issue in the creation of the IOM group reactivate via sql query.

Similar Questions

• How to identify columns that have the same data in a SQL query or function?

  Deal all,

  How to identify columns that have the same data in a SQL query or function? I have the sample data as below

  DEPT_ID

  EMP_ID

  Come on

  !CITYSTATECOUNTRY111 June 1983DELHIHUMAN RESOURCESIndia1218 January 1987DELHIHUMAN RESOURCESIndia1328 November 1985DELHIHUMAN RESOURCESIndia144 June 1985DELHIHUMAN RESOURCESIndia255 June 1983MUMBAIHDIndia266 June 1983MUMBAIHDIndia277 June 1983MUMBAIHDIndia288 Jun. 1983MUMBAIHDIndia399. June 1983GURGAONDLIndia31010 June 1983GURGAONDLIndia

  Now, I want to Indify columns that have the same data for the same Department ID.

  Is it possible in sql unique or do I have to write the function for this? Pls Help how to write?

  Thanks in advance.

  You can try this?

  WITH T1)

  DEPT_ID, EMP_ID, DATE OF BIRTH, CITY, STATE, COUNTRY

  ), ()

  SELECT 1, 1, TO_DATE('1.) June 1983', 'JJ. LUN. (YYYY'), 'DELHI', 'HR', 'INDIA' OF THE DUAL UNION ALL

  SELECT 1, 2, TO_DATE('18.) January 1987', 'JJ. LUN. (YYYY'), 'DELHI', 'HR', 'INDIA' OF THE DUAL UNION ALL

  SELECT 1, 3, TO_DATE('28.) November 1985', 'JJ. LUN. (YYYY'), 'DELHI', 'HR', 'INDIA' OF THE DUAL UNION ALL

  SELECT 1, 4, TO_DATE('4.) June 1985', 'JJ. LUN. (YYYY'), 'DELHI', 'HR', 'INDIA' OF THE DUAL UNION ALL

  SELECT 2.5, TO_DATE('5.) June 1983', 'JJ. LUN. (YYYY'), 'BOMBAY', 'HD', 'INDIA' OF THE DUAL UNION ALL

  SELECT 2.6, TO_DATE('6.) June 1983', 'JJ. LUN. (YYYY'), 'BOMBAY', 'HD', 'INDIA' OF THE DUAL UNION ALL

  SELECT 2.7, TO_DATE('7.) June 1983', 'JJ. LUN. (YYYY'), 'BOMBAY', 'HD', 'INDIA' OF THE DUAL UNION ALL

  SELECT 2.8, TO_DATE('8.) June 1983', 'JJ. LUN. (YYYY'), 'BOMBAY', 'HD', 'INDIA' OF THE DUAL UNION ALL

  SELECT 3, 9, TO_DATE('9.) June 1983', 'JJ. LUN. (YYYY'), 'GURGAON', 'DL', 'INDIA' OF THE DUAL UNION ALL

  SELECT 3.10, TO_DATE('10.) June 1983', 'JJ. LUN. (YYYY'), 'GURGAON', 'DL', 'INDIA' OF THE DOUBLE)

  SELECT DEPT_ID,

  RTRIM (XMLAGG (XMLELEMENT(A,VALS||',')). Extract ('//Text ()'), ',') COLUMNS_WITH_DUPLICATE

  DE)

  SELECT * FROM)

  SELECT DEPT_ID,

  EMP_ID,

  Date of birth

  CITY,

  STATE,

  COUNTRY

  DE)

  SELECT DEPT_ID,

  EMP_ID,

  Date of birth

  CITY,

  STATE,

  COUNTRIES,

  COUNT (*) OVER(PARTITION BY DEPT_ID ORDER BY EMP_ID DESC,DOB DESC,CITY DESC,STATE DESC, COUNTRY DESC) RN

  DE)

  SELECT DEPT_ID,

  CASE WHEN(CEID>1) AND THEN 'YES' ELSE 'NO' END AS EMP_ID.

  CASE WHEN(CDOB>1) THEN 'YES' ELSE 'NO' END AS DATE OF BIRTH,

  CASE WHEN(CCITY>1) AND THEN 'YES' ELSE 'NO' END AS CITY.

  CASE WHEN(CSTATE>1) AND THEN 'YES' ELSE 'NO' END AS STATE.

  CASE WHEN(CCOUNTRY>1) THEN 'YES' ELSE 'NO' END AS A COUNTRY

  DE)

  SELECT DISTINCT

  DEPT_ID,

  CEID,

  CDOB,

  CITY,

  CSTATE,

  CCOUNTRY

  DE)

  SELECT DEPT_ID,

  COUNT (*) TO THE CEID (DEPT_ID PARTITION, EMP_ID),.

  COUNT (*) ON CDOB (DEPT_ID SCORE, DATE OF BIRTH),

  COUNT (*) ON THE CITY (DEPT_ID PARTITION, CITY),

  COUNT (*) ON CSTATE (DEPT_ID PARTITION, STATE).

  COUNT (*) ON CCOUNTRY (DEPT_ID, COUNTRY PARTITION)

  FROM T1)))

  WHERE RN = 1)

  UNPIVOT (CLO FOR (VALS) IN (EMP_ID, DATE OF BIRTH, CITY, STATE, COUNTRY)))

  WHERE COLS = "YES".

  DEPT_ID GROUP;

  OUTPUT:

  DEPT_ID COLUMNS_WITH_DUPLICATE
  --------- ------------------------

  1 CITY, COUNTRY, STATE
  2 CITY, COUNTRY, STATE
  3 CITY, COUNTRY, STATE

  Post edited by: Parth272025

• Calculation of the time wall of a SQL query.

  Hello

  While trying to discover the time of running a SQL query (wall time) I read in one place that CPU_TIME/EXECUTIONS of v$ SQLAREA, is the precise runtime we can come close to.

