I have a flutter activity that launches another activity with
Navigator.push(context, MaterialPageRoute(builder: (context) => Chat()));}
from the second activity I launch another activity using the same method
if from the third activity I press the back button it returns to the first activity and if I use
Navigator.pop(context);
it returns to the second activity
how do I make the back button return to the second activity?
Surround your second and third activities in the following way
WillPopScope(
onWillPop: () async {
Navigator.pop(context);
return false;
},
child: AnotherActivity(
…
),
);
Related
Hello,
From the main screen of my app, pressing a button switches to another screen.
IconButton(
onPressed: () {
Navigator.push(
context,
MaterialPageRoute(
builder: (context) {
return screen2();
},
),
);
},
When I am in screen2 and I want to go back or I can press a button.
Navigator.pop(context); or press the android back button.
How can I call a function as soon as I go back to the main screen?
Thank you.
From what I understood you want to trigger a function when you go to the previous page!
So,
Make the main screen as a Stateful widget and then initialize an init state and inside the init state put the function you want to trigger
Such as like this below code
#override
void initState() {
// TODO: implement initState
super.initState();
FunctionYouWantToTrigger();
}
I want to hide the first alert dialog when I call showDialog() from it. After that when I close the second dialog I want the first dialog was visible again. How I can achieve this?
Before you call second dialog, use Navigator.of(context).pop() to close first dialog. Then, in the second one, you have functions then((value) {...}) or whenComplete(() {...}), inside that you can use it to re-open first dialog.
That's strange that you want to close first one, why don't you just leave it alone and let the second lies on it?
You can create common dialog to show data. if its already showing then just update data only.
showDialog return a future and you can pass data from dialog. The concept is here passing some flag to open the second dialog.
onPressed: () async {
final data = await showDialog(
context: context,
builder: (context) {
return AlertDialog(
content: ElevatedButton(
onPressed: () {
Navigator.of(context)
.pop(true); // true for to show second dialog
},
child: Text("open Second dialog"),
),
);
});
if (data == true) {
showDialog(
context: context,
builder: (context) {
return AlertDialog(
title: Text("Second dialog"),
);
});
}
},
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