i am using tabbar view with my app that have same floating action button:
floatingActionButton: FloatingActionButton(
onPressed: () async{
print("this is the current Screen :$currentScreen");
if(currentScreen==0)
{
await showInformationDialog(context);
}
else
Navigator.push(
context,
MaterialPageRoute(
builder: (context) => AddNewVehical(1),
),
);
},
but when i tap on tab it works perfectly and the currentscreen variable changes from 0 to 1 as expected but problem arises when i slide from one tab to another the currentscreen variable print always zero due to which on the other tab it also shows dialogue
can anyone tell me why this is happening? and a solution for it ?
Thanks in advance <3
You can add TabController listener in initState().
_tabController!.addListener(_handleTabSelection);
In _handleTabSelection method, get the tabController index.
currentScreen = _tabController!.index;
You can use TabController. With the index property you can get the currentIndex.
reference : https://api.flutter.dev/flutter/material/TabController-class.html
Related
In my app, I have a homepage that has 5 tabs on the bottom. On each tabbed page, there is an app bar that has a '+' symbol as an action, which navigates you to a different page. The navigation with that '+' button to the new page is done with the following code, alongside the Flutter Platform Widgets package:
Navigator.of(context, rootNavigator: true)
.push(
platformPageRoute(
context: context,
builder: (context) => Page1(),
),
);
I use the platformPageRoute feature as an easy way to navigate with a native feel. Now, that works fine to navigate to a new page, but the issue comes when I use
Navigator.pop(context);
to navigate back to the original page. When I use that to navigate back to that original page, it pays no attention to the tab that was selected originally. For example, if I were originally on the second tab on the homepage and then use the '+' button on that tab and then finally use
Navigator.pop(context);
on that new page, it returns the first tab of the homepage. Is there any way of ensuring when I use the above command, it goes to the right tab? I have tried something along the lines of:
Navigator.popUntil(context, '/homepageTab2');
alongside a named route, to return to the correct tab on the homepage, although that returns a black screen. Why might that be? I have also tried using:
Navigator.pushAndRemoveUntil(
context,
platformPageRoute(
context: context,
builder: (context) =>
HomePage(selectedPage: 1),
),
(route) => false,
);
This does not work either, since it returns the selected/correct page tab content, but with the first tab selected. In addition, the other
'problem' for me is that the animation is a 'push' one and that doesn't 'match' with the animation when I have more often used
Navigator.pop(context);
to navigate back to a screen. Is there a way to maybe use pushAndRemoveUntil but then change the animation to match a pop animation?
Thanks!
EDIT:
I have just noticed that with the situation I have described above, it is actually returning the correct screen content when I use Navigator.pop(context); but the tab in the tab bar at the bottom is showing as the first tab, in the second tab's position, essentially duplicating the first tab, until I navigate to a new tab and back, at which time it shows the correct tab in the correct position. I hope that makes sense!
As it turns out, the issue wasn't related to Navigator.pop(context); being used. It was the way I was controlling the selected tab. I'm posting this as an answer incase it helps someone else.
Initially, I created late values for a tab controller and the current selected page, like so:
late TabController _tabController;
late ScrollController _scrollController;
late int _selectedPage;
Then, I created a list of widgets that represented the actual page to display for each selected tab:
List<Widget> _pageWidgets = <Widget>[
Page1();
Page2();
Page3();
Page4();
Page5();
];
Then (and I think this was the bit that wasn't working) I used initState() as follows:
void initState() {
super.initState();
// Initialising a value that allows the 'final' page selector to be changed
_selectedPage = widget.selectedPage;
// Initialising the tab controller
_tabController = TabController(
length: 5,
vsync: this,
initialIndex: _selectedPage,
);
// updating the tab index when a new item is selected
_tabController.addListener(() {
setState(() {
_selectedPage = _tabController.index;
//_tabIndex = _tabController.index;
});
});
// Creating the scroll controller
_scrollViewController = ScrollController();
// Scrolling view to top when a new tab is selected
_tabController.addListener(() {
setState(() {
_scrollViewController
.jumpTo(_scrollViewController.position.minScrollExtent);
});
});
}
I then controlled the page content like this:
body: _pageWidgets.elementAt(_selectedPage),
I'm not 100% sure why this wasn't working, although I believe it would have something to do with the fact that initState() would only be called during the build and therefore placing the functionality inside there would mean changes wouldn't be detected. Either way, my new method, which works perfectly, is:
/// Controls the screen to display first
int _index = 0;
/// Creating a navigation key to control tab bar navigation
final _navigationKey = GlobalKey<CurvedNavigationBarState>();
Then, within the Scaffold() I show the page content like this:
body: _pageWidgets.elementAt(_index),
And finally, within the navigation bar (which is the CurvedNavigationBar() package from pub.dev) I give it a key and the index:
key: _navigationKey,
index: _index,
And this controls it perfectly, showing the correct tab.
