DropdownButton

Question

我有这个下拉按钮

列表中有两项

当我 Select 值 2 时，我希望它显示一个不包含在列表中的值

像这样：

有可能吗？

看起来像这样：

DropdownButton<String>(
        value: dropdownValue,
        items: <String>[
          'Value 1',
          'Value 2',
        ].map<DropdownMenuItem<String>>((String value) {
          return DropdownMenuItem<String>(
            value: value,
            child: Text(value),
          );
        }).toList(),
        onChanged: (String? newValue) {
          setState(() {
            dropdownValue = newValue!;

            if (dropdownValue == 'Value 2') {
              dropdownValue = 'Value Not in  list of items';
            }
          });
        },
      ),

如您所见，如果我设置 dropdownValue = 'Value Not in list of items';，我当然会收到错误消息，因为它不在列表中 - 但我希望它能工作:)

另一方面，正如预期的那样有效：

if (dropdownValue == 'Value 2') {
  dropdownValue = 'Value 1';
}

因为 'Value 1' 在列表中 - 但我想在单击 Value 2

时显示不在列表中的值

希望有人能帮忙！ :)

[为了搜索引擎优化这里是错误]:

Assertion failed:
../…/material/dropdown.dart:880
items == null ||
              items.isEmpty ||
              value == null ||
              items.where((DropdownMenuItem<T> item) {
                    return item.value == value;
                  }).length ==
                  1
"There should be exactly one item with [DropdownButton]'s value: Value Not in  list of items. \nEither zero or 2 or more [DropdownMenuItem]s were detected with the same value"

Answer 1

我这样做过，问题是我们需要确保 Item[current] == selected item.

import 'package:flutter/material.dart';

class MyStatefulWidget extends StatefulWidget {
  const MyStatefulWidget({Key? key}) : super(key: key);

  @override
  State<MyStatefulWidget> createState() => _MyStatefulWidgetState();
}

class _MyStatefulWidgetState extends State<MyStatefulWidget> {
  String dropdownValue = 'Value 1';

  List<String> values = [
    'Value 1',
    'Value 2',
  ];

  @override
  Widget build(BuildContext context) {
    return Scaffold(
      body: DropdownButton<String>(
        value: dropdownValue,
        onChanged: (String? newValue) {
          setState(
            () {
              if (newValue == 'Value 2') {
                values[1] = 'Value Not in  list of items';
                dropdownValue = 'Value Not in  list of items';
              }
              if (newValue == 'Value 1') {
                dropdownValue = 'Value 1';
                values[1] = 'Value 2';
              }
            },
          );
        },
        items: values.map<DropdownMenuItem<String>>((String value) {
          return DropdownMenuItem<String>(
            value: value,
            child: Text(value),
          );
        }).toList(),
      ),
    );
  }
}

Answer 2

如果您只想在选择 Value 2 时显示一些文本，您可以尝试 hint 小部件。

  String hintValue = 'Select Value'; // ---- hint
  var dropdownValue;

  @override
  Widget build(BuildContext context) {
    return Scaffold(
        appBar: AppBar(
          title: Text(widget.title),
        ),
        body: Column(
          children: [
            ElevatedButton(child: Text('Press here'), onPressed: null), // ---- ignore
            DropdownButton<String>(
              value: dropdownValue,
              hint: Text(hintValue), // ---- hint
              isExpanded: true,
              items: <String>[
                'Value 1',
                'Value 2',
              ].map<DropdownMenuItem<String>>((String value) {
                return DropdownMenuItem<String>(
                  value: value,
                  child: Text(value),
                );
              }).toList(),
              onChanged: (newValue) {
                setState(() {
                  // ----- if Value 2 is selected
                  if (newValue == 'Value 2') {
                    hintValue = 'Value Not in list of items';
                  } else
                    hintValue = '$newValue';  
                });
              },
            ),
          ],
        ));
  }

输出

注意 : Documentation 用于使用 hint 小部件

DropdownButton - 设置值：值不包含在用户可以 select 的项目列表中 - Flutter

DropdownButton - Set value: with value NOT contained by The list of items the user can select - Flutter

flutter

flutter-dropdownbutton

输出