loading of several groups of users in a session system variable GROUP table
Hi, I found that the value of the variable of session system THAT GROUP can be assigned by defining a sql in initialization blocks.However, if there is some users with more than one group of users.
is it possible to table leads (using sql in the initialization blocks)?
Thank you!!
Session variable of the SEO GROUP in initialization for different types of authentication blocks
http://108obiee.blogspot.com/2009/10/referencing-group-session-variable-in.html
Users who belong to many groups of external table with semicolons, see example 2 and 3 of the post office.
Concerning
Goran
http://108obiee.blogspot.com
Tags: Business Intelligence
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Hello
Could someone tell me how I might have several group from the different count function?
Here's what I'm trying to do.
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By this motion, I am trying to accomplish the following:
For example, I have a transaction in which category A is passed to category B.
I want to count distinct users who moved from category:
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(2) A to any category (A2X)
(3) any to B (X2B)
(4) all for the whole (X2X)
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Sample Data create table final as ( select 1 user_id,2 product_id,A categ_id, to_Date('1/1/2009','MM/DD/YYYY') dt from dual union all select 1 user_id,3 product_id,B categ_id, to_Date('1/1/2009','MM/DD/YYYY') dt from dual union all select 1 user_id,4 product_id,C categ_id, to_Date('1/3/2009','MM/DD/YYYY') dt from dual union all select 1 user_id,5 product_id,D categ_id, to_Date('1/3/2009','MM/DD/YYYY') dt from dual union all select 1 user_id,6 product_id,E categ_id, to_Date('1/3/2009','MM/DD/YYYY') dt from dual union all select 1 user_id,7 product_id,F categ_id, to_Date('1/10/2009','MM/DD/YYYY') dt from dual union all select 1 user_id,8 product_id,G categ_id, to_Date('1/11/2009','MM/DD/YYYY') dt from dual union all select 2 user_id,2 product_id,A categ_id, to_Date('1/1/2009','MM/DD/YYYY') dt from dual union all select 2 user_id,3 product_id,B categ_id, to_Date('1/2/2009','MM/DD/YYYY') dt from dual union all select 2 user_id,4 product_id,C categ_id, to_Date('1/4/2009','MM/DD/YYYY') dt from dual union all select 2 user_id,5 product_id,F categ_id, to_Date('1/5/2009','MM/DD/YYYY') dt from dual union all select 2 user_id,6 product_id,H categ_id, to_Date('1/6/2009','MM/DD/YYYY') dt from dual union all select 2 user_id,7 product_id,F categ_id, to_Date('1/12/2009','MM/DD/YYYY') dt from dual union all select 2 user_id,8 product_id,G categ_id, to_Date('1/15/2009','MM/DD/YYYY') dt from dual union all select 3 user_id,2 product_id,A categ_id, to_Date('1/11/2009','MM/DD/YYYY') dt from dual union all select 3 user_id,3 product_id,C categ_id, to_Date('1/12/2009','MM/DD/YYYY') dt from dual union all select 3 user_id,4 product_id,B categ_id, to_Date('1/13/2009','MM/DD/YYYY') dt from dual union all ) ;
Could you also tell me how I could make the County be repeated? For example, I want to count 3 to print for the two A to B and a-CSample output Prev_categ | Next_categ | countprev2next | countprev2any | countany2next | countany2any --------------------------------------------------------------------------------------- A B 2 3 3 3 A C 1 - 3 3 B C 2 2 - 3 C B 1 3 - 3 C D 1 - 1 3 C F 1 - 2 3 D E 1 1 1 3 E F 1 1 - 3 F G 2 2 2 3 F H 1 - 1 3 H F 1 1 - 3
under column of prev2any.
I appreciate all help.
Thanks again,Hello
You can do it with the analytical COUNT function:
SELECT DISTINCT , prev_categ , next_categ , COUNT (DISTINCT user_id) OVER (PARTITION BY prev_id , next_id ) AS countprev2next , COUNT (DISTINCT user_id) OVER (PARTITION BY prev_id) AS countprev2any , COUNT (DISTINCT user_id) OVER (PARTITION BY next_id) AS countany2next , COUNT (DISTINCT user_id) OVER () AS countany2any FROM next_categ_data WHERE next_categ IS NOT NULL ORDER BY prev_categ , next_categ ;
Sorry, I'm not a database now, so I can't test it for 12 hours.
Looking at the code you posted, it seems as if you were on the right track with the partitions, only you were trying the wrong analytical function.
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http://Windows.Microsoft.com/en-us/Windows-Vista/fix-a-corrupted-user-profile
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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
http://windowshelp.Microsoft.com/Windows/en-AU/help/769495bf-035C-4764-A538-c9b05c22001e1033.mspx
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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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