Question

交互式弹出手势识别器应允许用户在滑过屏幕的一半以上（或这些行周围的某物）时返回导航堆栈中的上一个视图。在SwiftUI中，滑动距离不够时手势不会被取消。

问题：

使用SwiftUI视图时能否获得UIKit行为？

尝试

我试图将UIHostingController嵌入UINavigationController内，但其行为与NavigationView完全相同。

struct ContentView: View {
    var body: some View {
        UIKitNavigationView {
            VStack {
                NavigationLink(destination: Text("Detail")) {
                    Text("SwiftUI")
                }
            }.navigationBarTitle("SwiftUI", displayMode: .inline)
        }.edgesIgnoringSafeArea(.top)
    }
}

struct UIKitNavigationView<Content: View>: UIViewControllerRepresentable {

    var content: () -> Content

    init(@ViewBuilder content: @escaping () -> Content) {
        self.content = content
    }

    func makeUIViewController(context: Context) -> UINavigationController {
        let host = UIHostingController(rootView: content())
        let nvc = UINavigationController(rootViewController: host)
        return nvc
    }

    func updateUIViewController(_ uiViewController: UINavigationController, context: Context) {}
}

Answer 1

我最终覆盖了默认的NavigationView和NavigationLink来获得所需的行为。这看起来是如此简单，以至于我必须忽略默认SwiftUI视图所做的事情？

NavigationView

我将UINavigationController包装在一个超级简单的UIViewControllerRepresentable中，该UINavigationController作为环境对象，将NavigationLink赋予SwiftUI内容视图。这意味着.edgesIgnoringSafeArea(.top)以后可以抢到它，只要它在我们想要的同一个导航控制器中（显示的视图控制器不接收environmentObjects）即可。

注意：NavigationView需要struct NavigationView<Content: View>: UIViewControllerRepresentable { var content: () -> Content init(@ViewBuilder content: @escaping () -> Content) { self.content = content } func makeUIViewController(context: Context) -> UINavigationController { let nvc = UINavigationController() let host = UIHostingController(rootView: content().environmentObject(nvc)) nvc.viewControllers = [host] return nvc } func updateUIViewController(_ uiViewController: UINavigationController, context: Context) {} } extension UINavigationController: ObservableObject {}，但我还不知道如何在结构本身中进行设置。如果您的nvc在顶部中断，请参见示例。

selection

NavigationLink

我创建了一个自定义NavigationLink，用于访问环境UINavigationController来推送托管下一个视图的UIHostingController。

注意：我没有实现SwiftUI.NavigationLink拥有的isActive和struct NavigationLink<Destination: View, Label:View>: View { var destination: Destination var label: () -> Label public init(destination: Destination, @ViewBuilder label: @escaping () -> Label) { self.destination = destination self.label = label } /// If this crashes, make sure you wrapped the NavigationLink in a NavigationView @EnvironmentObject var nvc: UINavigationController var body: some View { Button(action: { let rootView = self.destination.environmentObject(self.nvc) let hosted = UIHostingController(rootView: rootView) self.nvc.pushViewController(hosted, animated: true) }, label: label) } }，因为我还不完全了解它们的作用。如果您想提供帮助，请评论/编辑。

struct ContentView: View {
    @State var isPresented = false

    var body: some View {
        NavigationView {
            VStack(alignment: .center, spacing: 30) {
                NavigationLink(destination: Text("Detail"), label: {
                    Text("Show detail")
                })
                Button(action: {
                    self.isPresented.toggle()
                }, label: {
                    Text("Show modal")
                })
            }
            .navigationBarTitle("SwiftUI")
        }
        .edgesIgnoringSafeArea(.top)
        .sheet(isPresented: $isPresented) {
            Modal()
        }
    }
}

这解决了后向滑动无法在SwiftUI上正常工作的问题，并且因为我使用了NavigationView和NavigationLink这两个名称，所以我的整个项目都立即切换为这些。

示例

在示例中，我也显示了模态表示。

struct Modal: View {
    @Environment(\.presentationMode) var presentationMode

    var body: some View {
        NavigationView {
            VStack(alignment: .center, spacing: 30) {
                NavigationLink(destination: Text("Detail"), label: {
                    Text("Show detail")
                })
                Button(action: {
                    self.presentationMode.wrappedValue.dismiss()
                }, label: {
                    Text("Dismiss modal")
                })
            }
            .navigationBarTitle("Modal")
        }
    }
}

NavigationLink(destination: Text("Detail").environmentObject(objectToSendOnToTheNextView)) {
    Text("Show detail")
}

