我在我的视图模型中得到了这个属性:
private PushPinModel selectedPushPinModel;
public PushPinModel SelectedPushPinModel
{
get { return selectedPushPinModel; }
set
{
selectedPushPinModel = value;
RaisePropertyChanged(() => SelectedPushPinModel);
}
}
我想绑定视图以显示选择了哪一个:
<ContentControl DataContext="{Binding SelectedPushPinModel}" VerticalAlignment="Top">
<ContentControl.ContentTemplate>
<DataTemplate>
<Grid Height="100" VerticalAlignment="Top">
<Grid.RowDefinitions>
<RowDefinition Height="20*"/>
<RowDefinition Height="38*"/>
</Grid.RowDefinitions>
<Grid.ColumnDefinitions>
<ColumnDefinition Width="91*"/>
<ColumnDefinition Width="389*"/>
</Grid.ColumnDefinitions>
<Border Opacity="0.95" Width="480" Padding="0,0,0,0" BorderThickness="0" HorizontalAlignment="Left" BorderBrush="Transparent" Background="White" Grid.ColumnSpan="2" Grid.RowSpan="2"/>
<Image Width="70" Height="70" HorizontalAlignment="Center" VerticalAlignment="Center" Source="{Binding Icon}" Grid.Column="0" Grid.Row="0" Grid.RowSpan="2" />
<TextBlock Text="{Binding Header}" Grid.Column="1" Grid.Row="0" Style="{StaticResource TextboxLabelStyle}"/>
<TextBlock Text="{Binding Body}" Grid.Column="1" Grid.Row="1" Style="{StaticResource DefaultTextBlockStyle}"/>
</Grid>
</DataTemplate>
</ContentControl.ContentTemplate>
</ContentControl >
但是我无法让它工作。绑定未显示在视图中,我没有收到任何绑定错误。这是绑定到单个对象的正确方法吗?我更喜欢这样,而不是直接与更脏的 {Binding SelectedPushPushPinModel.Body} 绑定。
任何建议如何做到这一点?谢谢