Looking up a deactivated widget's ancestor is unsafe

Question

I am new in Flutter and I am trying receive data with a Dialog. When a click in textField the error of image2 appear...

Layout's Image Error's Image

show(BuildContext context){

    var dialog = Dialog(
      child: Container(
        margin: EdgeInsets.all(8.0),
        child: Form(
          child: Column(
            mainAxisSize: MainAxisSize.min,
            children: <Widget>[
              TextFormField(
                decoration: InputDecoration(
                    labelText: "Insira o número de telefone",
                    border: OutlineInputBorder(
                        borderRadius: BorderRadius.all(Radius.circular(2.0)))),
              ),
              Row(
                mainAxisAlignment: MainAxisAlignment.end,
                children: <Widget>[
                  FlatButton(
                      onPressed: () {
                        Navigator.of(context).pop();
                      },
                      child: Text("Cancelar")),
                  FlatButton(
                      onPressed: () {
                        Navigator.of(context).pop();
                      },
                      child: Text("Aceitar"))
                ],
              )
            ],
          ),
        ),
      ),
    );

    showDialog(context: context,builder: (context){
      return dialog;
    });
  }

This is my code.

I/flutter (31032): Looking up a deactivated widget's ancestor is unsafe.
I/flutter (31032): At this point the state of the widget's element tree is no longer stable. To safely refer to a
I/flutter (31032): widget's ancestor in its dispose() method, save a reference to the ancestor by calling
I/flutter (31032): inheritFromWidgetOfExactType() in the widget's didChangeDependencies() method.
I/flutter (31032):

score 72 · Answer 1 · answered Sep 12 '19 at 14:53

Declare a global variable

    final GlobalKey<ScaffoldState> _scaffoldKey = GlobalKey<ScaffoldState>();

then register the key on your widget build's scaffold eg

    @override
    Widget build(BuildContext context) {
     return Scaffold(
       key: _scaffoldKey,
       ...

then on the dialog

show(BuildContext context){

var dialog = Dialog(
  child: Container(
    margin: EdgeInsets.all(8.0),
    child: Form(
      child: Column(
        mainAxisSize: MainAxisSize.min,
        children: <Widget>[
          TextFormField(
            decoration: InputDecoration(
                labelText: "Insira o número de telefone",
                border: OutlineInputBorder(
                    borderRadius: BorderRadius.all(Radius.circular(2.0)))),
          ),
          Row(
            mainAxisAlignment: MainAxisAlignment.end,
            children: <Widget>[
              FlatButton(
                  onPressed: () {
                    Navigator.of(context).pop();
                  },
                  child: Text("Cancelar")),
              FlatButton(
                  onPressed: () {
                    Navigator.of(context).pop();
                  },
                  child: Text("Aceitar"))
            ],
          )
        ],
      ),
    ),
  ),
);

Pass that scaffold context to the showDialog method

showDialog(context: _scaffoldKey.currentContext ,builder: (context){
  return dialog;
 });
}

thanks! this helped me with one of my dialogs, but i have another page which whenever loggs in successfully i pop using ```navigator.of(context)..pop()..pop()..pop("yes")``` but when dialog is shown i face this error, i used a scaffold key and also the builder. — Mahdi-Jafaree, Oct 18 '20 at 06:19
This works for me having a big calling project.. which has multiple roles types .. passing scaffold.currentContext! has solved the issues as I was using the function for multiple roles of pages. Kudos to solution !! — neon97, Aug 06 '21 at 11:42
Thanks for the answer! Can you explain why this works a bit? — batuhankrbb, Aug 17 '21 at 03:54
``Navigator.of(_scaffoldKey.currentContext!).pop();`` to close dialog — Mod, May 19 '22 at 03:52

