JSSenior
asked on
Combobox selection driving another combobox
Hi im using the following code which should basically only make Combo2 visible if Combo1's selection is not "insurance". This code is on the AfterUpdate but does not work can someone point out where im going wrong? Cheers
Also could I add some code which drives the values that are available in Combo2's dropdown based on the User Selection in Combo1.
If Combo1.value = "Insurance" Then
Combo2.Visible = False
Else
Combo2.Visible = True
End If
Also could I add some code which drives the values that are available in Combo2's dropdown based on the User Selection in Combo1.
If Combo1.value = "Insurance" Then
Combo2.Visible = False
Else
Combo2.Visible = True
End If
ASKER CERTIFIED SOLUTION
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
SOLUTION
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
ASKER
Great that works, thank you. How about the second part of the question can I also drive the list of values that will be available in the dropdown on Combo2 depending on the value in Combo1?
Pete's solution should answer that...
http:#a36501858
http:#a36501858
ASKER
Hi points value increased as my question had two parts.
As there is only a few options could I just list them in the VBA code?
As there is only a few options could I just list them in the VBA code?
SOLUTION
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
Most UI experts agree that hiding/showing controls based on user selections is confusing and disconcerting to users. You are generally far better off enabling or disabling controls in those instancees. The exception to this would be if a control should only be shown to specific "groups" - for example, members of a Managers Group should be able to see ButtonA, but no one else should see that button. In those cases it's fine to hid controls.