是否有一种方法可以在特定页面上禁用Android后退按钮?
class WakeUpApp extends StatelessWidget {
@override
Widget build(BuildContext context) {
return new MaterialApp(
title: "Time To Wake Up ?",
home: new WakeUpHome(),
routes: <String, WidgetBuilder>{
'/pageOne': (BuildContext context) => new pageOne(),
'/pageTwo': (BuildContext context) => new pageTwo(),
},
);
}
}
在第一页，我有一个按钮去第二页:
new FloatingActionButton(
onPressed: () {
Navigator.of(context).pushNamed('/pageTwo');
},
)
我的问题是，如果我按下android屏幕底部的后退箭头，我就会回到pageOne。我想这个按钮不显示在所有。
理想情况下，除非用户的手指持续按在屏幕上5秒钟，否则我不希望用户有任何办法离开屏幕。(我正在尝试为幼儿编写一个应用程序，希望只有父母能够导航出特定的屏幕)。
如果你需要系统返回按钮点击和应用程序栏返回按钮点击有不同的行为:你可以在调用Navigator.of(context).pop()之前删除onWillPop回调:
@override
Widget build(BuildContext context) {
return WillPopScope(
onWillPop: onWillPop,
child: Scaffold(
appBar: AppBar(
leading: IconButton(
onPressed: () {
ModalRoute.of(context)?.removeScopedWillPopCallback(onWillPop);
Navigator.of(context).pop();
},
icon: const Icon(Icons.arrow_back),
),
title: Text(context.l10n.searchResults),
),
body: MyBody(),
),
);
}
在这种情况下，当用户点击系统返回按钮时，onWillPop回调将决定是否弹出屏幕。但当用户点击应用栏后退按钮时，屏幕会立即弹出。
我把这个贴在这里，以防有人发现这个，并希望他们能找到一个简单的例子
https://gist.github.com/b-cancel/0ca372017a25f0c120b14dfca3591aa5
import 'package:flutter/material.dart';
import 'dart:async';
void main() => runApp(new BackButtonOverrideDemoWidget());
class BackButtonOverrideDemoWidget extends StatefulWidget{
@override
_BackButtonOverrideDemoWidgetState createState() => new _BackButtonOverrideDemoWidgetState();
}
class _BackButtonOverrideDemoWidgetState extends State<BackButtonOverrideDemoWidget> with WidgetsBindingObserver{
//-------------------------Test Variable
bool isBackButtonActivated = false;
//-------------------------Required For WidgetsBindingObserver
@override
void initState() {
super.initState();
WidgetsBinding.instance.addObserver(this);
}
@override
void dispose() {
WidgetsBinding.instance.removeObserver(this);
super.dispose();
}
//-------------------------Function That Triggers when you hit the back key
@override
didPopRoute(){
bool override;
if(isBackButtonActivated)
override = false;
else
override = true;
return new Future<bool>.value(override);
}
//-------------------------Build Method
@override
Widget build(BuildContext context) {
return new Directionality(
textDirection: TextDirection.ltr,
child: new Container(
color: (isBackButtonActivated) ? Colors.green : Colors.red,
child: new Center(
child: new FlatButton(
color: Colors.white,
onPressed: () {
isBackButtonActivated = !isBackButtonActivated;
setState(() {});
},
child: (isBackButtonActivated) ?
new Text("DeActive the Back Button") : new Text("Activate the Back Button"),
)
)
),
);
}
}
答案也许你知道使用WillPopScope，但不幸的是，在IOS上你不能滑动回上一页，所以让我们自定义你的MaterialPageRoute:
class CustomMaterialPageRoute<T> extends MaterialPageRoute<T> {
@protected
bool get hasScopedWillPopCallback {
return false;
}
CustomMaterialPageRoute({
@required WidgetBuilder builder,
RouteSettings settings,
bool maintainState = true,
bool fullscreenDialog = false,
}) : super(
builder: builder,
settings: settings,
maintainState: maintainState,
fullscreenDialog: fullscreenDialog,
);
}
现在你可以使用WillPopScope并向后滑动在IOS上工作。详细答案在这里:https://github.com/flutter/flutter/issues/14203#issuecomment-540663717
如果你需要系统返回按钮点击和应用程序栏返回按钮点击有不同的行为:你可以在调用Navigator.of(context).pop()之前删除onWillPop回调:
@override
Widget build(BuildContext context) {
return WillPopScope(
onWillPop: onWillPop,
child: Scaffold(
appBar: AppBar(
leading: IconButton(
onPressed: () {
ModalRoute.of(context)?.removeScopedWillPopCallback(onWillPop);
Navigator.of(context).pop();
},
icon: const Icon(Icons.arrow_back),
),
title: Text(context.l10n.searchResults),
),
body: MyBody(),
),
);
}
在这种情况下，当用户点击系统返回按钮时，onWillPop回调将决定是否弹出屏幕。但当用户点击应用栏后退按钮时，屏幕会立即弹出。
