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Publisher/OBIEE privilege group

HanncapJul 14 2014

Hi,

I'm using OBIEE, where I have OBIEE reports that reads from repository and I have also reports that are from Publisher and are imported on OBIEE.

What I've done is a privilege group that can see OBIEE reports and don't have acess to Publisher reports, but like that the people from this group can't see the Publisher reports that are on OBIEE.

If I give privilege to have acess to Publisher reports it will possible any persons from the group change the reports.

It is any way to give privilege to just see the Publisher reports on OBIEE?

Regards,

Comments

Aketi Jyuuzou
I used sense of Tabibitosan B-)
870646

In this case,I imagine four traveler ;-)
create table t(prod,order_date,order_nbr) as
select 'ABC', to_date('6/12/2009','fmmm/dd/yyyy'), 1 from dual union all
select 'ABC', to_date('6/13/2009','fmmm/dd/yyyy'), 1 from dual union all
select 'ABC', to_date('6/14/2009','fmmm/dd/yyyy'), 1 from dual union all
select 'ABC', to_date('6/15/2009','fmmm/dd/yyyy'), 0 from dual union all
select 'ABC', to_date('6/16/2009','fmmm/dd/yyyy'), 0 from dual union all
select 'ABC', to_date('6/17/2009','fmmm/dd/yyyy'), 1 from dual union all
select 'ABC', to_date('6/18/2009','fmmm/dd/yyyy'), 1 from dual union all
select 'ABC', to_date('6/19/2009','fmmm/dd/yyyy'), 1 from dual union all
select 'ABC', to_date('6/20/2009','fmmm/dd/yyyy'), 1 from dual union all
select 'ABC', to_date('6/21/2009','fmmm/dd/yyyy'), 1 from dual union all
select 'ABC', to_date('6/22/2009','fmmm/dd/yyyy'), 1 from dual union all
select 'DEF', to_date('6/18/2009','fmmm/dd/yyyy'), 1 from dual union all
select 'DEF', to_date('6/19/2009','fmmm/dd/yyyy'), 1 from dual union all
select 'DEF', to_date('6/20/2009','fmmm/dd/yyyy'), 0 from dual union all
select 'DEF', to_date('6/21/2009','fmmm/dd/yyyy'), 0 from dual union all
select 'DEF', to_date('6/22/2009','fmmm/dd/yyyy'), 0 from dual;

select prod,min(order_date),count(*)
from (select prod,order_date,order_nbr,
       Row_Number() over(partition by prod           order by order_date)
      -Row_Number() over(partition by prod,order_nbr order by order_date)
      as distance
        from T)
 where order_nbr = 1
group by prod,distance
order by min(order_date);

PRO  MIN(ORDE   COUNT(*)
---  --------  ---------
ABC  09-06-12          3
ABC  09-06-17          6
DEF  09-06-18          2
Frank Kulash
Hi,

Thanks for posting the CREATE TABLE and INSERT statements: that really helps!

Aketi seems to be assuming that order_nbr is always 0 or 1, and that the order_dates in your table are consecutive days (which happens to be true in your sample data).
If that's not the case, you can modify Aketi's solution like this:
WITH	got_grp	    AS
(
	SELECT	product_name
	,	order_date
	,	order_nbr
	,	ROW_NUMBER () OVER ( PARTITION BY  product_name
			      	     ORDER BY  	   order_date
				   ) + ( TRUNC (SYSDATE) 
				       - TRUNC (order_date)
				       )      		AS d_grp
	FROM	tt1
	WHERE	order_nbr	> 0
)
SELECT	  product_name
,	  MIN (order_date)	AS start_date
,	  COUNT (*)		AS consecutive_days
FROM	  got_grp
GROUP BY  product_name
,	  d_grp
ORDER BY  product_name
,     	  start_date
;
As written, this assumes that the combination (product_name, TRUNC (start_date)) is unique. If that's no the case, it can be adapted.
Aketi Jyuuzou
Oh I did not realize that I can use "where clause" in first select :-)
Likewise my previous solution,I imagine four traveler
select prod,min(order_date),count(*)
from (select prod,order_date,order_nbr,
      order_date-Row_Number() over(partition by prod order by order_date)
      as distance
        from T
       where order_nbr > 0)
group by prod,distance
order by min(order_date);

PRO  MIN(ORDE   COUNT(*)
---  --------  ---------
ABC  09-06-12          3
ABC  09-06-17          6
DEF  09-06-18          2
85972
Thank you. I worked :)
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