How do you get the control that was clicked to open a ContextMenuStrip?

Question

I\'m using a ContextMenuStrip for multiple controls and I\'m trying to figure out the best way to get the control that was actually clicked on to open the Conte

Answer 1

On VB.NET 2013 this work so fine:

Dim cms As ContextMenuStrip = CType(sender, ContextMenuStrip)
MessageBox.Show(cms.SourceControl.Name)

Answer 2

Your sender is a ToolStripMenuItem -- cast it.
Its owner is a ContextMenuStrip -- get it.

SourceControl is a property on the ContextMenuStrip and references the last control from which the ContextMenuStrip was displayed.

Answer 3

Private Sub cmsRightClick_Click(ByVal sender As Object, ByVal e As System.Windows.Forms.MouseEventArgs) Handles cmsRightClick.MouseClick
    Dim s As String = CType(sender, ContextMenuStrip).GetItemAt(CType(sender, ContextMenuStrip).DisplayRectangle.X, _
     CType(sender, ContextMenuStrip).DisplayRectangle.Y + e.Y).Text.Trim()


    MsgBox(s)
    Select Case s 
        Case Is = "Select Summary Total"
            Dim x = 0
        Case Is = "Select Collections"
            Dim x = 1
        Case Is = "UnSelect"
            Dim x = 2
        Case Is = "Reconcile"
            Dim x = 3
        Case Is = "Undo Reconciliation"
            Dim x = 4
    End Select
End Sub

Answer 4

Private Sub mnuWebCopy_Click(ByVal sender As Object, ByVal e As System.EventArgs) Handles mnuWebCopy.Click

Dim myItem As ToolStripMenuItem = CType(sender, ToolStripMenuItem)
Dim cms As ContextMenuStrip = CType(myItem.Owner, ContextMenuStrip)

MessageBox.Show(cms.SourceControl.Name)

End Sub

Answer 5

Private Sub kdgToolStripMenuItem_Click(sender As Object, e As EventArgs) Handles kdgToolStripMenuItem.Click
    Dim sms = (sender.GetCurrentParent()).SourceControl.name
    MsgBox(sms)
End Sub

'///Faster