Tracking Users by Workgroup User ID

  • Thread starter Thread starter charles.kendricks
  • Start date Start date
C

charles.kendricks

I have a split database consisting of the front end residing on an XP
desktop and the backend on a server running Server 2003 R2, and not
utilizing a domain controller. The necessary tables are simply linked
to the frontend from the server. If I secure the database with the
Workgroup Administrator and issue each user his own User ID, how can I
then track or audit which user makes specific changes and/or updates
to the database and when they are made. I specifically want to know
which user adds which records to the database.
 
I have a split database consisting of the front end residing on an XP
desktop and the backend on a server running Server 2003 R2, and not
utilizing a domain controller. The necessary tables are simply linked
to the frontend from the server. If I secure the database with the
Workgroup Administrator and issue each user his own User ID, how can I
then track or audit which user makes specific changes and/or updates
to the database and when they are made. I specifically want to know
which user adds which records to the database.

I'm not sure how you get the UserID, but if you've secured the database and are requiring logins, then the CurrentUser
function will return the currently logged-on user. From there you'd need an audit trial, and Allen Browne has a detailed
page on this:

http://allenbrowne.com/AppAudit.html

Scott McDaniel
scott@takemeout_infotrakker.com
www.infotrakker.com
 
I'm not sure how you get the UserID, but if you've secured the database and are requiring logins, then the CurrentUser
function will return the currently logged-on user. From there you'd need an audit trial, and Allen Browne has a detailed
page on this:

http://allenbrowne.com/AppAudit.html

Scott McDaniel
scott@takemeout_infotrakker.comwww.infotrakker.com

Thank You Scott...Your input gave all the information I needed...Being
a complete novice, I didn't even know about the CurrentUser function.
Thanks again.

Charles Kendricks
 
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