InStr

S = "[test]"
If InStr(S, "[") And Instr(S, "]") then
B = True
Else
B = False
EndIf

In VB6 B = True, in .Net B = False

Can someone explain why?

Thanks,
Robb

Did you try this and gives you the same result ?

VB Code:

S = "[test]"
If InStr(S, "[") And[B]Also[/B]  Instr(S, "]") then
B = True
Else
B = False
EndIf

take a look at the IndexOf property , it replaces the InStr of vb6 , eg:

VB Code:

Dim s As String = "[test]"
        Dim b As Boolean = False
        If Not s.IndexOf("[") = -1 AndAlso Not s.IndexOf("]") = -1 Then
            b = True
        Else
            b = False
        End If
 
        MessageBox.Show(b)

AndAlso makes it work.
I also found that:

If InStr(S, "[") > 0 And Instr(S, "]") > 0 then

works as well.

I didn't try the Not Index of -1 version. I am assuming it will work also because of the AndAlso instead of the And.

So can anyone tell me why the original works in VB but to get the same code to work in .net I have to explicitly say 'is greater than 0' or change the wording from And to AndAlso ?

The reason may become important in other scenarios down the road.
Thanks,
Robb

InStr returns the position of the character or -1 if it's not in the haystack.

in vb6

VB Code:

S = "[test]"
If InStr(S, "[") And Instr(S, "]") then
B = True
Else
B = False
EndIf

you'll find that B will always be false~

Also

Code:

---

If Not s.IndexOf("[") = -1 AndAlso Not s.IndexOf("]") = -1 Then

---

would work regardless of whether it's AndAlso or And

I see it now. :blush:

Thanks for setting me straight.:)

Check MSDN Help for the differences between And and AndAlso .