Sub-pages of a TabBarView cannot be navigated using Navigator.
You can use TabController to go to your desired tab page after awaiting Navigator.push():
await Navigator.of(context, rootNavigator: true)
.push(
platformPageRoute(
context: context,
builder: (context) => Page1(),
),
);
tabController.animateTo(<index of tab>);
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
I have the following flow Screen 1 -> Screen 2 -> Dialog (in a separate widget).
Screen 2 displays a dialog (Close? Yes or No). If someone presses Yes, I would like to return to the Screen 1, if they press No, just close the dialog and return to Screen 2. What I currently do is when Yes is tapped, I do Navigator.pop(context) twice. Is this a good practice? Is there a way to pass the context of Screen 2 to my dialog widget so I can pop that one directly?
Personally, I think it would be better to pass the response from the dialog back to the page, and let the page handle the rest.
You can do this:
//I'm using a raised button just to call the alert as an example...
RaisedButton(
child: Text('Press me'),
//This part here is the important part
onPressed: () async {
//You can return anything when you use Navigator.pop
//In this case I'm returning a bool indicating if the page should close or not.
//You have to await this because it depends on user input.
bool shouldPopResult = await showDialog<bool>(
context: context,
builder: (context) => AlertDialog(
//The content of your dialog
actions: <Widget>[
// The value you pass here in Navigator.of(context).pop
// is the value that will be stored in shouldPopResult,
// so if "Yes" is pressed, true will return...
// and if "No", false is returned.
FlatButton(
child: Text('Yes'),
onPressed: () => Navigator.of(context).pop(true),
),
FlatButton(
child: Text('No'),
onPressed: () => Navigator.of(context).pop(false),
)
],
),
);
// This is for if the user dismisses the dialog without pressing a button
// In that case shouldPopResult would be null, so I'm setting it to false.
// You can prevent the user from dismissing the dialog
// setting barrierDismissible to false in the showDialog method.
if (shouldPopResult == null) shouldPopResult = false;
// And finally with the dialog already dismissed, you can decide
// to go back or not.
if (shouldPopResult) Navigator.of(context).pop();
});
As usual you can extract the dialog as a Widget, or extract the function that handles the dialog response altogether or anything else.
You can see the example of returning data from a page in the flutter documentation here.
Screen 1: shows list of item with add button.
Screen 2: form to add a new item to the list.
Screen 2 >> Screen 1 - While calling navigator.pop() in screen 2, how to call method/setState (to update list) in screen 1?
Can anyone help me out?
I don't want to relaunch screen again. I just need to run a method to update my list after popping previous screen?
When you navigate from Screen 1 to Screen 2, you use the push method. The push method returns a Future object.
You can use the then method of Future to execute your code after Screen 2 was popped from the navigation stack.
Navigator.of(context)
.push(MaterialPageRoute(
builder: (context) => Screen2(),
))
.then((value) {
// you can do what you need here
// setState etc.
});
You can use Provider or BLoc or you can await for the result when you push the page
You can do it like this :
// this method waits for the pop to be called
var data = await Navigator.push(
context,
MaterialPageRoute(builder: (context) => LoginScreen()),
);
debugPrint(data);
// here the second argument is the data
Navigator.pop(context, "data"); // popped from LoginScreen().
output
data
Same method can also be done like below
// this method waits for the pop to be called
Navigator.push(
context,
MaterialPageRoute(builder: (context) => LoginScreen()),
).then((data){
// then will return value when the loginScreen's pop is called.
debugPrint(data);
});
Here is a good article to look into this
https://medium.com/flutter-community/flutter-push-pop-push-1bb718b13c31
Navigator.pop has a second argument called result.
You could then return the value that you need to the first page that will read it as the return value of Navigator.push
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