import SwiftUI
struct ContentView: View {
@State var showSecond = false
@State var showThird = false
var body: some View {
VStack(spacing: 50) {
Text("FirstView")
Button("to SecondView") {
self.showSecond = true
}
.sheet(isPresented: $showSecond) {
VStack(spacing: 50) {
Text("SecondView")
Button("to ThirdView") {
self.showThird = true
}
.sheet(isPresented: self.$showThird) {
VStack(spacing: 50) {
Text("ThirdView")
Button("back") {
self.showThird = false
}
Button("back to FirstView") {
self.showThird = false
self.showSecond = false
}
}
}
Button("back") {
self.showSecond = false
}
}
}
}
}
}
struct ContentView_Previews: PreviewProvider {
static var previews: some View {
ContentView()
}
}
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上面的代码从FirstView过渡到SecondView,从SecondView过渡到ThirdView。并且SecondView 和ThirdView 中的“后退”按钮正常返回上一屏幕。
但是,如果您点击 ThirdView 中的“返回 FirstView”按钮,则将显示 SecondView,而不会返回 FirstView。而在这个操作之后,当你点击SecondView的“返回”按钮时,它不会返回到FirstView。
如何直接从 ThirdView 返回到 FirstView?
2020 年 2 月 19 日添加
我已经根据答案添加了解决方案代码。
解决方案 1:基于 Asperi 的计划 A。
struct ContentView: View {
@State var showSecond = false
@State var showThird = false
var body: some View {
VStack(spacing: 50) {
Text("FirstView")
Button("to SecondView") {
self.showSecond = true
}
.sheet(isPresented: $showSecond) {
VStack(spacing: 50) {
Text("SecondView")
Button("to ThirdView") {
self.showThird = true
}
.sheet(isPresented: self.$showThird) {
VStack(spacing: 50) {
Text("ThirdView")
Button("back") {
self.showThird = false
}
Button("back to FirstView") {
DispatchQueue.main.async {
self.showThird = false
DispatchQueue.main.async {
self.showSecond = false
}
}
}
}
}
Button("back") {
self.showSecond = false
}
}
}
}
}
}
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解决方案2:基于Asperi的plan B。
struct ContentView: View {
@State var showSecond = false
@State var showThird = false
@State var backToFirst = false
var body: some View {
VStack(spacing: 50) {
Text("FirstView")
Button("to SecondView") {
self.showSecond = true
}
.sheet(isPresented: $showSecond) {
VStack(spacing: 50) {
Text("SecondView")
Button("to ThirdView") {
self.showThird = true
}
.sheet(isPresented: self.$showThird, onDismiss: {
if self.backToFirst {
self.showSecond = false
}
}) {
VStack(spacing: 50) {
Text("ThirdView")
Button("back") {
self.showThird = false
self.backToFirst = false
}
Button("back to FirstView") {
self.showThird = false
self.backToFirst = true
}
}
}
Button("back") {
self.showSecond = false
}
}
}
}
}
}
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解决方案3:基于约瑟夫的建议。
struct ContentView: View {
@State var showSecond = false
@State var showThird = false
var body: some View {
GeometryReader { geometry in
ZStack {
VStack(spacing: 50) {
Text("FirstView")
Button("to SecondView") {
self.showSecond = true
}
}
.frame(width: geometry.size.width, height: geometry.size.height)
.background(Rectangle().foregroundColor(.white))
if self.showSecond {
VStack(spacing: 50) {
Text("SecondView")
Button("to ThirdView") {
self.showThird = true
}
Button("back") {
self.showSecond = false
}
}
.frame(width: geometry.size.width, height: geometry.size.height)
.background(Rectangle().foregroundColor(.white))
if self.showThird {
VStack(spacing: 50) {
Text("ThirdView")
Button("back") {
self.showThird = false
}
Button("back to FirstView") {
self.showThird = false
self.showSecond = false
}
}
.frame(width: geometry.size.width, height: geometry.size.height)
.background(Rectangle().foregroundColor(.white))
}
}
}
}
}
}
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现在根本没有办法(IMO 永远不会 - 这是两个模态会话)...我发现了两种可能值得考虑的可能方法:
A. 从一处顺序关闭
Button("back to FirstView") {
DispatchQueue.main.async {
self.showThird = false
DispatchQueue.main.async {
self.showSecond = false
}
}
}
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B.从不同地方顺序关闭
.sheet(isPresented: self.$showThird, onDismiss: {
self.showSecond = false // with some additional condition for forced back
})...
...
Button("back to FirstView") {
self.showThird = false
}
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