vb1992 Well-Known Member Licensed User Longtime User Dec 13, 2011 #1 How would you move a panel with an image to the left Randomly with dpi ? Panel1.left = Panel1.left - Rnd(0,5) As in Panel1.Left = Panel1.left - 5dpi
How would you move a panel with an image to the left Randomly with dpi ? Panel1.left = Panel1.left - Rnd(0,5) As in Panel1.Left = Panel1.left - 5dpi
stevel05 Expert Licensed User Longtime User Dec 14, 2011 #2 Do you mean DIP? in which case you can use B4X: Panel1.Left = Panel1.Left-DipToCurrent(RND(0,5)) Upvote 0
vb1992 Well-Known Member Licensed User Longtime User Dec 14, 2011 #3 Yes, I meant dip, thanks.! Upvote 0