那么更喜欢 EnvironmentValues.dismiss property over the deprecated 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 和目标 iOS 15 进行构建，请选择此线程中提出的解决方法之一。

关闭包含在 UIHostingController 中的 SwiftUI 视图

Dismiss a SwiftUI View that is contained in a UIHostingController

uikit

ios

swiftui