我有一个用户控件,其中包含一个来自ObservableCollection的简单项目的ListView。我希望ListView的ContextMenu根据ListView中选择的内容来包含项目。如果没有选择项目,一些MenuItems不应该是可见的。绑定ContextMenu的MenuItem对ListView选择的可见性
我打开ContextMenu时甚至不会调用我的转换器。的结合似乎是错的,我觉得这在输出窗口:
System.Windows.Data Error: 4 : Cannot find source for binding with reference 'ElementName=listView'. BindingExpression:Path=SelectedItem; DataItem=null; target element is 'MenuItem' (Name=''); target property is 'Visibility' (type 'Visibility')
我不明白什么是错的,不能在网上搜索看着办吧。
下面是一些简单的代码:
<UserControl x:Class="MyApp.DatabaseControl"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
xmlns:l="clr-namespace:MyApp"
Height="Auto"
Width="Auto">
<UserControl.Resources>
<l:ValueToVisibilityConverter x:Key="valueToVisibility" />
</UserControl.Resources>
<Grid>
<ListView x:Name="listView" ItemsSource="{Binding Persons}">
<ListView.View>
<GridView>
<GridViewColumn Width="140" Header="First Name" DisplayMemberBinding="{Binding FirstName}"/>
<GridViewColumn Width="140" Header="Last Name" DisplayMemberBinding="{Binding LastName}" />
</GridView>
</ListView.View>
<ListView.ContextMenu>
<ContextMenu>
<MenuItem
Header="Open"
Visibility="{Binding SelectedItem, ElementName=listView, Converter={StaticResource valueToVisibility}}"/>
<Separator/>
<MenuItem Header="Add..."/>
<MenuItem Header="Remove"/>
</ContextMenu>
</ListView.ContextMenu>
</ListView>
</Grid>
非常感谢!
谢谢,这个伎俩! – fury 2009-06-21 03:20:13