When I left click the image the menu should display in wpf [duplicate]

Question

This code is working, when the image is right clicked the menu is displayed.

I need the menu to display when I left click.

Xamal:

            <Image.ContextMenu>
                <ContextMenu>
                    <MenuItem Header="Reset password" />
                    <MenuItem Header="Edit Profile"/>
                    <MenuItem Header="About us"/>
                </ContextMenu>
            </Image.ContextMenu>
        </Image>

stackoverflow.com/questions/4305565/…

zmbq
– zmbq

2015-10-12 11:33:22 +00:00
Commented Oct 12, 2015 at 11:33 — zmbq
– zmbq, Commented Oct 12, 2015 at 11:33

fahimalizain · Accepted Answer · 2015-10-12 11:37:56Z

0

You can do this by using the MouseLeftButtonUp event of an Image like this

<Image ... MouseDown="Image_MouseLeftButtonUp">
    <Image.ContextMenu>
        <ContextMenu>
            <MenuItem .../>
            <MenuItem .../>
        </ContextMenu>
    </Image.ContextMenu>
</Image>

And then show the ContextMenu in the EventHandler in code behind

private void Image_MouseLeftButtonUp(object sender, MouseButtonEventArgs e)
{
     Image image = sender as Image;
     ContextMenu contextMenu = image.ContextMenu;
     contextMenu.PlacementTarget = image;
     contextMenu.IsOpen = true;         
}

edited Oct 12, 2015 at 11:37

answered Oct 12, 2015 at 11:33

fahimalizain

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3 Comments

Clemens Over a year ago

Better handle the MouseLeftButtonDown event.

fahimalizain Over a year ago

Yeah :) Changed it :)

don Anand Over a year ago

Yes perfect . .Thanks

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