Get x , y of drag moded picture box

[joey o.]

I can't believe I can't do this! Any help? I take a picture box and set dragmode to automatic but can't get my Picbox.cursor-position unless I right click! I want to left click and set X to OldX and Y to OldY. I need this to smoothly drop it in the DragDrop sub using:

source.move (X - OldX),(Y - OldY)

Please help!

[joey o.]

I'm bringing this to the top as I was quite late posting. Thanks, hope you all can help.
A friend of mine has just informed me he is having similiar problems with left click on drag of treeview nodes.I really hope you can help us.
Thanks,
Joey O.

[joey o.]

C'mon guys, do you just answer the easy ones? This is a problem you will eventually run into.Please help me out.

If you set the DragMode to Manual, it might be easier.

Dim OldX As Single
Dim OldY As Single

Private Sub Form_DragDrop(Source As Control, X As Single, Y As Single)

    Source.Left = X - OldX
    Source.Top = Y - OldY
    
End Sub

Private Sub Picture1_MouseDown(Button As Integer, Shift As Integer, X As Single, Y As Single)

    If Button = 1 Then
        OldX = X
        OldY = Y
        Picture1.Drag
    End If
    
End Sub