CODEWITHSUNDEEP
windowsvbscript
BruceHill
BruceHill

Reputation: 7164

vbscript for creating registry entries was working, now it isn't. Any ideas?

I have a vbscript that creates a registry entry on a Windows Server 2003 machine. This script has been working fine for about a year now, but recently it just stopped working. I am thinking that a windows update must have changed something, maybe a security setting, whereby this script is no longer permitted to execute. The script uses the following function to create an entry in HKEY_LOCAL_MACHINE\SOFTWARE\ODBC\ODBC.INI\ for new ODBC connections:

Function CreateRegKey (sComputer, hTree, sKey)
  Dim oRegistry
  Dim lResult   
  Set oRegistry = GetObject("winmgmts:{impersonationLevel=impersonate}//" & sComputer & "/root/default:StdRegProv")
  lResult = oRegistry.CreateKey(hTree, sKey)
  If (lResult = 0) And (Err.Number = 0) Then
     CreateRegKey = 0
  Else
     CreateRegKey = 1
     msgbox("Create Key " & sKey & " Failed")
  End If
  Set oRegistry = Nothing
End Function

This function is called as follows:

Const HKEY_LOCAL_MACHINE    = &H80000002
sPath = "SOFTWARE\ODBC\ODBC.INI\" & DataSourceName 
'Create ODBC entry
If (0 = CreateRegKey(sComputer, HKEY_LOCAL_MACHINE, sPath)) Then
  ....
Else    
  ....
End If

Does anyone know of a windows update that could have caused this script to suddenly stop working? The script stops on the following line:

lResult = oRegistry.CreateKey(hTree, sKey)

It does not give an error or anything. It just stops on that line.

Anyone got an idea what is going wrong here and how I could fix it? Thanks.

EDIT: I now get the error number returned by CreateKey. It returns the following:

Err.Number: -2147023533

Err.Description: Cannot start a new logon session with an ID that is already in use

Does anyone know what is causing this and how to work around it? Thanks.

Upvotes: 0

Views: 1958

Answers (1)

Helen
Helen

Reputation: 97677

The description for the error code -2147023533 (0x80070553) is:

Cannot start a new logon session with an ID that is already in use.

A search for this code and description reveals:

Give these a try and see if it helps.

Upvotes: 1

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