M
monoco
Hi everyone,
I need to find a way to check the status (locked-out/disabled) of an
UserAccount if a Windows 2000 based domain forest.
I've seen that in the .Net Framework 2.0 and above the
method .InvokeGet("userLockedOut") is used and does work.
I'm trying to do this in an enterprise application that was coded
in .Net Framework 1.1 and there's no chance of using newer versions of
the .Net Framework
(just if you thought of suggesting so)
I've tried using the "userAccountControl" property but it does not
seem to change "a byte" when I force an account to get locked (by
trying to log on to a workstation providing the wrong password for
that specific user, it always display its bitmask set on decimal value
512 -> instead of having it set to 528).
I can tell by using ADExplorer.exe utility made by semi-god -> Mr.
Russinovich
(http://live.sysinternals.com/adexplorer.exe)
*** Please have in mind that has to be a solution entirely suited
for .Net Framework 1.1 ****
docs:
http://support.microsoft.com/kb/305144
Gratefully,
Monoco.
I need to find a way to check the status (locked-out/disabled) of an
UserAccount if a Windows 2000 based domain forest.
I've seen that in the .Net Framework 2.0 and above the
method .InvokeGet("userLockedOut") is used and does work.
I'm trying to do this in an enterprise application that was coded
in .Net Framework 1.1 and there's no chance of using newer versions of
the .Net Framework
(just if you thought of suggesting so)
I've tried using the "userAccountControl" property but it does not
seem to change "a byte" when I force an account to get locked (by
trying to log on to a workstation providing the wrong password for
that specific user, it always display its bitmask set on decimal value
512 -> instead of having it set to 528).
I can tell by using ADExplorer.exe utility made by semi-god -> Mr.
Russinovich
(http://live.sysinternals.com/adexplorer.exe)
*** Please have in mind that has to be a solution entirely suited
for .Net Framework 1.1 ****
docs:
http://support.microsoft.com/kb/305144
Gratefully,
Monoco.