WPF Datatemplate with same viewmodel

Question

Please check below Xaml code

  <Window.Resources>  
        
        <DataTemplate  DataType="{x:Type local:MainViewModel}">  
            <view1:View1_1_Usr/>  
        </DataTemplate>  
        <DataTemplate  DataType="{x:Type local:MainViewModel}">  
            <view2:View_2_Usr/>  
        </DataTemplate>  
   
    </Window.Resources>  

<grid>
<ContentControl x:Name="Display" Content="{Binding CurrentViemModel}"/>

</Grid>
Error :The key is already defined in this scope

The View1_1_Usr and View_2_Usr need the same viewmodel. if added with different Key, problem will be able to go away.

How can i switch View in ContentControl by loading MainViewModel?

Answer 1

You could try to refer to the code below.
App.xaml.cs:

<Application.Resources>  
        <local:ViewModel x:Key="SharedViewModel" />  
     </Application.Resources>  

MainWindow.xaml and ViewModel:
264596-content-control-two-view-on-viewmodel.txt

View1:

<UserControl   
...  
  DataContext="{StaticResource SharedViewModel}">  
     <Canvas>  
    <TextBlock Text="{Binding Name}" Width="100" Height="50" Background="LightPink"  Margin="50"/>  
    <Line x:Name="lin1" Stroke="Gray" StrokeThickness="2" X1="30" X2="30" Y1="20" Y2="40"/>  
    <Line x:Name="lin2" Stroke="Blue" StrokeThickness="2" X1="40" X2="40" Y1="20" Y2="40"/>  
    <Line x:Name="lin3" Stroke="Red" StrokeThickness="2" X1="50" X2="50" Y1="20" Y2="40"/>  
</Canvas>  
</UserControl>  

View2:

<UserControl    
...  
DataContext="{StaticResource SharedViewModel}" >  
    <Canvas>  
    <Button Content="{Binding Address}" Width="100" Height="50" Background="AliceBlue"  Margin="70"/>  
    <Line x:Name="lin1" Stroke="Gray" StrokeThickness="2" X1="30" X2="130" Y1="40" Y2="40"/>  
    <Line x:Name="lin2" Stroke="Blue" StrokeThickness="2" X1="30" X2="120" Y1="50" Y2="50"/>  
    <Line x:Name="lin3" Stroke="Red" StrokeThickness="2" X1="30" X2="120" Y1="60" Y2="60"/>  
</Canvas>  
</UserControl>  

The result:

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