编辑：我从“这看起来很简单，我必须忽略某些东西”开始，我想我找到了它。这似乎没有将EnvironmentObjects转移到下一个视图。我不知道默认的NavigationLink是如何做到的，所以现在我手动将对象发送到需要它们的下一个视图。

NavigationView

编辑2：

这通过执行@EnvironmentObject var nvc: UINavigationController将导航控制器暴露给wsgi.py内部的所有视图。解决此问题的方法是使用于管理导航的environmentObject成为fileprivate类。我已将其修正为要点：https://gist.github.com/Amzd/67bfd4b8e41ec3f179486e13e9892eeb

Answer 2

您可以通过进入UIKit并使用自己的UINavigationController来实现。

首先创建一个SwipeNavigationController文件：

import UIKit
import SwiftUI

final class SwipeNavigationController: UINavigationController {

    // MARK: - Lifecycle

    override init(rootViewController: UIViewController) {
        super.init(rootViewController: rootViewController)
    }

    override init(nibName nibNameOrNil: String?, bundle nibBundleOrNil: Bundle?) {
        super.init(nibName: nibNameOrNil, bundle: nibBundleOrNil)

        delegate = self
    }

    required init?(coder aDecoder: NSCoder) {
        super.init(coder: aDecoder)

        delegate = self
    }

    override func viewDidLoad() {
        super.viewDidLoad()

        // This needs to be in here, not in init
        interactivePopGestureRecognizer?.delegate = self
    }

    deinit {
        delegate = nil
        interactivePopGestureRecognizer?.delegate = nil
    }

    // MARK: - Overrides

    override func pushViewController(_ viewController: UIViewController, animated: Bool) {
        duringPushAnimation = true

        super.pushViewController(viewController, animated: animated)
    }

    var duringPushAnimation = false

    // MARK: - Custom Functions

    func pushSwipeBackView<Content>(_ content: Content) where Content: View {
        let hostingController = SwipeBackHostingController(rootView: content)
        self.delegate = hostingController
        self.pushViewController(hostingController, animated: true)
    }

}

// MARK: - UINavigationControllerDelegate

extension SwipeNavigationController: UINavigationControllerDelegate {

    func navigationController(_ navigationController: UINavigationController, didShow viewController: UIViewController, animated: Bool) {
        guard let swipeNavigationController = navigationController as? SwipeNavigationController else { return }

        swipeNavigationController.duringPushAnimation = false
    }

}

// MARK: - UIGestureRecognizerDelegate

extension SwipeNavigationController: UIGestureRecognizerDelegate {

    func gestureRecognizerShouldBegin(_ gestureRecognizer: UIGestureRecognizer) -> Bool {
        guard gestureRecognizer == interactivePopGestureRecognizer else {
            return true // default value
        }

        // Disable pop gesture in two situations:
        // 1) when the pop animation is in progress
        // 2) when user swipes quickly a couple of times and animations don't have time to be performed
        let result = viewControllers.count > 1 && duringPushAnimation == false
        return result
    }
}

here与SwipeNavigationController相同，并附加了pushSwipeBackView()函数。

此功能需要一个SwipeBackHostingController，我们将其定义为

import SwiftUI

class SwipeBackHostingController<Content: View>: UIHostingController<Content>, UINavigationControllerDelegate {
    func navigationController(_ navigationController: UINavigationController, didShow viewController: UIViewController, animated: Bool) {
        guard let swipeNavigationController = navigationController as? SwipeNavigationController else { return }
        swipeNavigationController.duringPushAnimation = false
    }

    override func viewWillDisappear(_ animated: Bool) {
        super.viewWillDisappear(animated)

        guard let swipeNavigationController = navigationController as? SwipeNavigationController else { return }
        swipeNavigationController.delegate = nil
    }
}

然后，我们将应用的SceneDelegate设置为使用SwipeNavigationController：

    if let windowScene = scene as? UIWindowScene {
        let window = UIWindow(windowScene: windowScene)
        let hostingController = UIHostingController(rootView: ContentView())
        window.rootViewController = SwipeNavigationController(rootViewController: hostingController)
        self.window = window
        window.makeKeyAndVisible()
    }

最后在您的ContentView中使用它：

struct ContentView: View {
    func navController() -> SwipeNavigationController {
        return UIApplication.shared.windows[0].rootViewController! as! SwipeNavigationController
    }

    var body: some View {
        VStack {
            Text("SwiftUI")
                .onTapGesture {
                    self.navController().pushSwipeBackView(Text("Detail"))
            }
        }.onAppear {
            self.navController().navigationBar.topItem?.title = "Swift UI"
        }.edgesIgnoringSafeArea(.top)
    }
}

如何在SwiftUI中返回与UIKit（interactivePopGestureRecognizer）相同的行为

2 个答案:

NavigationView

NavigationLink

示例