2011-08-10 40 views
1

我有一个用于viewmodel的数据模板中定义的usercontrol的viewmodel。我想将usercontrol的'GridViewData'属性绑定到viewmodel的'Data'属性。绑定多个视图模型

我还是新的WPF和可怕的绑定,所以请善待:对

XAML:

<Window x:Class="ReportUtility.MainWindow" 
    xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation" 
    xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml" 
    xmlns:lite="clr-namespace:ReportUtility.Controls.LiteGrid" 
    xmlns:vm="clr-namespace:ReportUtility.ViewModels" 
    xmlns:wf="clr-namespace:System.Windows.Forms;assembly=System.Windows.Forms" 
    Title="MainWindow" Height="350" Width="525"> 

<!--This is the view model I want to bind to variable name is Grid: hosted in the content control below--> 
<Window.Resources> 
    <DataTemplate DataType="{x:Type vm:LiteGridViewModel}"> 
     <lite:LiteGrid GridViewData="{Binding ??}"/> 
    </DataTemplate> 
</Window.Resources> 

<Grid> 
    <Grid.RowDefinitions> 
     <RowDefinition Height="auto"/> 
     <RowDefinition Height="50"/> 
     <RowDefinition Height="*"/> 
    </Grid.RowDefinitions> 
    <Grid.ColumnDefinitions> 
     <ColumnDefinition/> 
     <ColumnDefinition Width="150"/> 
    </Grid.ColumnDefinitions> 

    <Button Grid.Row="0" Content="TestButton" HorizontalAlignment="Stretch" Command="{Binding ExecuteQueryCommand}"/> 
    <TextBox Grid.Row="1" Text="{Binding Path=SqlCommandText, UpdateSourceTrigger=PropertyChanged}"/> 
    <ContentControl Content="{Binding Grid}" Grid.Row="2" Background="Red"/> 

    <Border Background="Aqua" Grid.Column="1" Grid.RowSpan="3"/> 
    <GridSplitter Grid.Column="0" Width="3" Grid.RowSpan="3"/> 
</Grid> 

+0

如果数据是一个属性在你的viewmodel中,将GridViewData =“{绑定数据}”不工作? –

+0

我也有一个视图模型,它是主窗口的datacontext ...它不会试图绑定到那个呢? – ChandlerPelhams

回答

4

你会绑定到Data

<DataTemplate DataType="{x:Type vm:LiteGridViewModel}"> 
    <lite:LiteGrid GridViewData="{Binding Data}"/> 
</DataTemplate> 

LiteGrid中做这个绑定甚至会更好UserControl而不是依赖使用控件设置值的XAML。

<UserControl GridViewData="{Binding Data}"> 
    ... 
</UserControl> 

您的绑定始终引用当前对象的DataContext。由于您的DataTemplate适用于LiteGridViewModel类型,因此DataTemplate中的DataContext始终为LiteGridViewModel

举例来说,如果你有一个像

public class MyClassA 
{ 
    MyClassB ClassB {get; set;} 
} 

public class MyClassB 
{ 
    MyClassC ClassC {get; set;} 
} 

类和MyClassA ClassA一个ViewModel财产(这意味着你可以参考ClassA.ClassB.ClassC),你可以不喜欢

<ContentControl Content="{Binding ClassA}"> 
    <ContentControl Content="{Binding ClassB}"> <!-- DataContext is MyClassA --> 
     <ContentControl Content="{Binding ClassC}"> <!-- DataContext is MyClassB --> 
      <!-- DataContext is MyClassC --> 
     </ContentControl> 
    </ContentControl> 
</ContentControl> 
+0

清除它谢谢! – ChandlerPelhams