score 50 · Answer 2 · answered Aug 28 '19 at 13:50

Try This

Give different context name for dialog

 showDialog(context: context,builder: (dialogContex){
              return Dialog(
                child: Container(
                  margin: EdgeInsets.all(8.0),
                  child: Form(
                    child: Column(
                      mainAxisSize: MainAxisSize.min,
                      children: <Widget>[
                        TextFormField(
                          decoration: InputDecoration(
                              labelText: "Insira o número de telefone",
                              border: OutlineInputBorder(
                                  borderRadius: BorderRadius.all(Radius.circular(2.0)))),
                        ),
                        Row(
                          mainAxisAlignment: MainAxisAlignment.end,
                          children: <Widget>[
                            FlatButton(
                                onPressed: () {
                                  Navigator.of(dialogContex).pop();
                                },
                                child: Text("Cancelar")),
                            FlatButton(
                                onPressed: () {
                                  Navigator.of(context).pop();
                                },
                                child: Text("Aceitar"))
                          ],
                        )
                      ],
                    ),
                  ),
                ),
              );
            });

this answer is technically right as we need to remove dialog so we should give diffrent name to dialog context — Dhananjay pathak, Dec 23 '21 at 05:37

Andy King · Answer 3 · 2020-07-21T16:31:15.190

25

I got the same error when attempting to open a dialog and I found a solution here: github flutter issues. Specifically, I followed the poster's recommendation, which was to create a GlobalKey and associate it with the Scaffold widget, and use the context from that key when creating the dialog. In my case, I have a globally accessible object which holds the GlobalKey:

MyGlobals myGlobals = MyGlobals();
class MyGlobals {
  GlobalKey _scaffoldKey;
  MyGlobals() {
    _scaffoldKey = GlobalKey();
  }
  GlobalKey get scaffoldKey => _scaffoldKey;
}

In the Scaffold widget constructor call:

Scaffold(
  appBar: ...,
  body: ...,
  drawer: ...,
  key: myGlobals.scaffoldKey,
)

And in the showDialog call:

showDialog<String>(
  barrierDismissible: ...,
  builder: ...,
  context: myGlobals.scaffoldKey.currentContext,
);

edited Jul 21 '20 at 16:31

answered May 12 '19 at 22:00

Andy King

1,562
1
20
28

@ajs.sonawane Do you mean that it doesn't work when you create a dialog from within a dialog, or it doesn't work when you try to use it a second time with the same dialog, or that it doesn't work if you use it for one dialog, close that dialog, and then open a different dialog? – Andy King Jul 11 '20 at 17:17
2

nope, calling dialog then popping it and then calling it again – ajs.sonawane Jul 11 '20 at 18:03
@ajs.sonawane It seems to be working for my situation ... perhaps you will need to post a new question. Sorry. – Andy King Jul 12 '20 at 06:19

score 17 · Answer 4 · answered Feb 10 '19 at 15:13

17

You’re trying to access a context that isn’t probably available. That happens because you’ve assigned your Dialog to a var and afterwards use a different context (the one from your dialog builder).

Either create your dialog directly after your return in the builder or make it a method instead that returns a Dialog and pass it a BuildContext parameter.

Widget myDialog(BuildContext context) => Dialog(/*your dialog here*/);

This is also a more convenient Flutter practice. You should use methods that return widgets instead of assigning it to variables.

answered Feb 10 '19 at 15:13

Miguel Ruivo

13,004
5
46
78

You need to provide more code context/info then. At first glance that's what it seems, because when you are popping the `Dialog` you may be referring to an invalid widget context. – Miguel Ruivo Feb 10 '19 at 23:01
1

This is **not** a convenient flutter practice. See https://stackoverflow.com/a/53234826/10122791 – Augustin R Nov 25 '19 at 09:00

score 11 · Answer 5 · answered Mar 19 '21 at 04:38

11

use this:

Navigator.of(context,rootNavigator: true).pop();

instead of

Navigator.of(context).pop();

answered Mar 19 '21 at 04:38

biruk IS

1,029
6
11

5

Can you elaborate a bit on this? Why / in which cases is the latter preferable? – dgilperez Apr 09 '21 at 13:05
For Explaination please visit : https://stackoverflow.com/questions/60349741/what-is-the-use-of-rootnavigator-in-navigator-ofcontext-rootnavigator-true – Sayed Muhammad Idrees Nov 16 '21 at 18:59
This will not necessarily help. If your context is outdated, this might still fail, as even with the `rootNavigator: true` part you are still accessing a `BuildContext` that is no longer viable. Also: you might discover unwanted navigation behaviour, e.g., when you use tab bars. – Matthias Schicker May 27 '22 at 12:34

score 8 · Answer 6 · edited Feb 20 '21 at 09:08

8

This might happen while you are popping from the context and trying to open new content on the context you are popping.

