Not sure if you will be able to use a wilder card, because to evaluate a string you have to know what you are looking for... exact characters.

Maybe you could just use the "Left" command to do this...

Dim MyString As String
MyString = Left("http://www.website.com/loggedin.asp ???????", 34)

If MyString = "http://www.website.com/loggedin.asp" then
label.caption = "LOGGED IN"
End If

Using LEFT() will only return all the characters you are sure will always be there. "34" is the number of char to get from the string.

Don't know if that will help you in this case or not.