当我点击弹出菜单按钮时,如何删除项目列表?但是,我的列表和弹出菜单位于两个单独的文件中。我需要知道根据哪个列表项被按下哪一个我删除。
pop_up_menu.dart:
import 'package:flutter/material.dart';
class PopUpMenu extends StatelessWidget {
void showMenuSelection(String value) {
print("pressed");
}
@override
Widget build(BuildContext context) {
return PopupMenuButton<String>(
padding: EdgeInsets.zero,
icon: Icon(Icons.more_vert),
onSelected: showMenuSelection,
itemBuilder: (BuildContext context) => <PopupMenuEntry<String>>[
const PopupMenuItem<String>(
value: 'Create another',
child: ListTile(
leading: Icon(Icons.add), title: Text('Create another'))),
const PopupMenuItem<String>(
value: 'Delete',
child: ListTile(
leading: Icon(Icons.delete), title: Text('Delete')))
],
);
}
}
所以在list_tile.dart中导入了这个弹出菜单,每当我点击弹出菜单按钮“删除”时,我需要删除按下弹出菜单的所选列表项
List_tile.dart:
import 'package:flutter/material.dart';
import '../pop_up_menu/pop_up_menu.dart';
var levelsData = [];
class ListTile extends StatefulWidget {
@override
ListTileState createState() => new ListTileState();
}
class ListTileState extends State<ListTile> {
@override
Widget build(BuildContext context) {
return ListView.builder(
itemBuilder: (BuildContext context, int index) => Card(
child: SingleChildScrollView(
child: StuffInTiles(
levelsData[index]['user_id'],
levelsData[index]['price'],
),
),
),
itemCount: levelsData.length,
);
}
}
class StuffInTiles extends StatelessWidget {
final String userId;
final double price;
StuffInTiles(this.userId, this.price);
@override
Widget build(BuildContext context) {
List<dynamic> _getChildren() {
List<Widget> children = [];
levelsData.forEach(
(element) {
if (price.toString() == element['price'].toString()) {
children.add(ListTile(
title: Text("@" + element['price'].toString(),
style: TextStyle(color: Colors.lightGreen)),
subtitle: Text(
"Created on 01 Jun 2018 at 06:24AM \n when price was " +
element['price'].toString()),
trailing: PopUpMenu()));
}
},
);
return children;
}
}
}
因此,此列表中的每个项目都有一个弹出菜单,其中包含该菜单中的删除选项。按下删除选项时,必须删除触发它的项目。
示例:当按下用户2的弹出菜单按钮删除时,应删除user2。
在PopUpMenu
类中添加一个回调函数:
class PopUpMenu extends StatelessWidget {
VoidCallback onDelete;
PopUpMenu({this.onDelete});
void showMenuSelection(String value) {
switch (value) {
case 'Delete':
onDelete();
break;
// Other cases for other menu options
}
}
然后在原始类中创建它:
...
trailing: PopUpMenu(
onDelete: () {
levelsData.removeWhere((element) => element == element);
}
)));
}
Flutter中的一般经验法则是将回调传递给子项而不是尝试访问父项中的数据。
你可能还需要让你的StuffInTiles
Widget有状态并将setState(() {});
添加到你的onDelete
,因为简单地删除该值实际上不会用新列表更新你的视图。