Error: Only static members can be accessed in initializers what does this mean?

匿名 (未验证) 提交于 2019-12-03 01:10:02

问题:

I have something like this. I am having difficulty understanding this error. Why does accessing filterController here give this error here ? but it doesnt give this error if I move the current entire TextFormField creation (between comments A and B) inside the build method ? How does moving the entire TextFormField inside the build method make filterController static then and resolve this issue ?

class AppHomeState extends State<AppHome> with SingleTickerProviderStateMixin {      TabController _tabController;     final filterController = new TextEditingController(text: "Search");         //----A         TextFormField email = new TextFormField(         keyboardType: TextInputType.emailAddress,         controller: filterController,    ------>ERROR : Error: Only static members can be accessed in initializers         );        //----B    @override     Widget build(BuildContext context)     {         return new Scaffold(                 appBar: new AppBar(..),         );     } } 

How can I resolve this issue ?

回答1:

class AppHomeState extends State<AppHome> with SingleTickerProviderStateMixin {      TabController _tabController;     final filterController = new TextEditingController(text: "Search");     TextFormField email = ... 

... is an initializer and there is no way to access this at this point. Initializers are executed before the constructor, but this is only allowed to be accessed after the call to the super constructor (implicit in your example) was completed. Therefore only in the constructor body (or later) access to this is allowed.

This is why you get the error message:

controller: filterController, 

accesses this.filterController (this is implicit if you don't write it explicit).

To work around your issue (assuming email needs to be final) you can use a factory constructor and a constructor initializer list:

class AppHomeState extends State<AppHome> with SingleTickerProviderStateMixin {   factory SingleTickerProviderStateMixin() =>        new SingleTickerProviderStateMixin._(new TextEditingController(text: "Search"));    SingleTickerProviderStateMixin._(TextEditingController textEditingController) :        this.filterController = textEditingController,          this.email = new TextFormField(         keyboardType: TextInputType.emailAddress,         controller: textEditingController);    TabController _tabController;   final filterController;   final TextFormField email; 

or when the email field does not need to be final email can be initialized in the constructor initializer list:

class AppHomeState extends State<AppHome> with SingleTickerProviderStateMixin {    SingleTickerProviderStateMixin() {     email = new TextFormField(         keyboardType: TextInputType.emailAddress,         controller: filterController,     );   }    TabController _tabController;   final filterController = new TextEditingController(text: "Search");   TextFormField email; 

but in Flutter widgets initState is usually used for that

class AppHomeState extends State<AppHome> with SingleTickerProviderStateMixin {    @override   void initState() {     super.initState();     email = new TextFormField(         keyboardType: TextInputType.emailAddress,         controller: filterController,     );   }    TabController _tabController;   final filterController = new TextEditingController(text: "Search");   TextFormField email;  


回答2:

You can convert this variable to a function and you can take context in this function parameters.

Example

Widget myDialog (BuildContext context) {   return new Scaffold(     backgroundColor: Colors.white,     body: new Center(       child: new Column(         children: <Widget>[           new Text("Invalid Username/Password"),           new Text("Please verify your login credentials"),           new RaisedButton(             child: new Text("Ok"),             onPressed:() {               Navigator.pop(context);//Error : Only static members can be accessed in initializers             }           ),         ],       ),     )   ); }  // Using if you are doing in a class this.myDialog(context);  // Using if you are using a global function myDialog(context); 

But, i think you want to show a error message. So, you can do it with a dialog not an page. It's more efficient because you can specify your dialog box with buttons or messages and you can use this error dialog everywhere. Let's look my global helper function for showing error messages.

void showError(BuildContext context, String error) {   showSnackBar(     context,     new Text(       'Error',       style: new TextStyle(color: Theme.of(context).errorColor),     ),     content: new SingleChildScrollView(       child: new Text(error)     ),     actions: <Widget>[       new FlatButton(         child: new Text(           'Ok',           style: new TextStyle(             color: Colors.white           ),         ),         onPressed: () {           Navigator.of(context).pop();         },         color: Theme.of(context).errorColor,       ),     ]   ); }  // Using in everywhere showError(context, 'Sample Error'); 


标签
易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!