首先,我了解这个主题:How to make context menu work for windows phone?
但这种方式太复杂了...... 所以我有这个XAML代码:
<StackPanel Name="friendsGrid" Margin="0,0,0,0" Background="Transparent">
<ListBox Name="friendsListBox" FontSize="32" Tap="friendsListBox_Tap">
<toolkit:ContextMenuService.ContextMenu>
<toolkit:ContextMenu Name="MyContextMenu" Opened="MyContextMenu_Opened">
<toolkit:MenuItem Header="action" Click="contextMenuAction_Click"/>
</toolkit:ContextMenu>
</toolkit:ContextMenuService.ContextMenu>
</ListBox>
</StackPanel>
我正在填写这样的列表:
this.friendsListBox.Items.Add(friend.serviceName);
但是,当然,当我执行longtap时,会出现上下文菜单并选择整个List,而不仅仅是一个项目。
是否有一些简单的方法来识别项目被点击? 感谢
顺便说一下,我找到了这个方法,但是contextMenuListItem没有收到任何东西,它仍然是null:ListBoxItem contextMenuListItem = friendsListBox.ItemContainerGenerator.ContainerFromItem((sender as ContextMenu).DataContext) as ListBoxItem;
答案 0 :(得分:14)
您应该将ContextMenu块放入ItemTemplate(而不是ListBox块)。 这里是简短的样本 XAML:
<ListBox Name="TestList" Margin="26,0,26,0" Height="380" >
<ListBox.ItemTemplate>
<DataTemplate>
<TextBlock Text="{Binding}">
<toolkit:ContextMenuService.ContextMenu>
<toolkit:ContextMenu Name="ContextMenu" >
<toolkit:MenuItem Name="Edit" Header="Edit" Click="Edit_Click"/>
<toolkit:MenuItem Name="Delete" Header="Delete" Click="Delete_Click"/>
</toolkit:ContextMenu>
</toolkit:ContextMenuService.ContextMenu>
</TextBlock>
</DataTemplate>
</ListBox.ItemTemplate>
</ListBox>
代码:
public List<string> Items = new List<string>
{
"Item1",
"Item2",
"Item3",
"Item4",
"Item5",
};
// Constructor
public MainPage()
{
InitializeComponent();
TestList.ItemsSource = Items;
}
private void Edit_Click(object sender, RoutedEventArgs e)
{
if (TestList.ItemContainerGenerator == null) return;
var selectedListBoxItem = TestList.ItemContainerGenerator.ContainerFromItem(((MenuItem) sender).DataContext) as ListBoxItem;
if (selectedListBoxItem == null) return;
var selectedIndex = TestList.ItemContainerGenerator.IndexFromContainer(selectedListBoxItem);
MessageBox.Show(Items[selectedIndex]);
}
希望这有帮助。