Service logon failiure 2

[strangetimes]

Hi
I have a network of 8 W2K servers and about 400 2K/XP workstations. I'm having an slowly increasing problem where if I have to stop a service it will not start up again until I retype the username and password of the account used to start the service. Sometimes the service will not restart until I bounce the server. This is happening on more than 1 server btw. I am getting an error message like 'Logon failiure, account currently disabled' even on the system account.

I have been getting errors in the event log for a while to say that 'Security policies were propagated with warning. 0x534 : No mapping between account names and security IDs was done'(Event 1202), on most workstations. I presumed this was down to an Exchange system account that was deleted and recreated and let it ride.

Any help much appreciated

Wayne

 

[Albion]

http://www.eventid.net

is a great resource. There are quite a few entries there for that Event. I would paste them here, but I think you can handle going there. Just type the event number (1202) in the box that is labeled "Evend ID:" and click Go. The next page will show up, scroll down until you see the proper source object "SceCli" and click on "Comments and links for event ID 1202 from source SceCli"

Good luck.

-Al

 

[strangetimes]

Fixed the problem in the second paragraph, it was a deleted user account referenced in group policy. It's just the services thing worrying me now..

Strangetimes

 

[ouzojd]

I know its a while since this has being posted, but it helped me as well so a star for Albion.

Strangetimes - with the services not starting, I found that in the local security settings for the server I had the users added to log on as a service, but I still had the same problem as you, having to renter the password. I found that in the GPO for the DC OU, they werent given that right and as such the effective policy on the server was that they didnt have the logon right for the service. Once I added the users to the DC's Group Policy, all OK.