Question

我有两个相互排斥的下拉按钮。设置另一个时，如何清除（或设置）一个的值？

谢谢

Answer 1

对于第一个下拉列表：

onChanged: (String newValue) {
      setState(() {
        dropdownValueFirst = newValue;
         dropdownValueSecond = "Bangladesh";
      });
    },

第二个下拉菜单：

onChanged: (String newValue) {
      setState(() {
        dropdownValueSecond = newValue;

        dropdownValueFirst ="One";
      });
    },

请参见以下代码：

import 'package:flutter/material.dart';
import 'package:font_awesome_flutter/font_awesome_flutter.dart';

void main() => runApp(MyApp());

class MyApp extends StatelessWidget {
  // This widget is the root of your application.
  @override
  Widget build(BuildContext context) {
    return MaterialApp(
      title: 'Flutter Demo',
      theme: ThemeData(

        primarySwatch: Colors.blue,
      ),
      home: MyHomePage(title: 'Flutter Demo Home Page'),
    );
  }
}

class MyHomePage extends StatefulWidget {
  MyHomePage({Key key, this.title}) : super(key: key);

  // This widget is the home page of your application. It is stateful, meaning
  // that it has a State object (defined below) that contains fields that affect
  // how it looks.

  // This class is the configuration for the state. It holds the values (in this
  // case the title) provided by the parent (in this case the App widget) and
  // used by the build method of the State. Fields in a Widget subclass are
  // always marked "final".

  final String title;

  @override
  _MyHomePageState createState() => _MyHomePageState();
}

class _MyHomePageState extends State<MyHomePage> {
  int _counter = 0;
 String  dropdownValueFirst="One";
String  dropdownValueSecond="Bangladesh";
  void _incrementCounter() {
    setState(() {

      _counter++;
    });
  }

  @override
  Widget build(BuildContext context) {

    return Scaffold(
      appBar: AppBar(

        title: Text(widget.title),
      ),
      body: Center(

        child: Row(
          mainAxisAlignment: MainAxisAlignment.center,
          children: <Widget>[
            DropdownButton<String>(
    value: dropdownValueFirst,
    icon: Icon(Icons.arrow_downward),
    iconSize: 24,
    elevation: 16,
    style: TextStyle(
      color: Colors.deepPurple
    ),
    underline: Container(
      height: 2,
      color: Colors.deepPurpleAccent,
    ),
    onChanged: (String newValue) {
      setState(() {
        dropdownValueFirst = newValue;
         dropdownValueSecond = "Bangladesh";
      });
    },
    items: <String>['One', 'Two', 'Free', 'Four']
      .map<DropdownMenuItem<String>>((String value) {
        return DropdownMenuItem<String>(
          value: value,
          child: Text(value),
        );
      })
      .toList(),
  ),
            const Padding(padding: EdgeInsets.only(left: 8)),
          DropdownButton<String>(
    value: dropdownValueSecond,
    icon: Icon(Icons.arrow_downward),
    iconSize: 24,
    elevation: 16,
    style: TextStyle(
      color: Colors.deepPurple
    ),
    underline: Container(
      height: 2,
      color: Colors.deepPurpleAccent,
    ),
    onChanged: (String newValue) {
      setState(() {
        dropdownValueSecond = newValue;

        dropdownValueFirst ="One";
      });
    },
    items: <String>['Bangladesh', 'India', 'China']
      .map<DropdownMenuItem<String>>((String value) {
        return DropdownMenuItem<String>(
          value: value,
          child: Text(value),
        );
      })
      .toList(),
  ),
          ],
        ),
      ),
      floatingActionButton: FloatingActionButton(
        onPressed: _incrementCounter,
        tooltip: 'Increment',
        child: Icon(Icons.add),
      ), // This trailing comma makes auto-formatting nicer for build methods.
    );
  }
}

Answer 2

按下第一个下拉菜单后，尝试在setState事件的onChanged内部重置第二个下拉菜单的值，反之亦然，

onChanged: (String newValue) {
        setState(() {
          dropdownValueFirst = newValue;
           dropdownValueSecond='Initial Value of second',// remeber this value must be same as initial value of 2nd dropdown =>value: 'Initial Value of second',
        });
      },

如何以编程方式清除Flutter DropdownButton？

2 个答案: