导航弹出到索引1

Use popUntil method as mentioned in the docs

Typical usage is as follows:

Navigator.popUntil(context, ModalRoute.withName('/login'));

//========================================================
new ListTile(
title: new RaisedButton(
child: new Text("POP until"),
onPressed: () {
var route = new MaterialPageRoute(
builder: (BuildContext context) =>
new NextPage3(value:"hi there from 3"),
);
//Navigator.pop(context, ModalRoute.withName('/'));
Navigator.popUntil(context,ModalRoute.withName('/'));
},
),
),
//========================================================

replace .pop with .popUntil, actually works very elegantly.

If you are using MaterialPageRoute to create routes, you can use this command:

Navigator.popUntil(context, ModalRoute.withName(Navigator.defaultRouteName))

Navigator.defaultRouteName reflects the route that the application was started with. Here is the piece of code that illustrates it in more detail:

child: InkWell(
child: Row(
mainAxisAlignment: MainAxisAlignment.spaceEvenly,
children: <Widget>[
Image(
image: AssetImage('assets/img/ic_reset.png'),),
Text('Change my surgery details',
style: TextStyle(color: Colors.blue, decoration: TextDecoration.underline),),
],
),
onTap: () =>
Navigator.popUntil(context, ModalRoute.withName(Navigator.defaultRouteName))
),

Hope this helps.

If you do not use named routes, you can use

Navigator.of(context).popUntil((route) => route.isFirst);

For me I used this when pushing a new page:

widget = MyWidget();
Route route = CupertinoPageRoute(builder: (context) => widget, settings:RouteSettings(name: widget.toStringShort()));
Navigator.push(context, route);

Then to go back to specific page:

Navigator.of(context).popUntil((route) => route.settings.name == "MyWidget");

In case you know exactly how many pops should be performed:

For example for 2 pops:

count = 0;
Navigator.popUntil(context, (route) {
return count++ == 2;
});

You can also do it like this

Navigator.of(context)
.pushNamedAndRemoveUntil('/Destination', ModalRoute.withName('/poptillhere'),arguments: if you have any);

The use case is to go the desired screen and pop the screens in between as you require.

For more info, you can check this Post Explaining other Solutions

I tried other answers in this post, and somehow they causing the following exception.

To safely refer to a widget's ancestor in its dispose() method, save a reference to the ancestor by calling dependOnInheritedWidgetOfExactType() in the widget's didChangeDependencies() method.


The relevant error-causing widget was
MaterialApp
lib/main.dart:72
When the exception was thrown, this was the stack
#0      Element._debugCheckStateIsActiveForAncestorLookup.<anonymous closure>
package:flutter/…/widgets/framework.dart:3781
#1      Element._debugCheckStateIsActiveForAncestorLookup
package:flutter/…/widgets/framework.dart:3795
#2      Element.dependOnInheritedWidgetOfExactType
package:flutter/…/widgets/framework.dart:3837
#3      Theme.of
package:flutter/…/material/theme.dart:128
#4      XXxxXX.build.<anonymous closure>
package:xxx/widgets/xxx.dart:33
...
════════════════════════════════════════════════════════════════════════════════

The following answer fixed the issue. https://stackoverflow.com/a/52048127/2641128

Navigator.pushNamedAndRemoveUntil(context, '/', (_) => false);

Here Dashboard() is the screen name. So this will pop out all the screens and goto Dashboard() screen.

Navigator.of(context).pushAndRemoveUntil(
MaterialPageRoute(builder: (c) => Dashboard()),
(route) => false)

This always gets me the expected result. And will pop to route of current Navigator stack

 Navigator.of(context, rootNavigator: true).pop();

This will pop all the routes until the main default route and push to your destination route.

Navigator.pushNamedAndRemoveUntil(context, "destination_route", ModalRoute.withName('/'));

Example for two pops, using cascade operator:

Navigator.of(context)..pop()..pop();