Flutter:选定的值未显示在下拉列表中

问题描述

我正在从SQLite数据库填充城市名称，并尝试显示为下拉列表.我通过遵循一个教程使它起作用，但是有一个小问题.所选值未显示在下拉列表中，而是继续显示默认提示值.但是，我可以分配和检索正确的选定值.

I'm populating cities name from SQLite database and trying to display as a drop down list. I make it work by following a tutorial, but having a small issue. The selected value is not displayed in dropdown, it keep displaying default hint value. However, I was able to assign and retrieve correct selected value.

这是我的代码:

cities.dart

class Cities {
  int id;
  String name;

  Cities(this.id, this.name);

  Cities.fromMap(Map<String, dynamic> json) {
    this.id = json["id"];
    this.name = json["name"];
  }

  Map<String, dynamic> toMap() => {
        'id': null,
        'name': name,
      };
}

Function从db检索并返回值:

Function that retrieve and returns value from db:

Future<List<Cities>> getCitiesList() async {
Database db = await instance.database;

final citiesData = await db.query('cities');

if (citiesData.length == 0) return null;

List<Cities> citiesList = citiesData.map((item) {
  return Cities.fromMap(item);
}).toList();

return citiesList;
}

构建的代码在Widget build内的下拉列表中:

The code which builds drop down, inside Widget build:

//these are defined above in the code
Cities _city;
final databaseHelper = DatabaseHelper.instance;

FutureBuilder<List<Cities>>(

    future: databaseHelper.getCitiesList(),
    builder: (BuildContext context, AsyncSnapshot<List<Cities>> snapshot) {
      if (!snapshot.hasData) return CircularProgressIndicator();
      return DropdownButton<Cities>(

        items: snapshot.data
            .map((city) => DropdownMenuItem<Cities>(
                  child: Text(city.name),
                  value: city,
                ))
            .toList(),
        onChanged: (Cities value) {
          setState(() {
            _city = value;
          });
        },
        isExpanded: true,
        // value: _city, //uncommenting this line breaks the layout
        hint: Text('Select City'),
      );
    },
  ),

控制台中的错误:

'package:flutter/src/material/dropdown.dart': Failed assertion: line 620 pos 15: 'items == null || items.isEmpty || value == null || items.where((DropdownMenuItem<T> item) => item.value == value).length == 1': is not true.

取消对此value: _city,的注释会在显示中添加相同的错误(显示8次错误，而不是下拉列表).

Un-commenting this value: _city, add same error in display (displays error 8 times, instead of dropdown list).

问题:

1. 如何解决此问题?
2. 如何从列表中设置默认值? (默认情况下会选中)

推荐答案

您可以通过简单的方式完成操作，只需创建一个简单的字符串列表并将该列表传递到下拉菜单即可.

You can do it in simple way, just create a simple list of strings and pass that list to dropdown menu.

方法如下:

1. 更新您的getCitiesList()函数:

Future<List<String>> getCitiesList() async {
  Database db = await instance.database;

  final citiesData = await db.query(tblCities);

  if (citiesData.length == 0) return null;

  return citiesData.map((Map<String, dynamic> row) {
    return row["name"] as String;
  }).toList();
}

将此添加到您的表单页面中:

Add this inside your form page:

//initialize these at top
List<String> _citiesList = <String>[];
String _city;

void _getCitiesList() async {
  final List<String> _list = await databaseHelper.getCitiesList();
  setState(() {
    _citiesList = _list;
  });
}

在initState()内部调用_getCitiesList();.

Call _getCitiesList(); inside initState().

将此添加到您的build方法中:

Add this inside your build method:

DropdownButtonHideUnderline(
    child: DropdownButton<String>(
      value: _city,
      items: _citiesList.map((String value) {
        return DropdownMenuItem<String>(
          value: value,
          child: Text(value),
        );
      }).toList(),
      onChanged: (String newValue) {
      setState(() {
          _city = newValue;
      });
      },
)),

这篇关于Flutter:选定的值未显示在下拉列表中的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！