()async{
    Navigator.of(context).pop();
    _alertPopUp(); // shows a dialog
    // might do some work after
}

if alert dialog is created on current context then it throws an error because context doesn't exist anymore

edited Feb 20 '21 at 09:08

Pang

9,073
146
84
117

answered Jan 03 '21 at 15:15

enumerator

147
3
2

It didn't work for me. – SalahAdDin Mar 26 '22 at 15:17

score 4 · Answer 7 · answered Sep 24 '21 at 00:14

4

My problem was that I was using hot reload for pretty long time, I think at some point everything got messed up, doing a normal run of the app fixed the problem.

answered Sep 24 '21 at 00:14

Ray Rojas

101
1
8

Seems like this is the answer which worked i dont know if flutter could be notified of this issue – ditso k May 14 '22 at 11:36

score 0 · Answer 8 · answered Feb 11 '19 at 08:37

Try this:

    Future<AlertDialog> myDialog(BuildContext context) {
    return showDialog<AlertDialog>(
      context: context,
      builder: (BuildContext context) {
        return AlertDialog(
          content: Container(
            margin: EdgeInsets.all(8.0),
            child: Form(
              child: Column(
                mainAxisSize: MainAxisSize.min,
                children: <Widget>[
                  TextFormField(
                    decoration: InputDecoration(
                        labelText: "Insira o número de telefone",
                        border: OutlineInputBorder(
                            borderRadius:
                                BorderRadius.all(Radius.circular(2.0)))),
                  ),
                ],
              ),
            ),
          ),
          actions: <Widget>[
            FlatButton(
                onPressed: () {
                  Navigator.of(context).pop();
                },
                child: Text("Cancelar")),
            FlatButton(
                onPressed: () {
                  Navigator.of(context).pop();
                },
                child: Text("Aceitar"))
          ],
        );
      },
    );
  }

That's akward, I've tested it myself and worked fine – Fellipe Malta Feb 11 '19 at 23:19 — Fellipe Malta, Feb 11 '19 at 23:19
Can you post here the complete example, please ? – a2x Feb 12 '19 at 20:53 — a2x, Feb 12 '19 at 20:53

score 0 · Answer 9 · answered Feb 08 '22 at 11:30

0

declare dialog and set in initState

  late Dialog dialog;

  @override
  void initState() {
    super.initState();

    dialog = Dialog(
      ...
    );
  }

answered Feb 08 '22 at 11:30

Linar

468
3
7

score 0 · Answer 10 · answered Apr 04 '22 at 04:05

0

first : declare a FormKey.

 GlobalKey<FormState>myFormKey=GlobalKey<FormState>();

second : add the FormKey to your Form widget.

Form( 

 key:myFormKey,

 child:child

 )

answered Apr 04 '22 at 04:05

Ahmed Taha

1

score 0 · Answer 11 · edited Jun 03 '22 at 12:53

0

In my case i was calling

setState(() {
   Navigator.pop(context);
});

edited Jun 03 '22 at 12:53

Ravindra S. Patil

6,553
2
6
23

answered Apr 30 '22 at 08:29

Muhammad Kashif

111
1
5

score 0 · Answer 12 · answered May 12 '22 at 18:19

0

removing application from emulator and run below commands

flutter clean
flutter pub get

works for me

answered May 12 '22 at 18:19

mayfall

21
4

score -2 · Answer 13 · answered Dec 08 '21 at 21:14

I simply solved this by wrapping the showDialog with a Builder widget, though for me the error came from a stream builder I simply wrap the stream builder with a builder widget and the remove the notify listeners from the a stream am calling in the stream builder, but in your case wrap the showDialog with a Builder widget and it will use the context from the builder, problem solved

Looking up a deactivated widget's ancestor is unsafe

13 Answers13

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