如何在颤动中动态setState变量？

Question

我刚刚开始学习flutter，我很好奇如何在flutter中动态地setState()

在React Native中，我通常会使用下面这样的函数来动态setState：

class foo extends React.Component{
  state={
    username:null,
    password:null
  }

  toggleSetState(whatState, value){
    this.setState({ [whatState]: value })
  }

  render() {
    return (
      <View>
        <TextInput
          value={this.state.username}
          onChangeText={(text)=>{toggleSetState(username, text)}
        />
        <TextInput
          value={this.state.password}
          onChangeText={(text)=>{toggleSetState(password, text)}
        />
      </View>
    );
  }
}

在Flutter中，上述代码的等价物是什么

我已经在Flutter中尝试过了，但似乎不起作用

class _LoginFormState extends State<LoginForm> {
  String username, password;

  void ToogleState(typedata, text){
    setState(() {
      typedata = text;
    });
  }

  //Widget
  TextField(
    onChanged: (text){ToogleState(username, text); print(username);},
    decoration: InputDecoration(
      hintText: 'input username', labelText: 'Username'
    ),
  ),
}

Answer 1

经过一些研究和尝试，我可以用下面的代码实现我想要的：

class _LoginFormState extends State<LoginForm> {
  String username, password;
  //create an object
  var loginForm = {};
  final myController = TextEditingController();

  void ToogleState(typedata, text){
    setState(() {
      //i can assign any different variable with this code
      loginForm[typedata] = text;
      //output of LoginForm: {username: foo, password: bar}
    });
  }

  //widget
  Padding(
    padding: const EdgeInsets.all(16.0),
    child: Column(
      children: <Widget>[
        TextField(
          onEditingComplete: (){print(loginForm);},
          onChanged: (text){ToogleState("username", text); print(loginForm['username']);},
          decoration: InputDecoration(
            hintText: 'input username', labelText: 'Username'
          ),
        ),
        TextField(
          onEditingComplete: (){print(loginForm);},
          onChanged: (text){ToogleState("password", text); print(loginForm['password']);},
          decoration: InputDecoration(
            hintText: 'input password', labelText: 'Password'
          ),
        ),
      ],
    ),
  );
}

Answer 2

你只需要创建一个变量来保存这个值。我搞不懂为什么在不修改短暂状态的情况下还要调用setState

下面是一些有用的文档https://flutter.dev/docs/development/data-and-backend/state-mgmt/ephemeral-vs-app

class _LoginFormState extends State<LoginForm> {
  String _username = "";
  String __password = "";
  void ToogleState( text){
    setState(() {
      _username = text;
    });
  }

  //Widget
  TextField(
    onChanged: (text){ToogleState( text); print(username);},
    decoration: InputDecoration(
      hintText: 'input username', labelText: 'Username'
    ),
  ),
}