WPF Context menu on left click - wpf

Question

I have a WPF application..In which I have an Image control in Xaml file.
On right click of this image I have a context menu.
I would like to have same to be displayed on "Left click" also.
How do I do this in MVVM way ?

Answer 1

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">
<Setter.Value>
<ContextMenu>
<MenuItem />
<MenuItem />
</ContextMenu>
</Setter.Value>
</Setter>
</Style>
</Button.Style>
</Button>

Answer 2

You can do this by using the MouseDown event of an Image like this
<Image ... MouseDown="Image_MouseDown">
<Image.ContextMenu>
<ContextMenu>
<MenuItem .../>
<MenuItem .../>
</ContextMenu>
</Image.ContextMenu>
</Image>
And then show the ContextMenu in the EventHandler in code behind
private void Image_MouseDown(object sender, MouseButtonEventArgs e)
{
if (e.ChangedButton == MouseButton.Left)
{
Image image = sender as Image;
ContextMenu contextMenu = image.ContextMenu;
contextMenu.PlacementTarget = image;
contextMenu.IsOpen = true;
e.Handled = true;
}
}

Answer 3

You can invent your own DependencyProperty which opens a context menu when image is clicked, just like this:
<Image Source="..." local:ClickOpensContextMenuBehavior.Enabled="True">
<Image.ContextMenu>...
</Image.ContextMenu>
</Image>
And here is a C# code for that property:
public class ClickOpensContextMenuBehavior
{
private static readonly DependencyProperty ClickOpensContextMenuProperty =
DependencyProperty.RegisterAttached(
"Enabled", typeof(bool), typeof(ClickOpensContextMenuBehavior),
new PropertyMetadata(new PropertyChangedCallback(HandlePropertyChanged))
);
public static bool GetEnabled(DependencyObject obj)
{
return (bool)obj.GetValue(ClickOpensContextMenuProperty);
}
public static void SetEnabled(DependencyObject obj, bool value)
{
obj.SetValue(ClickOpensContextMenuProperty, value);
}
private static void HandlePropertyChanged(
DependencyObject obj, DependencyPropertyChangedEventArgs args)
{
if (obj is Image) {
var image = obj as Image;
image.MouseLeftButtonDown -= ExecuteMouseDown;
image.MouseLeftButtonDown += ExecuteMouseDown;
}
if (obj is Hyperlink) {
var hyperlink = obj as Hyperlink;
hyperlink.Click -= ExecuteClick;
hyperlink.Click += ExecuteClick;
}
}
private static void ExecuteMouseDown(object sender, MouseEventArgs args)
{
DependencyObject obj = sender as DependencyObject;
bool enabled = (bool)obj.GetValue(ClickOpensContextMenuProperty);
if (enabled) {
if (sender is Image) {
var image = (Image)sender;
if (image.ContextMenu != null)
image.ContextMenu.IsOpen = true;
}
}
}
private static void ExecuteClick(object sender, RoutedEventArgs args)
{
DependencyObject obj = sender as DependencyObject;
bool enabled = (bool)obj.GetValue(ClickOpensContextMenuProperty);
if (enabled) {
if (sender is Hyperlink) {
var hyperlink = (Hyperlink)sender;
if(hyperlink.ContextMenu != null)
hyperlink.ContextMenu.IsOpen = true;
}
}
}
}

Answer 4

If you want to do this just in Xaml without using code-behind you can use Expression Blend's triggers support:
...
xmlns:i="schemas.microsoft.com/expression/2010/interactivity"
...
<Button x:Name="addButton">
<Button.ContextMenu>
<ContextMenu ItemsSource="{Binding Items}" />
<i:Interaction.Triggers>
<i:EventTrigger EventName="Click">
<ei:ChangePropertyAction TargetObject="{Binding ContextMenu, ElementName=addButton}" PropertyName="PlacementTarget" Value="{Binding ElementName=addButton, Mode=OneWay}"/>
<ei:ChangePropertyAction TargetObject="{Binding ContextMenu, ElementName=addButton}" PropertyName="IsOpen" Value="True"/>
</i:EventTrigger>
</i:Interaction.Triggers>
</Button.ContextMenu>
</Button>

Answer 5

you only need add the code into function Image_MouseDown
e.Handled = true;
Then it will not disappear.

Answer 6

Hey I came across the same problem looking for a solution which I didn't find here.
I don't know anything about MVVM so it's probably not MVVM conform but it worked for me.
Step 1: Give your context menu a name.
<Button.ContextMenu>
<ContextMenu Name="cmTabs"/>
</Button.ContextMenu>
Step 2: Double click the control object and insert this code. Order matters!
Private Sub Button_Click_1(sender As Object, e As Windows.RoutedEventArgs)
cmTabs.StaysOpen = True
cmTabs.IsOpen = True
End Sub
Step 3: Enjoy
This will react for left & right click. It's a button with a ImageBrush with a ControlTemplate.

Answer 7

you can bind the Isopen Property of the contextMenu to a property in your viewModel like "IsContextMenuOpen".
but the problem is your can't bind directly the contextmenu to your viewModel because it's not a part of your userControl hiarchy.So to resolve this you should bing the tag property to the dataontext of your view.
<Image Tag="{Binding DataContext, ElementName=YourUserControlName}">
<ContextMenu IsOpen="{Binding PlacementTarget.Tag.IsContextMenuOpen,Mode=OneWay}" >
.....
</ContextMenu>
<Image>
Good luck.

Answer 8

Interactivity is old and not support any more. The new approach for the implementation is:
xmlns:b="http://schemas.microsoft.com/xaml/behaviors"
<Button x:Name="ConvertVideoButton">
<Button.ContextMenu>
<ContextMenu VerticalContentAlignment="Top" >
<MenuItem Header="Convert 1" Command="{Binding ConvertMkvCommand}" />
<MenuItem Header="Convert 2" Command="{Binding ConvertMkvCommand}" />
</ContextMenu>
</Button.ContextMenu>
<b:Interaction.Triggers>
<b:EventTrigger EventName="Click">
<b:ChangePropertyAction TargetObject="{Binding ContextMenu, ElementName=ConvertVideoButton}" PropertyName="PlacementTarget" Value="{Binding ElementName=ConvertVideoButton, Mode=OneWay}"/>
<b:ChangePropertyAction TargetObject="{Binding ContextMenu, ElementName=ConvertVideoButton}" PropertyName="IsOpen" Value="True"/>
</b:EventTrigger>
</b:Interaction.Triggers>
</Button>

Answer 9

XAML
<Button x:Name="b" Content="button" Click="b_Click" >
<Button.ContextMenu >
<ContextMenu >
<MenuItem Header="Open" Command="{Binding OnOpen}" ></MenuItem>
<MenuItem Header="Close" Command="{Binding OnClose}"></MenuItem>
</ContextMenu>
</Button.ContextMenu>
</Button>
C#
private void be_Click(object sender, RoutedEventArgs e)
{
b.ContextMenu.DataContext = b.DataContext;
b.ContextMenu.IsOpen = true;
}