Detemining the "owner" of a ContextMenuStrip

Detemining the "owner" of a ContextMenuStrip

• scoman81

  I was trying to avoid being critical of your approach, you wouldn't have to use a Rube Goldberg style approach if your design wasn't flawed in the first place. A better design would be to create a context menu for each different type of control. At design time add the menu items specific to each type of control to their respective contextmenus. At runtime you could copy menu items that are common to all controls onto all of the contextmenus. The problem with your approach is that one control could have focus and the user could right-click over another control and the common context menu would appear, the user could click one of the menu items thinking that he is activating the menu on the control that his mouse is over but he would actually be activating the menu over the focused control.

  
  

• infrandom

  Try this:

  For Each c As Control In Me.Controls

  If c.ContextMenuStrip.Name Is CType(sender, ToolStripMenuItem).Owner.Name Then

  'Your code here

  Exit For

  End If

  Next

  
  

• GlennZarb

  Use the 'sender' object ...it can be cast to the correct object type to determine which object fired the event

  
  
  

• JBlackburn__

  That definitely helped me out quite a bit, thanks.
  
  

• Damien Morton

  Even if you have 20 controls of the same type, the sender object once cast will give you the exact control that you right clicked.

  
  

• NitinAgarwal

  Something like this in the click event

  Dim TheControl as Object
  TheControl = CType(CType(sender, ToolStripMenuItem).Owner, ContextMenuStrip).SourceControl
  

  Then you can inspect TheControl for .Type, .Name, etc. in a Select statement.

  
  

• Jubber

  In C# it would be this in your ToolStripMenuItem click event:
  
  

  ToolStripMenuItem myItem = (ToolStripMenuItem)sender;

  ContextMenuStrip myStrip = (ContextMenuStrip)myItem.Owner;

  Button clickedButton = (Button)(myStrip.SourceControl);

  I hope it helps a little.

  
  

• Benke

  The issue I was faced with was that I was trying to keep my code within the Click event of the ToolStripItem. It's sender was the ContextMenuStrip, which had no obvious property to tell me what opened it.
  
  

• Hossam Abdel Wahab

  That would work if every control had a unique ContextMenuStrip, however, it's each *type* of control that has one. If I have 20 controls of each type on the TableLayoutPanel then it will execute code against all 20 controls, where I only want it against the single control that I right-clicked on
  
  

• GrandpaB

  I've come up with an indirect workaround that serves my needs--but would still like to know if there is a more direct way.
  
  I declare a class-variable:
  dim mpntRightClick as point
  
  Then I add a MouseDown handler to every control in the TableLayoutPanel
  for each ctl as Control in TableLayoutPanel.Control
  addhandler ctl.mousedown, addressof ctlMouseDown_Handler
  next
  
  Then in the handler I check if it's a right-click and save the point of it if so
  private sub ctlMouseDown_Handler(byval sender as object, e as mouseeventargs)
  if e.button = right then
  mpntRightClick = e.location
  endif
  end sub
  
  Now when I get to the click handler for the items int he ContextMenuStrip I can referance mpntRightClick to determine what control I'm over.
  
  It's ugly, but it works.
  
  

• Pratyush

  I've tried doing something like that. I can get to the ContextMenuStrip itself, but can't find a property that tells me what control it was opened on..
  
  DirectCast(sender, ToolStripMenuItem).Owner.Location
  
  *I've been using the ContextMenuStrip's Location property for the time being, which works unless the control I'm opening it on is located at the bottom of the screen.
  
  

• bala_excel

  How about this:

  Code Snippet

  Dim ac As String

  Dim mbutton As Boolean

  Dim mlistbox As Boolean

  Dim mcombobox As Boolean

  Private Sub TestToolStripMenuItem_Click(ByVal sender As Object, ByVal e As System.EventArgs) Handles TestToolStripMenuItem.Click

  Select Case ac

  Case "Button1"

  MsgBox(Button1.Name)

  Case "ListBox1"

  MsgBox(ListBox1.Name)

  Case "ComboBox1"

  MsgBox(ComboBox1.Name)

  End Select

  End Sub

  Private Sub ContextMenuStrip1_Opening(ByVal sender As Object, ByVal e As System.ComponentModel.CancelEventArgs) Handles ContextMenuStrip1.Opening

  'set variable to active control name

  ac = Me.ActiveControl.Name

  'if the mouse is not over the active control do not show the menu

  Select Case ac

  Case "Button1"

  If Not mbutton Then

  e.Cancel = True

  End If

  Case "ListBox1"

  If Not mlistbox Then

  e.Cancel = True

  End If

  Case "ComboBox1"

  If Not mcombobox Then

  e.Cancel = True

  End If

  End Select

  End Sub

  Private Sub Button1_MouseEnter(ByVal sender As Object, ByVal e As System.EventArgs) Handles Button1.MouseEnter

  mbutton = True

  End Sub

  Private Sub Button1_MouseLeave(ByVal sender As Object, ByVal e As System.EventArgs) Handles Button1.MouseLeave

  mbutton = False

  End Sub

  Private Sub ListBox1_MouseEnter(ByVal sender As Object, ByVal e As System.EventArgs) Handles ListBox1.MouseEnter

  mlistbox = True

  End Sub

  Private Sub ListBox1_MouseLeave(ByVal sender As Object, ByVal e As System.EventArgs) Handles ListBox1.MouseLeave

  mlistbox = False

  End Sub

  Private Sub ComboBox1_MouseEnter(ByVal sender As Object, ByVal e As System.EventArgs) Handles ComboBox1.MouseEnter

  mcombobox = True

  End Sub

  Private Sub ComboBox1_MouseLeave(ByVal sender As Object, ByVal e As System.EventArgs) Handles ComboBox1.MouseLeave

  mcombobox = False

  End Sub

  
  

• Steve Graber

  I haven't taken the time to go through this solution, but just wanted to say my intentions of this post were mainly to find out if I was overlooking some obvious property within the Click event of the ToolStripItem...
  
  However, if I could give you a Rube Goldberg award I would :-P -- Seriously though, thank you for the response and solution :-)
  
  
• Detemining the "owner" of a ContextMenuStrip