Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
2.0k views
in Technique[技术] by (71.8m points)

flutter - Looking up a deactivated widget's ancestor is unsafe

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): 
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

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;
 });
}

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...