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Client Request

komments

Message from Greg Kellogg <gkellogg@opentext.com> via eLink

I have a client that has requested a Live Report that showsall users with System Admin privileges. I figured someone here has alreadybuilt the query (MS SQL Server) that identifies which bitmasks contain thatprivilege. I seem to think Bhupinder Singh actually produced a document thathas all the possible bitmask combinations but I can’t find the documentnow. Any help is greatly appreciated.

GregKellogg

OpenText Global Services

CertifiedLivelink Administrator

(361)815-8950

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Donna Nalls

Hi Greg,How about this:select id, name, firstname, lastname, userprivileges from kuaf where type = 0 and deleted = 0 and userprivileges = (select max(userprivileges) from kuaf) order by namethe above makes the assumption that system privs are the highest in the table...which i am pretty sure they are; this saves you from having to know the bitmask specifically.Regards,Donna

komments

Message from Greg Kellogg <gkellogg@opentext.com> via eLink

eLink

I think the problem with that is that some folks may have ALLprivileges and some may have SysAdmin without the User/Group privileges. Byknowing which bitmasks contain SysAdmin, I can put together an IN clause thatcaptures all of them. I believe (but am not sure) that your approach will onlyget those users with ALL privileges.

GregKellogg

(361) 815-8950

From: eLink Discussion:Open Text Live Reports Discussion[mailto:livereportsdiscussion@elinkkc.opentext.com]
Sent: Monday, April 12, 2010 9:29 PM
To: eLink Recipient
Subject: Hi Greg,

Hi Greg,		Posted by dnalls (Nalls, Donna) on 2010/04/12 22:26
In reply to: Client Request		Posted by gkellogg (Kellogg, Gregory) on 2010/04/12 22:08

Hi Greg,

How about this:

select id, name, firstname, lastname, userprivileges
from kuaf
where type = 0 and deleted = 0
and userprivileges = (select max(userprivileges) from kuaf)
order by name

the above makes the assumption that system privs are the highest in the table...which i am pretty sure they are; this saves you from having to know the bitmask specifically.

Regards,
Donna

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Donna Nalls

Ah, good point….How about this: I created a user and selected login enabled and public access enabled and system administration rights….the bitmask for that user is 2319. Any other combination other than all options selected seems to be less that 2319…so, I think you could do where userprivileges >= 2319

komments

Message from Greg Kellogg <gkellogg@opentext.com> via eLink

eLink

Thanks Donna – I will give that to the client to test.

GregKellogg

(361) 815-8950

From: eLink Discussion:Open Text Live Reports Discussion[mailto:livereportsdiscussion@elinkkc.opentext.com]
Sent: Monday, April 12, 2010 10:05 PM
To: eLink Recipient
Subject: Ah, good point….

Ah, good point….		Posted by dnalls (Nalls, Donna) on 2010/04/12 23:03
In reply to: RE Hi Greg,		Posted by gkellogg (Kellogg, Gregory) on 2010/04/12 22:44

Ah, good point….

How about this: I created a user and selected login enabled and public access enabled and system administration rights….the bitmask for that user is 2319.

Any other combination other than all options selected seems to be less that 2319…so, I think you could do where userprivileges >= 2319

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Evert_Kuiken_(ekuiken_-_(deleted))

This is the Oracle version of what you are looking for.SELECT name AS login, firstname||' '||+lastname AS nameFROM KUAFWHERE TYPE = 0 AND deleted = 0 AND BITAND (userprivileges, 270) = 270ORDER BY NAME ASC/* TYPE=0 means "User", userprivileges 270 means "System administration rights" */This one is even better. It's shows all users with more than Login enabled and Public Access:select LLUser.LastName, LLUser.FirstName, LLUser.name "UserID", Department.Name "Department",(CASE WHEN trunc(to_char(mod((LLUser.UserPrivileges/1), 2))) = 1 then 'Yes' else ' ' END) "LoginEnabled",(CASE WHEN trunc(to_char(mod((LLUser.UserPrivileges/2048),2))) = 1 then 'Yes' else ' ' END) "PublicAccess",(CASE WHEN trunc(to_char(mod((LLUser.UserPrivileges/32),2))) = 1 then 'Yes' else ' ' END) "CreateUsers",(CASE WHEN trunc(to_char(mod((LLUser.UserPrivileges/64),2))) = 1 then 'Yes' else ' ' END) "CreateGroups",(CASE WHEN trunc(to_char(mod((LLUser.UserPrivileges/16),2))) = 1 then 'Yes' else ' ' END) "UserAdmin",(CASE WHEN trunc(to_char(mod((LLUser.UserPrivileges/256),2))) = 1 then 'Yes' else ' ' END) "SystemAdmin"FROM kuaf LLUser INNER JOIN kuaf Department ON LLUser.GroupID = Department.ID WHERE LLUser.Deleted = 0 AND (trunc(to_char(mod((LLUser.UserPrivileges/32),2))) = 1 OR trunc(to_char(mod((LLUser.UserPrivileges/64),2))) = 1 OR trunc(to_char(mod((LLUser.UserPrivileges/16),2))) = 1OR trunc(to_char(mod((LLUser.UserPrivileges/256),2))) = 1)ORDER BY LLUser.UserPrivileges DESC, LLUser.LASTNAME

Bhupinder_Singh

Message from Bhupinder Singh <bsingh@opentext.com> via eLink

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">eLink

Select name from kuaf where ID in (select ID from kuaf where (bitand(userprivileges, 2) = 2 and bitand(userprivileges, 4) = 4 and bitand(userprivileges, 8) = 8 and bitand(userprivileges, 256) = 256 )) and type=0 and deleted=0;Explanation: the above query contains a sub-query that checks for all users who have the system administration privilege (bits 8,3,2, and 1 being "on" in the bitmask), and then the main query asks to list all users that fall into that privileged group of users. The main query checks for type=0 and deleted=0 ensure you only get back users (and not groups) that have not been deleted. The advantage of using this approach is that you need not take into consideration permutations and combinations of other privileges being present in addition to the system administration privilege.Please let me know if that helps to provide an answer to your question, i.e. which users have the system administration privilege.- Bhupinder
----------------------------------------------
Bhupinder Singh, B.Math, B.Ed.
Senior Systems Analyst, Information Technology
Open Text, Waterloo, Ontario, Canada
----------------------------------------------

From: eLink Discussion: Open Text Live Reports Discussion [mailto:livereportsdiscussion@elinkkc.opentext.com]
Sent: Monday, April 12, 2010 10:09 PM
To: eLink Recipient
Subject: Client Request

Client Request

Posted by gkellogg (Kellogg, Gregory) on 2010/04/12 22:08

Message from Greg Kellogg <gkellogg@opentext.com> via eLink

I have a client that has requested a Live Report that shows all users with System Admin privileges. I figured someone here has already built the query (MS SQL Server) that identifies which bitmasks contain that privilege. I seem to think Bhupinder Singh actually produced a document that has all the possible bitmask combinations but I can’t find the document now. Any help is greatly appreciated. -- Greg Kellogg Open Text Global Services Certified Livelink Administrator (361) 815-8950

Appu_Nair

BitAnd is oracle specific for sql you would have to take Bhupinder's and check for '8' for SA because decimal is 270 and binary is100001110 (going from the left first position is zero we arrive at the last bit set which is eight)select a.name,a.id,a.userprivileges,a.type,a.deleted,FLOOR(a.userprivileges/Power(2,8))%2 from les971.livelink.kuaf a where a.type=0 and a.deleted=0 Hope that helps and is covered in the LR & Schema class as well