SwiftUI中Menu未关闭时点击Button无法弹出sheet的问题及解决需求

问题详情 #

开发SwiftUI视图时遇到异常：视图包含Menu组件和触发Sheet弹窗的Button，当Menu处于展开状态时快速点击按钮，Sheet无法弹出，且后续点击该按钮也完全失效。控制台日志报错：

Attempt to present <TtGC7SwiftUI29PresentationHostingControllerVS_7AnyView: 0x113811a00> on <TtGC7SwiftUI19UIHostingControllerGVS_15ModifiedContentVS_7AnyViewVS_12RootModifier_: 0x112825600> (from <TtGC7SwiftUI19UIHostingControllerGVS_15ModifiedContentVS_7AnyViewVS_12RootModifier_: 0x112825600>) which is already presenting <_UIContextMenuActionsOnlyViewController: 0x10c1183c0>.

仅快速点击会触发该问题，长按按钮则正常。测试环境为Xcode 15，问题覆盖iOS 15、16、17版本。

复现代码：

import SwiftUI

@main
struct menuAndButtonApp: App {
    var body: some Scene {
        WindowGroup {
            ContentView()
        }
    }
}

struct ContentView: View {
    @State var modalIsPresented = false

    var body: some View {
        HStack(spacing: 28) {
            Menu("Open Menu") {
                Text("Menu Item")
            }
            Button("Present Modal") {
                modalIsPresented = true
            }
        }.sheet(isPresented: $modalIsPresented) {
            Text("Here is a sheet")
        }
    }
}

目前已验证用DispatchQueue.main.asyncAfter(.now() + 0.2) { }包裹按钮操作可解决，但该方案需无条件延迟，无法精准判断Menu是否展开。

可行Workaround方案 #

方案1：基于UIKit控制器状态检测 #

通过UIViewControllerRepresentable获取底层宿主控制器，仅当无正在展示的控制器时触发Sheet，避免冲突：

import SwiftUI

struct PresentationChecker: UIViewControllerRepresentable {
    @Binding var isTriggering: Bool
    var onSafePresent: () -> Void
    
    func makeUIViewController(context: Context) -> UIViewController {
        let vc = UIViewController()
        vc.view.backgroundColor = .clear
        return vc
    }
    
    func updateUIViewController(_ uiViewController: UIViewController, context: Context) {
        guard isTriggering else { return }
        // 检查当前是否有正在弹出的控制器
        if uiViewController.presentingViewController == nil {
            onSafePresent()
        } else {
            isTriggering = false
        }
    }
}

struct ContentView: View {
    @State var modalIsPresented = false
    @State var triggerModal = false

    var body: some View {
        HStack(spacing: 28) {
            Menu("Open Menu") {
                Text("Menu Item")
            }
            Button("Present Modal") {
                triggerModal = true
            }
        }
        .overlay(PresentationChecker(isTriggering: $triggerModal) {
            modalIsPresented = true
        })
        .sheet(isPresented: $modalIsPresented) {
            Text("Here is a sheet")
        }
    }
}

方案2：追踪Menu展开状态 #

通过监听Menu的点击和消失事件，判断是否需要延迟触发Sheet：

struct ContentView: View {
    @State var modalIsPresented = false
    @State var isMenuExpanded = false

    var body: some View {
        HStack(spacing: 28) {
            Menu("Open Menu") {
                Text("Menu Item")
                    .onDisappear {
                        isMenuExpanded = false
                    }
            }
            .onTapGesture {
                isMenuExpanded = true
            }
            
            Button("Present Modal") {
                if isMenuExpanded {
                    DispatchQueue.main.asyncAfter(deadline: .now() + 0.15) {
                        modalIsPresented = true
                    }
                } else {
                    modalIsPresented = true
                }
            }
        }
        .sheet(isPresented: $modalIsPresented) {
            Text("Here is a sheet")
        }
    }
}

方案3：临时替换为Alert（应急用） #

若上述方案仍有问题，可临时用Alert替代Sheet，规避SwiftUI的Sheet与Menu的冲突：

struct ContentView: View {
    @State var showAlert = false

    var body: some View {
        HStack(spacing: 28) {
            Menu("Open Menu") {
                Text("Menu Item")
            }
            Button("Present Modal") {
                showAlert = true
            }
        }
        .alert(isPresented: $showAlert) {
            Alert(title: Text("Here is a sheet"), dismissButton: .default(Text("OK")))
        }
    }
}

内容的提问来源于stack exchange，提问作者Andrew Bennet