Thanks Naomi, that took care of that error - now a new one, but it's dealing with the substring, so I need to check that I'm getting what I expect to get.


>>I'm getting a 'Already as open DataReader of that name' error on the following code:
>>
>>

>>    Using sconn As New SqlConnection(ConnString)
>>      Try
>>        PswrdTries = PswrdTries + 1
>>        If PswrdTries > 2 Then
>>          Me.Close()
>>        End If
>>        sconn.Open()
>>        Dim msql1 As String = "Select * from CATDecode where DecName = 'PASSWORD'"
>>        Using ms1 As New SqlCommand(msql1, sconn)
>>          Using ds1 = ms1.ExecuteReader
>>            If ds1.Read Then
>>              If Trim(ds1.Item("DecValue").ToString.ToUpper) <> Trim(txtPassWord.Text.ToUpper) Then
>>                MsgBox("Password does not match" & vbCrLf & "One try left", vbOKOnly + vbCritical, "Incorrect Password")
>>              Else
>>                Dim msql2 As String = "Select * from CATDecode where DecName = 'SUBBANK'"
>>                Using ms2 As New SqlCommand(msql2, sconn)
>>** Error occurs on next statement **
>>                  Using ds2 = ms2.ExecuteReader
>>                    If ds2.Read() Then
>>                      TestSub = ds2.Item("DecValue").ToString.Substring(4, ds2.Item("DecValue").ToString.Length)
>>                    End If
>>                  End Using
>>                End Using
>>              End If
>>            End If
>>          End Using
>>        End Using
>>        sconn.Close()
>>      Catch ex As Exception
>>        sconn.Close()
>>        MsgBox("Sql Exception:  " & ex.Message)
>>      End Try
>>    End Using
>>

>>
>>What am I missing?
>
>I think you have too many using statements. After you opened the connection, don't do any more using, just define a variable for data reader and read the results, then close the reader. Then do another one as a normal variable also.