I have 3 screens A -> B -> C. Screen B is pushed from screen A, screen C is pushed from B. My goal now is to call this method of navigator:
Navigator.popUntil(
context,
(route) => route.isFirst,
);
But apply custom animation like I was doing when pushing those screens. Example of animation I was using when pushing screen was this (where screen slides from the top):
Navigator.push(
context,
SlideInRoute(
page: ScreenB(),
direction: SlideInDirection.Top,
),
);
End result now is that I've made FlipRoute class that flips 2 screens, and I want flip to root.
Related
I have three screens: 1, 2, and 3.
I want to switch from screen 2 to screen 3, but when I pop screen 2 and push screen 3, then it shows screen 1 for a while, then it shows screen 3, so is there any way to switch screens using pop after push or not?
this code not expectations
Navigator.push(
context,
MaterialPageRoute<Widget>(
builder: (context) => widget,
),
);
Navigator.pop(context);
Using pushReplacement insted of push
Navigator.pushReplacement(
context,
MaterialPageRoute<Widget>(
builder: (context) => widget,
),
);
Using pushReplacement insted of push
Navigator.pushReplacement(
context,
MaterialPageRoute(
builder: (context) => widget,
),
);
The behavior you're describing is likely because when you call Navigator.pop(context), it removes the current screen (screen 2) from the navigation stack, but the framework hasn't yet had a chance to build and display the new screen (screen 3) that you pushed. So, for a brief moment, the previous screen (screen 1) is visible.
One way to switch screens without that flicker is to use Navigator.pushReplacement instead of Navigator.push followed by Navigator.pop. Navigator.pushReplacement removes the current screen from the navigation stack and replaces it with the new screen, all in one step.
Example:
Navigator.pushReplacement(
context,
MaterialPageRoute<Widget>(
builder: (context) => widget3,
),
);
This will directly switch from screen 2 to screen 3.
It is generally not recommended to use both Navigator.push() and Navigator.pop() together in the same function call, as this can cause unexpected behavior. Instead, you can use Navigator.replace() to replace the current screen with a new one, or use Navigator.pushNamedAndRemoveUntil() to push a new screen and remove all the screens that come before it.
There are four tabbars, and the ui is created when you attempt to reset the page ui in the tabbarview number 2. I want to create a tabbarview again while maintaining four tabbars.
Navigator.pushReplacement(context,
MaterialPageRoute(builder: (BuildContext context) =>tabbarview() ));
or
Tabview with an index of 2 on a tab bar with 4 pages was reset to its initial state with a button.
If you press the button, all the tab bars disappear...
Navigator. push Replacement (context,
Page RouteBuilder (
pageBuilder: (_, __, ___) => hintPage(),
위 button을 실행하면 TABBAR---> disappear
I pushed three screens: ScreenOne > ScreenTwo(1) > ScreenTwo(2)
I'm at the second instance of ScreenTwo now, but I want to remove the first instance of ScreenTwo from the stack, so it should be ScreenOne > ScreenTwo(2).
When launching ScreenTwo(2) I know I shouldn't remove ScreenTwo(1) from the stack yet, so I can't just call Navigator.replace(). I really need to have ScreenOne > ScreenTwo(1) > ScreenTwo(2) for some time, and then remove the first instance of ScreenTwo(1).
How I can handle it? Navigator.pop() and similars only take into account the screen or screens on top of the stack.
If someone needs more context, this is for a phone app. Not an app for phones, but an app that mimics the behavior of a phone. So in reality, we have HomeScreen > CallScreen(Caller1) > CallScreen(Caller2). As the app can handle different calls at a time the first approach has been to map every call to a CallScreen and let every screen handle their own call events, so the first call can finish while the user is talking in the second one.
As discussed in this post:
How can I pop to specific screen in flutter
I quote:
If you didn't define any route in the MaterialApp then you need to define at the time of push.
Navigator.of(context).push(
MaterialPageRoute(builder: (_) {
return SecondPage();
},
settings: RouteSettings(name: 'SecondPage',),
));
You need to define the same route name
Navigator.of(context).popUntil((route){
return route.settings.name == 'SecondPage';
})
;
Or as an alternative if you did define the routes you can use that:
pushNamedAndRemoveUntil(
'/BottomNavigation', (Route<dynamic> route) => false);
In this case it is most suitable that you go from screen1 > Screen2 > replace previous with Screen3
Use this navigation from screen 2 to screen 3.
Navigator.pushReplacementNamed(context,"Third Screen");
I have a list of Screens in a custom drawer.
Screen 1,
Screen 2,
Screen 3
What is the proper way of navigating from Screen 1 to Screen 2 on push of a button. Currently, I am losing the hamburger (the drawer button) option when I push or push replace.
The code I have was not written by me and I do not have access to the person who wrote the code.
Any thoughts?
Just use the standard navigation:
TextButton(
onPressed: (){
Navigator.push(context, MaterialPageRoute(builder: (context) => Screen2()));
}
child:
//child goes here
)
I have updated my Sqlite Table in Screen B and want to reflect that change in Screen A Also.
So when Pop back from Screen B> Screen A nothing updated.
In Android, we have onResume() to make changes reflect in Activity A.
I have used Widgets Binding Observer but it works if app background, foreground.
Navigator.of(context).pop();
You can await Navigator.push, when pop is called anything below Navigator.push will be called. Ex:
await Navigator.push(SomePage());
/// This will be called after SomePage cals Navigator.pop();
doSomething();
You can get back from the second page with data like this:
// Second Page
Navigator.pop(context, 'result');
Then by the below code on the first page, you are able to lunch the second page and wait until return from it and get data.
// First Page
var result = await Navigator.push(
context,
MaterialPageRoute(builder: (context) => SecondPage()),
);