寻找一些指导,以一种简单的方法在SwiftUI中从导航堆栈弹出多个视图。
我使用NavigationLink将4个视图链接在一起。在最后一个视图中,我想跳回到初始的ContentView,将所有其他视图弹出堆栈。我不想使用每个视图的NavigationBar上的“后退”按钮来实现此目的。
提前致谢。
鲍勃
'''
import SwiftUI
struct ContentView: View {
var body: some View {
NavigationView {
VStack {
NavigationLink(destination: BView()) {
Text("This is View A, now go to View B.")
}
}
}
}
}
struct BView: View {
var body: some View {
NavigationLink(destination: CView()) {
Text("This is View B, now go to View C.")
}
}
}
struct CView: View {
var body: some View {
NavigationLink(destination: DView()) {
Text("This is View C, now go to View D.")
}
}
}
struct DView: View {
var body: some View {
// The following line adds ContentView onto the existing navigation stack. Instead, I want to pop the previous views off the stack, leaving me back at ContentView.
NavigationLink(destination: ContentView()) {
Text("This is View D, now jump back to View A.")
}
}
}
'''
答案 0 :(得分:1)
这并不是真正从堆栈中“弹出”视图,但是您的SceneDelegate
可以将rootViewController
设置为所需的任何视图(请参见默认SceneDelegate.swift
的第28行)。就您而言,您希望它再次成为ContentView
。
例如,在您的SceneDelegate
中添加以下内容:
func toContentView() {
let contentView = ContentView()
window?.rootViewController = UIHostingController(rootView: contentView)
}
然后在DView
中,将NavigationLink
更改为仅能执行以下操作的Button
:
(UIApplication.shared.connectedScenes.first?.delegate as? SceneDelegate)?.toContentView()
答案 1 :(得分:1)
让Cenk Bilgen的答案更通用。
struct RootView {
static func change(to view: AnyView) {
guard let windowScene = UIApplication.shared.connectedScenes.first as? UIWindowScene,
let sceneDelegate = windowScene.delegate as? SceneDelegate else {
return
}
let contentView = view
sceneDelegate.window?.rootViewController = UIHostingController(rootView: contentView)
}
}
用法:
RootView.change(to: AnyView(DashboardView()))