我有一个对象列表,在窗口打开时会动态创建。例如:
//Set content for listview sentitems
inbox.ItemsSource = from email in _dataDC.emails
where email.from == _username
orderby email.time descending
select email;
我的XAML:
<TabItem Header="Inbox" Height="30">
<TabItem.Content>
<ListView Name="inbox" BorderThickness="2" Margin="5,0,-5,0">
<ListView.View>
<GridView>
<GridViewColumn Header="Van" Width="70" DisplayMemberBinding="{Binding from}" />
<GridViewColumn Header="Onderwerp" Width="120" DisplayMemberBinding="{Binding subject}" />
<GridViewColumn Header="Op" Width="130" DisplayMemberBinding="{Binding time}" />
</GridView>
</ListView.View>
</ListView>
</TabItem.Content>
</TabItem>
当列表中的项目被双击时,我只想打开一个新窗口。将对象传递给新窗口,在那里我会对其进行操作。有简单的解决方案吗?
item
转换为Email类型。例如:NewModal s = new NewModal((Email)item);
- John Stritenberger