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From: Mikhail Bogorad on 14 May 2010 13:57
hi
i have 3 combos (user chooses Yes or No from a drop down) and a text
field that is disabled by default and i'm trying to enable if either
one of these 3 combos has "NO" value.

So here is what i have and it's not working so far.

Private Sub Form_current()
If Me.Combo1 = "NO" Then
Me.Text.Enabled = True
Else
If Me.Combo2 = "NO" Then
Me.Text.Enabled = True
Else
If Me.Combo3 = "NO" Then
Me.Text = True
Else
Me.Text.Enabled = False
End If
End If
End If

End Sub

Any suggestions?
Thanks
From: Linq Adams via AccessMonster.com on 14 May 2010 17:02
Text is a Reserved Word in Access and is, in fact, the name of a textbox
property, and

Me.Text.Enabled = True

may confusing the Access gnomes! I'd modify the name of the textbox.

Also, your code for thee third combobox is

If Me.Combo3 = "NO" Then
Me.Text = True

and I expect you mean it to be like the rest

If Me.Combo3 = "NO" Then
Me.Text.Enabled = True

--
There's ALWAYS more than one way to skin a cat!

Answers/posts based on Access 2000/2003

Message posted via AccessMonster.com
http://www.accessmonster.com/Uwe/Forums.aspx/access-forms/201005/1

From: John W. Vinson on 14 May 2010 17:44
On Fri, 14 May 2010 10:57:58 -0700 (PDT), Mikhail Bogorad
<mikhail.bogorad(a)gmail.com> wrote:

>hi
>i have 3 combos (user chooses Yes or No from a drop down) and a text
>field that is disabled by default and i'm trying to enable if either
>one of these 3 combos has "NO" value.
>
>So here is what i have and it's not working so far.
>
>Private Sub Form_current()
>If Me.Combo1 = "NO" Then
> Me.Text.Enabled = True
> Else
> If Me.Combo2 = "NO" Then
> Me.Text.Enabled = True
> Else
> If Me.Combo3 = "NO" Then
> Me.Text = True
>Else
>Me.Text.Enabled = False
>End If
> End If
> End If
>
>End Sub
>
>Any suggestions?
>Thanks

A single line will do it:

Me.Text.Enabled = (Me.Combo1 = "NO" OR Me.Combo2 = "NO" OR Me.Combo3 = "NO")

Either your code or mine assumes that the *bound column* of the combo contains
the text string NO; if the combo has a concealed numeric ID field as the bound
column (which the wizard will probably generate for you) you'll need to either
test for that numeric value or use the combo's Column() property to select the
correct column. It's zero based so Me.Combo2.Column(1) would be the second
column.
--

John W. Vinson [MVP]
From: Mikhail Bogorad on 17 May 2010 09:11
Thanks a lot!!!!
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