InstrLike

Public Function InStrLike(Optional Start, Optional String1, Optional String2, Optional intCompareMethod As VbCompareMethod = vbTextCompare) As Variant
On Error Goto err_InStrLike
Dim intPos As Integer
Dim intLength As Integer
Dim strBuffer As String
Dim blnFound As Boolean
Dim varReturn As Variant
If Not IsNumeric(Start) And IsMissing(String2) Then
String2 = String1
String1 = Start
Start = 1
End If
If IsNull(String1) Or IsNull(String2) Then
varReturn = Null
Goto exit_InStrLike
End If
If Left(String2, 1) = "*" Then
err.Raise vbObjectError + 2600, "InStrLike", "Comparison mask cannot start With '*' since a start position cannot be determined."
Exit Function
End If
For intPos = Start To Len(String1) - Len(String2) + 1
If InStr(1, String2, "*", vbTextCompare) Then
For intLength = 1 To Len(String1) - intPos + 1
strBuffer = Mid(String1, intPos, intLength)
If strBuffer Like String2 Then
blnFound = True
Goto done
End If
Next intLength
Else
strBuffer = Mid(String1, intPos, Len(String2))
If strBuffer Like String2 Then
blnFound = True
Goto done
End If
End If
Next intPos
done:
If blnFound = False Then
varReturn = 0
Else
varReturn = intPos
End If
exit_InStrLike:
InStrLike = varReturn
Exit Function
err_InStrLike:
Select Case err.Number
Case Else
varReturn = Null
MsgBox err.Description, vbCritical, "Error #" & err.Number & "(InStrLike)"
Goto exit_InStrLike
End Select
End Function

Example:
InStrLike("Test String 123abc45 Stuff","###*##")
returns 13, because 123abc45 matches the mask and
it starts at character 13. Hope this is useful to somebody.
Inputs:
Start=Position to start searching, optional
String1=String To search
String2=Mask To search For
intCompareMethod=vbCompareMethod To use, optional