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From: CG Rosen on 2 May 2010 11:05 Good evening, The code below, found in this group, works fine when the multiselect listbox has 1 column, Struggling to get it work when the listbox has 2 columns but without success. Grateful for some hints. Brgds CG Rosen ------------------------------------------------------------ Private Sub SpinButton1_SpinUp() If ListBox1.ListIndex <> -1 Then i = ListBox1.ListIndex If i = 0 Then Exit Sub s = ListBox1.List(i) ListBox1.RemoveItem i ListBox1.AddItem s, i - 1 ListBox1.ListIndex = i - 1 ListBox1.Selected(i - 1) = True SpinButton1.Value = 0 End If End Sub
From: JLGWhiz on 2 May 2010 14:36
There is some good basic information at this site: http://support.microsoft.com/kb/829070 "CG Rosen" <carlgran.rosen(a)telia.com> wrote in message news:972B75F5-0D0D-4640-A759-DB28B0298C37(a)microsoft.com... > Good evening, > > The code below, found in this group, works fine when > the multiselect listbox has 1 column, Struggling to > get it work when the listbox has 2 columns but without > success. Grateful for some hints. > > Brgds > > CG Rosen > ------------------------------------------------------------ > Private Sub SpinButton1_SpinUp() > > If ListBox1.ListIndex <> -1 Then > i = ListBox1.ListIndex > If i = 0 Then Exit Sub > s = ListBox1.List(i) > ListBox1.RemoveItem i > ListBox1.AddItem s, i - 1 > ListBox1.ListIndex = i - 1 > ListBox1.Selected(i - 1) = True > SpinButton1.Value = 0 > End If > > End Sub > |