1) Do not call Play() from the view model. Raise an event in the view model instead (for instance PlayRequested) and listen to this event in the view: view model: public event EventHandler PlayRequested; … if (this.PlayRequested != null) { this.PlayRequested(this, EventArgs.Empty); } view: ViewModel vm = new ViewModel(); this.DataContext = vm; vm.PlayRequested += (sender, … Read more
This should work. The idea is you have this attached property that you will attach to the DataGrid. In the xaml where you attach it, you’ll bind it to a property on your ViewModel. Whenever you want to programmatically assign a value to the SelectedItem, you also set a value to this property, which the … Read more
I don’t think you can do this from XAML, for exactly the reasons you describe. I ended up doing it in the code-behind. Although it’s code, it’s only a single line of code, and still rather declarative, so I can live with it. However, I’d really hope that this is solved in the next version … Read more
Here is a XAML only solution. Just add this style to your button. This will cause the context menu to open on both left and right click. Enjoy! <Button Content=”Open Context Menu”> <Button.Style> <Style TargetType=”{x:Type Button}”> <Style.Triggers> <EventTrigger RoutedEvent=”Click”> <EventTrigger.Actions> <BeginStoryboard> <Storyboard> <BooleanAnimationUsingKeyFrames Storyboard.TargetProperty=”ContextMenu.IsOpen”> <DiscreteBooleanKeyFrame KeyTime=”0:0:0″ Value=”True”/> </BooleanAnimationUsingKeyFrames> </Storyboard> </BeginStoryboard> </EventTrigger.Actions> </EventTrigger> </Style.Triggers> <Setter Property=”ContextMenu”> … Read more
Frustratingly, I had exactly this error and spent forever trying to work out the cause. For me, it was once working but then I made some very minor changes to the XAML of the derived control, and the compiler started giving that error message. Short solution, cutting out many hours of trying to figure it … Read more
A very simple “plug and play” spinner could be one of the spinning icons from the Font Awesome Wpf Package (Spinning icons). The usage is quite simple, just install the nuget package: PM> Install-Package FontAwesome.WPF Then add the reference to the namespace xmlns:fa=”http://schemas.fontawesome.io/icons/” and use the ImageAwesome control. Set the Spin=”True” property and select one … Read more
What do you think, have I broken the MVVM-Pattern? No. The view model have no dependency upon the view, it only knows about an interface that you could easily mock in your unit tests. So this doesn’t really break the pattern as long as “View” is just an abstraction of something. For type-safety reasons you … Read more
The following is one of many different ways to approach the problem. I prefer view models which are much more lightweight. I then add a class whose responsibility is to compose the view model from one or more sources (e.g. a repository). This doesn’t eliminate the problem of cascading dependencies, but it does free up … Read more
Part 1. Display the properties of the control in MVVM As I said in comments: In MVVM ViewModel should not know about the controls, which are located. In such cases, use the attached behavior or leave the same side logic in View ViewModel is not directly associated with a View, so just refer to the … Read more
Rather than writing code to explicitly add things to the ResourceDictionary, how about just generating the right view on demand? You can do this with a ValueConverter. Your resources would look like this: <Views:ConventionOverConfigurationConverter x:Key=”MyConverter”/> <DataTemplate DataType=”{x:Type ViewModels:ViewModelBase}”> <ContentControl Content=”{Binding Converter={StaticResource MyConverter}}”/> </DataTemplate> You still need a DataTemplate resource, but as long as your ViewModels … Read more