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"),
);
});
}
},
Related
Using the mockingjay Flutter package, how do I verify in a widget test that an alert dialog has been displayed after e.g. tapping a button? The alert dialog is displayed using the showDialog function.
This is very similar to verifying page navigation behaviour, the thing that caught me out was the showDialog function must have the parameter useRootNavigator set to false, as documented in the example project. Without it the test will fail to detect that the alert dialog was displayed.
showDialog(
context: context,
useRootNavigator: false, // By default this is true, which won't work with Mockingjay
routeSettings: RouteSettings(name: 'verify_checkout_failure'),
builder: (context) {
return AlertDialog(title: Text('Checkout failed'));
});
Then in your test:
await tester.tap(find.byType(PlaceOrderButton));
verify(
() => mockNavigator.push(
any(
that: isRoute<void>(
whereName: equals('verify_checkout_failure'),
),
),
),
);
I am fairly new to state management in Flutter, and I have an application where the user will be interacting with dialogs, and a dialog may have a button to open another dialog which can result in the following:
https://i.stack.imgur.com/VBSzr.png
As shown, multiple issues arise when more than one is opened at once -- most notably the previous dialog is still visible in the background. So when a user opens a new dialog, is there any method where I can hide the dialog behind it, and when the frontmost dialog closes, show the previous one which is hidden? Thanks in advance.
Use Navigator.pop(context); method to close the dialog on your button's onPressed before opening the second dialog.
And then while closing your second dialog you can reopen your previous dialog.
Example:
showDialog(
context: context,
builder: (context) {
return AlertDialog(
title: Text("Hello World"),
content: Text("This is content"),
actions: [
TextButton(
onPressed: () {
Navigator.pop(context); //Close your current dialog
//TODO:open your second dialog
},
child: Text("Open"))
],
);
});
I would first try to re-think the logic for opening multiple dialogs at the same time.
Maybe there is a quick workaround for that, but if you keep adding more and more dialogs, then it would become a mess.
Since showDialog<T> function returns Future<T?>, you could potentially use await and wait until the first dialog is closed and then open the second one and so on:
bool canContinue = await showDialog<bool>(
context: context,
builder: (context) => YourWidget(),
);
Alternatively, you could pop by using Navigator but as said, this will be visible to the user that something weird is going on (pushing and popping dialogs programatically).
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.
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
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(
…
),
);