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)。
答案是WillPopScope。它将防止页面被系统弹出。你仍然可以使用Navigator.of(context).pop()
@override
Widget build(BuildContext context) {
return new WillPopScope(
onWillPop: () async => false,
child: new Scaffold(
appBar: new AppBar(
title: new Text("data"),
leading: new IconButton(
icon: new Icon(Icons.ac_unit),
onPressed: () => Navigator.of(context).pop(),
),
),
),
);
}
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"),
)
)
),
);
}
}
虽然Remi的回答是正确的，但通常情况下，您不希望简单地阻止返回按钮，而是希望用户确认退出。
你可以用类似的方式从确认对话框中获取答案，因为onWillPop是一个未来。
@override
Widget build(BuildContext context) {
return WillPopScope(
child: Scaffold(...),
onWillPop: () => showDialog<bool>(
context: context,
builder: (c) => AlertDialog(
title: Text('Warning'),
content: Text('Do you really want to exit'),
actions: [
FlatButton(
child: Text('Yes'),
onPressed: () => Navigator.pop(c, true),
),
FlatButton(
child: Text('No'),
onPressed: () => Navigator.pop(c, false),
),
],
),
),
);
}
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
这里有一个替代的解决方案，如果你是用零安全编码。您需要禁用默认的后退按钮，并将其替换为IconButton。在本例中，当用户在退出前单击后退按钮确认时，我将按下AlertDialog。您可以替换此函数并将用户发送到任何其他页面
return WillPopScope(
onWillPop: () async => false,
child: Scaffold(
appBar: AppBar(
automaticallyImplyLeading: true,
title: Text(),
leading: IconButton(
icon: Icon(Icons.arrow_back),
onPressed: () => showDialog<bool>(
context: context,
builder: (c) => AlertDialog(
title: Text('Warning'),
content: Text('Are you sure you want to exit?'),
),
actions: [
TextButton(
child: Text('Yes'),
onPressed: () {
Navigator.pop(c, true);
Navigator.pop(context);
}),
TextButton(
child: Text('No'),
onPressed: () => Navigator.pop(c, false),
),
],
),
),
),
),
尝试这将杀死你的应用程序状态
@override
Widget build(BuildContext context) {
return WillPopScope(
////////////////
onWillPop: () => showDialog<bool>(
context: context,
builder: (c) => AlertDialog(
title: Text(
'Warning',
textAlign: TextAlign.center,
),
content: Text('Are you sure to exit?'),
actions: [
TextButton(
style: TextButton.styleFrom(
primary: Colors.green,
),
onPressed: () async {
exit(0);// kill app
},
child: Text('Yes'),
),
TextButton(
style: TextButton.styleFrom(
primary: Colors.red,
),
onPressed: () => Navigator.pop(c, false),
child: Text('No'),
)
],
),
),
/////////////////////
child: Scaffold(),
);
}
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。
