Flutter TextField文本不打印到变量？

Question

我的文本框文本onChanged:我试图改变我在同一个类中声明的变量，并让它在无效函数中打印。请解释如何打印我的文本字段文本。

class LoginPageState extends State<LoginPage> { 
    @override 

    var _email; 
    var _password; 
    var _username; 

    final TextEditingController controller = new TextEditingController(); 

void _loginButton() { 

    print("Login from Page"); 
    print(_password); 
    print(_username); 
    print(_email); 
    } 

Widget build(BuildContext context) { 
    return new Scaffold(
     backgroundColor: Colors.white70, 
     appBar: new AppBar(
     backgroundColor: Colors.amber, 
     title: new Text("Login/Signup"), 
    ), 
     body: new Container(
     child: new Center(
      child: new Column(
      mainAxisAlignment: MainAxisAlignment.center, 
      children: <Widget>[ 
       new TextField(
       controller: controller, 
       decoration: 
       new InputDecoration(labelText: "E M A I L A D D R E S S"), 
       onChanged: (String newString){ 
        setState((){ 
        _email = controller.text; 

        }); 
       }, 
      ), 

Answer 1

嗯，我不知道你在这里做什么。但根据我的理解，您试图在TextField中保存用户的输入。

我已经而不需要setState()

如果您需要获得用户的输入，如电子邮件，您应该获取用户提交后的数据来实现这一点，所以我不知道为什么你想在你的情况下使用onChanged。

基本上你需要做的是向用户提供一个按钮，在完成数据填充后应该按下按钮，然后使用你的TextEditingController来获得用户输入的内容，并且你可以做你想做的事那里。

这是展示如何做到这一点的代码，我想你正在尝试做类似的东西，以一个新会员注册功能，你的应用程序中：

import 'package:flutter/material.dart'; 

void main() { 
    runApp(new MaterialApp(
    home: new MyApp(), 
)); 
} 

class MyApp extends StatefulWidget { 
    @override 
    _MyAppState createState() => new _MyAppState(); 
} 

class _MyAppState extends State<MyApp> { 
    var _email; 
    var _password; 
    var _username; 

    final TextEditingController _nameController = new TextEditingController(); 
    final TextEditingController _emailController = new TextEditingController(); 
    final TextEditingController _passController = new TextEditingController(); 

    void _loginButton({String name, String pass, String email}) { 
    print("Login from Page"); 
    this._username = name; 
    this._email = email; 
    this._password = pass; 
    print(_username); 
    print(_email); 
    print(_password); 
    } 


    @override 
    Widget build(BuildContext context) { 
    return new Scaffold(
     body: new Column (
      mainAxisAlignment: MainAxisAlignment.center, 
      children: <Widget>[ 
      new TextField(controller: this._nameController, decoration: new InputDecoration(hintText: "Type your user name"),), 
      new TextField(controller: this._emailController,decoration: new InputDecoration(hintText: "Type your email")), 
      new TextField(controller: this._passController,decoration: new InputDecoration(hintText: "Type your password")), 

      new RaisedButton(child: new Text ("Sign In"), 
       onPressed:() { 
        _loginButton(name: this._nameController.text, 
         email: this._emailController.text, 
         pass: this._passController.text 
       ); 
       }) 
      ], 

     )); 
    } 
} 

如果你不想使用一个按钮，然后使用onSubmitted属性在键盘上按下返回按钮时保存用户输入，而onChanged将在编辑TextField时更新该值。

new TextField(controller: this._nameController, 
       decoration: new InputDecoration(hintText: "Type your user name"), 
       onSubmitted: (String s) { //when the user press the return button do this 
       _loginButton(name: this._nameController.text); 
       }, 
       onChanged: (String s) { //when the user is editing the text do this 
       _loginButton(name: this._nameController.text); 
       },), 

，不过也许这将是更好地为您，因为您有多个TextField需要立刻被保存使用的按钮。

Answer 2

这是我如何管理与onChanged:来完成这个...

new TextField(
       controller: _emailController, 
       decoration: 
        new InputDecoration(labelText: "E M A I L A D D R E S S"), 
       onChanged: (str) { 
        setState(() { 
        str = _emailController.text; 
        }); 
       }, 
      ), 

Future<Null> _loginButton() async { 
    try { 
     FirebaseAuth.instance.signOut(); 

     print("Login from Page"); 

     _email = _emailController.text.toString(); 
     _password = _passController.text.toString(); 
     _username = _nameController.text.toString(); 

     print(_email); 
     print(_username); 
     print(_password); 

     await FirebaseAuth.instance 
      .signInWithEmailAndPassword(email: _email, password: _password); 

     var userid = FirebaseAuth.instance.currentUser.uid; 

     if (FirebaseAuth.instance.currentUser.photoUrl == null) { 
     fb.child("users/${userid}/image").onValue.listen((Event event) { 
      print('Child added: ${event.snapshot.value}'); 
     }); 
     } 

     print(FirebaseAuth.instance.currentUser); 
    } catch (error) { 
     print(error); 
    } 
    }