Flutter Dropdown Button赢得了滚动后的答案

Question

我正在将用户从一个问题列表中做出不同选择的应用程序，只是在用户向下滚动到其他问题之后，下拉列表不会保留所选答案

我使用提供商状态管理系统，这是代码：

import 'package:flutter/material.dart';

class ChoiceHandler extends ChangeNotifier {
  final List<String> _dropdownElements = ['Not Done', 'Partially Done', 'Done'];
  List<String> get dropdownElement => _dropdownElements;
  late String _selectedItemValue;
  String get selected => _selectedItemValue;

  selectedValues(String s) {
    _selectedItemValue = s;
    notifyListeners();
  }
}

这是下拉列表button widget代码：

Expanded(
            child: ListView.builder(
              itemCount: propositions.length,
              itemExtent: 50.0,
              itemBuilder: (BuildContext context, index) {
                String dropdownValue = "Not Done";

                return ListTile(
                    title: Text(propositions[index]),
                    trailing: Consumer<ChoiceHandler>(
                      builder: (_, provider, __) {
                        return DropdownButton<String>(
                          value: dropdownValue,
                          onChanged: (newValue) {
                            dropdownValue = newValue as String;
                            Provider.of<ChoiceHandler>(context, listen: false)
                                .selectedValues(dropdownValue);

                            print((propositions[index]) + "  " + newValue);
                            dropdown_answer.add(dropdownValue);
                          },
                          items: provider.dropdownElement
                              .map<DropdownMenuItem<String>>((String value) {
                            return DropdownMenuItem<String>(
                              value: value,
                              child: Text(value),
                            );
                          }).toList(),
                        );
                      },
                    ) //_dropdown(index),

                    );
              },
            ),

Answer 1

请勿在此字符串下拉值=“不完成”之类的变量；
在ListView Builder中。
只需在列表对象中添加状态变量

class Document {
 String title;
 String data;
 String status;
  Document({
    required this.title,
    required this.data,
    this.status="Not Done",
  });
}

使用并维护状态。

 List<Document> propositions=[];

一起

Expanded(
            child: ListView.builder(
              itemCount: propositions.length,
              itemExtent: 50.0,
              itemBuilder: (BuildContext context, index) {

               // now don't need to user this variable
               // String dropdownValue = "Not Done";

                return ListTile(
                    title: Text(propositions[index]),
                    trailing: Consumer<ChoiceHandler>(
                      builder: (_, provider, __) {
                        return DropdownButton<String>(
                          // instead use status property from propositions[index] object
                          value: propositions[index].status,
                          onChanged: (newValue) {
                            propositions[index].status = newValue as String;
                            Provider.of<ChoiceHandler>(context, listen: false)
                                .selectedValues(propositions[index].status);

                            print((propositions[index]) + "  " + newValue);
                            dropdown_answer.add(propositions[index].status);
                          },
                          items: provider.dropdownElement
                              .map<DropdownMenuItem<String>>((String value) {
                            return DropdownMenuItem<String>(
                              value: value,
                              child: Text(value),
                            );
                          }).toList(),
                        );
                      },
                    ) //_dropdown(index),

                    );
              },
            ),

，即使小部件重建时，也将与状态值