On Fri, 11 Sep 2009 17:14:02 -0700, HarryisTrying
wrote:

This works but it doesn't reject input with more than 4 octets
xxx.xxx.xxx.xxx.yyy

Thanks and I will use this to try and learn some more
--
Thank You

Here's another approach that will reject

1.1.1.1.1

but accept

1.1.1.1.1 1.1.1.1

since the latter "contains" a valid IP address that is separated from the
invalid construct:

===============================
Option Explicit
Function containsIP(str As String) As Boolean
Dim re As Object

Set re = CreateObject("vbscript.regexp")
re.Pattern = _
"(?:^|\s)\b(?:(?:25[0-5]|2[0-4]\d|[01]?\d\d?)\.){3}(?:25[0-5]|2[0-4]\d|[01]?\d\d?)\b(?:\s|$)"

containsIP = re.Test(str)
Set re = Nothing
End Function
===========================
--ron