Re: Very Confused On IIF

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From: Jonathan Wood (jwood_at_softcircuits.com)
Date: 03/24/04 

Date: Tue, 23 Mar 2004 18:08:26 -0700

While Left$( 

-- 
Jonathan Wood
SoftCircuits
http://www.softcircuits.com
Available for consulting: http://www.softcircuits.com/jwood/resume.htm
"Michael Harrington" <mikeh@inetprovider.com> wrote in message
news:Oa9z4LSEEHA.240@tk2msftngp13.phx.gbl...
> Okay everyone, here's one for ya..... I have a statement using the "IIF"
> function and it fails, but if I break it apart into a traditional If-Then
> statement, it works fine.  Can anyone tell me what the difference is, or
if
> this is a bug in VB6?  I get "Invalid procedure call or argument
>
> ---- NON-WORKING IIF VERSION ----
>
> Public Sub Main()
>     Debug.Print BracketsRemove("123456") 'Works
>     Debug.Print BracketsRemove("[123456]") 'Works
>     Debug.Print BracketsRemove("[1]") 'Works
>     Debug.Print BracketsRemove("1") 'THE ERROR IS HERE
> End Sub
>
> Private Function BracketsRemove(ByVal sString As String) As String
>     BracketsRemove = IIf(Mid$(sString, 1, 1) = "[", Mid$(sString, 2,
> Len(sString) - 2), sString)
> End Function
>
> ---- WORKING IF-THEN VERSION ----
>
> Public Sub Main()
>     Debug.Print BracketsRemove("123456") 'Works
>     Debug.Print BracketsRemove("[123456]") 'Works
>     Debug.Print BracketsRemove("[1]") 'Works
>     Debug.Print BracketsRemove("1") 'NO ERROR NOW!
> End Sub
>
> Private Function BracketsRemove(ByVal sString As String) As String
>     If Mid$(sString, 1, 1) = "[" Then
>         BracketsRemove = Mid$(sString, 2, Len(sString) - 2)
>     Else
>         BracketsRemove = sString
>     End If
> End Function
>
>

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