i have app with two screens,one fingerprint screen and another home screen.once
fingerprint is authenticated i will get routed to home screen(using navigator.push()),but if I click back button in phone now it is getting routed to fingerprint screen again.Now how to close the app instead of routing back to finger print screen?
Use Navigator.pushReplacement. It will replace the current topmost widget on the stack with the new one instead of simply pushing it on top.
Example usage -
Navigator.pushReplacement(
context, MaterialPageRoute(builder: (BuildContext context) => MyHomePage()));
More documentation about this method can be found here.
put your homepage scaffold inside WillPopScope as a child
WillPopScope(
child: Scaffold(...),
onWillPop: () async {
return false;
},
),
Related
I am using showcaseview package for displaying a message in a screen. It is dismissed when we click anywhere on the screen. But if the user came to that page and use the backbutton for going back, the showcase is not getting dismissed. Is there any way to dismiss the showcaseview other than touching the screen.
I highly think from what I got from your question that the show case view starts every time you go to that page, so you can pass a parameter to choose to show case view or not when open the page like :
Navigator.push<void>(
context,
MaterialPageRoute<void>(
builder: (_) => const Detail(showCaseView:false),
),
);
and in Details screen where you use show case view :
#override
void initState() {
super.initState();
if(isShowCaseView){
WidgetsBinding.instance
ambiguate(WidgetsBinding.instance)?.addPostFrameCallback(
(_) => ShowCaseWidget.of(context)
.startShowCase([_one, _two, _three, _four, _five]),
);
}
When I press the device back button, it navigate to the homepage but not to the previous page.
I would like to know how to navigate to the previous page using the the device back button.
Thank you
You might have used Navigator.pushReplacement which destroys the previous page and creates the new page.However if you use Navigator.push pressing the back button will navigate you to the previous screen.
Example:
Container(
child: Center(
child: TextButton(child: Text("Next page"),onPressed: (){
Navigator.push(context, MaterialPageRoute(builder: (context) => Page2(),)); // back button will navigate to previous screen
//Navigator.pushReplacement(context, MaterialPageRoute(builder: (context) => Page2(),)); // back button will not navigate to previous page
},),),),
It was a Materialapp widget problem.
I solved the problem by deleting the Materialapp widget that was in one of the pages.
This occurs because the previous navigation was not saved in the route stack.
What you should do is this:
Widget A
Navigator.push(context, WidgetB());
Inside Widget B, press the device button and you will return to widget A.
It is important that in widget A you are not using replacement or another that removes widget A from the stack.
You can use:
Navigator.pop(context);
or
Navigator.pushNamed(context, '/routeName');
I try to build a flutter app and want to navigate trough the pages.
I build a drawer with a menu with several items to open new pages - that works fine.
onTap: () {
Navigator.of(context).pop();
Navigator.push(
context,
new MaterialPageRoute(
builder: (BuildContext context) => new Page1(),
),
);
},
By pressing on the Page1 menuitem page1 opens fine.
Inside page1 I have a slidable list with "edit" and "details". I open this with
onTap: () {
Navigator.of(context).pop();
Navigator.push(
context,
MaterialPageRoute(
builder: (BuildContext context) => new DetailsPage(),
),
);
},
In the details Page I see all the things i wanna see, but when I press the Back button in the appBar i am back at home screen with closed drawer.
I tried to put onPressed: () => Navigator.pop(context); inside the appBar section but this doesn´t work either.
Now I got what I want by using always Navigator.push but I am sure that this is not right, so that navigator stack gets bigger and bigger.
What I do wrong? Why .pop does not bring me to the last page?
Any ideas?
Thx in advance
Patrick
Just remove the line with the Navigator.of(context).pop();in the Page1 onTap.
you can use pushReplacement in any cases you need to remove current page from stack and route to new page.
Replace the current route of the navigator that most tightly encloses
the given context by pushing the given route and then disposing the
previous route once the new route has finished animating in.
here is the documentation
When calling Navigator.pop() a result can be passed to the previous screen. Is there a way to set the result so that if the user navigates back on their own the result is still returned to the previous page?
I could pass an object to the second page and then modify it so that the first page can check it when the second page returns, but I'd rather use the result returned from the Navigator as it's more readable.
Overriding the back button's tap detector as I've seen suggested elsewhere is not an acceptable solution because the user may navigate back in some other way such as swiping or pressing the Android back button.
Yes, you can pass data between screens both ways.
While popping back, send the data you wish to send like this from the second screen
RaisedButton(
onPressed: () {
// The Nope button returns "data" as the result.
Navigator.pop(context, 'data');
},
child: Text('Nope!'),
);
And catch the result in the your first screen like this
final result = await Navigator.push(
context,
MaterialPageRoute(builder: (context) => SecondScreen()),
);
source
For the other concern where the user is able to go back to the previous screen by pressing the back button, you can wrap your second screen with WillPopScope widget and provide the onWillPop callback. This will override the popping of the second screen and execute the callback where you can define the value you wish to return from the second screen.
#override
Widget build(BuildContext context) {
return WillPopScope(
child: Scaffold(), // or your widget
onWillPop: () {
return Future.delayed(Duration(microseconds: 0), () {
return Navigator.pop(context, "return-data");
});
},
);
}
Yes, this is possible, when navigating to Screen B from Screen A, make the onTap() functionasynchronous and await the result from Screen B.
Check the code below: It works perfectly:
On Screen A, put this code:
onTap: () async {
// receive the data you are sending from screen B here
final result = await Navigator.push( context);
},
On Screen B, put this code:
onTap: (){
// pass the data you want to use in screen A as the second paramter
Navigator.pop(context,number);
},
I hope this helps
Now I'm using the showModalBottomSheet,
when I click on the semi-transparent bottomsheet, it pops this view out, but what I want is that it doesn't respond to the user's click and still stays there, how do I do that?
How to create a full screen mask view?
If user click on - semi-transparent area - ModalBottomSheet won't close using - WillPopScope
showModalBottomSheet(
context: context,
builder: (context) {
return WillPopScope( // add this
onWillPop: () async => false,
child: Container()); // your code continues
});