  I can't use "set timing on ' or 'DBMS_UTILITY. GET_TIME' that I need to extract the execution time of the story because the query will be drawn to an end, and I need to know how long it took to DB level and compare it with the end time before calculating the % of time used in the DB level total.

  Maybe another way to track sessions and the user TKPORPOF but now I donot want to take the help of the ADMINISTRATOR at this initial stage.

  Is CPU_TIME/EXECUTIONS of v$ SQLAREA where (SQL_TEXT) AS "SELECT... OF... %'; should be enough?

  user2925917, yes as Brian already answered your understanding as posted above seems correct.  Except in the case where there has been only a running query you will download an average time.  The problem with averages is that one or two unusual executions that can skew the average, but in most cases the average will be probably fairly accurate.

  - -

  HTH - Mark D Powell.

• Find the name of the view based on a sql query

  Can anyone suggest me how to find the name of the view based on a sql query? When I try to the following select statement:

  select view_name from user_views where text like '%SELECT * from TABLE%';

  but I get this error:
  

  SQL Error: ORA-00932: inconsistent datatypes: expected NUMBER got LONG
  00932. 00000 -  "inconsistent datatypes: expected %s got %s"

  But as I notice that the TEXT column is really LONG to type... Are there any other table system that store information about the text using each view?

  See this example, using DBMS_METADATA. GET_DDL:

  Re: Search text in column

• Eliminate the duplicate based on the condtion in Select of SQL query.

  Hi all

  I write the SQL query where I have to select values based on the condition in the column.

  Lets say I have 3 columns position, description, used, there are different values in the position but for some positions of the column description of the lines is the same and if column Description is the same and employee is null then that there should be only one row returned and if the description is the same but the employee column is not null then it should be several lines.

  I can't use Group by that we have around 35 columns in the select query.

  Please suggest any Solution.

  Hi Michael,

  I adds a column to the t2 to get the good understanding of my needs.

  Level	Employee	From Date	to_date
  1	Test2	21.03.2014	21.04.2014
  2	Test4	21.02.2014	20.03.2014
  2	Test1	21.03.2014	21.04.2014
  2	Test3	21.04.2014
  3	MgrTest	21.03.2014

  Now, the result should look like this.

  Level	Employee	From Date	TO Date
  1	Test2	21.03.2014	21.04.2014
  2	Test3	21.04.2014
  2	Test1	21.03.2014	21.04.2014
  3	Mgrtes	21.03.2014
  4

  There was an addition more as if this day is not null for the given level, then the query must return a single line of balnk more with the same position, I am reached using any Union and works very well I'm stuck with the point above.

• Problem with LDAP in the APEX and not in sql query * more

  Hello everyone.
  
  Hereby, I refer to an existing thread: Query LDAP APEX
  
  I have a problem using LDAP in the APEX (DB version: 11.2.0.2.0;) APEX version: 4.0)
  
  I get "Authentication failed" by their SUMMIT. However, when I run it in SQL * more (SQL Developer) (I created it as seen in the referenced forum thread) it works! Can I use my own function, but that looks like reinventing the wheel.
  

      l_ldap_host := 'oursite.be';
      l_ldap_port := '389';
      l_ldap_domn := 'oursite';
      l_ldap_user := i_username;
      l_ldap_pass := i_pw;
      l_ldap_base := 'ou=oursite,dc=oursite,dc=be';
  
  
      dbms_ldap.use_exception := true;
      
      l_session  := dbms_ldap.init(l_ldap_host,l_ldap_port);
      l_retval   := dbms_ldap.simple_bind_s(l_session, l_ldap_domn||'\'||l_ldap_user, l_ldap_pass);    
      l_attrs(1) := 'name';
      l_attrs(2) := 'title';
      l_retval   := dbms_ldap.search_s(
                      l_session, 
                      l_ldap_base, 
                      dbms_ldap.scope_subtree, 
                      '(sAMAccountName='||l_ldap_user||')',
                      l_attrs,
                      0,
                      l_message
                    );
  
      l_retval := dbms_ldap.count_entries(l_session, l_message);

  We must search the sAMAccountName because that contains our login credentials (dennis.surname). The common name is just our full name (Dennis Surname)
  
  
  At the SUMMIT, I have these settings:
  
  * LDAP host: oursite.be
  * Port: 389
  Use SSL: No SSL
  Use exact DN: No.
  * String DN: ou = oursite, dc = oursite, dc = be
  * Search filter: sAMAccountName = % LDAP_USER %
  
  
  When I try to test it I get "Authentication failed" but I don't know why. It works very well in sql * more (in the the same pattern of course!) so I have really no idea what I'm doing wrong. In addition, the message comes instantly and sql * more it takes about a second to authenticate.
  
  I tried so many things! remove the 'or '. Connect with my name, change 'cn = % LDAP_USER %' filter, connect with dennis.surname and Dennis Surname, using exact DN,... and all the possible combinations of them... Nothing works.
  
  I can go further by using my own function, but I really want to use the settings of the APEX, because it's so much easier.
  
  Thanks in advance for help out me!
  Dennis

  Hi Dennis,

  Try this

  The exact use DN Yes value
  Change your DN string to

  %LDAP_USER%@domain
  

  or

  domain\%LDAP_USER%
  

  The authentication uses a simple_bind_s. You must use the same syntax in these text boxes. You actually do a single with bond

  dbms_ldap.simple_bind_s(l_session, 'sAMAccountName=' || l_ldap_user, l_ldap_pass);
  

  It does not work. It's the syntax to use in the search for search_s filter.

  Please keep in mind that the apex_040100 (for apex 4.1) user must connect the rights on the domain server.

• Get the count of the two different statuses in SQL query

  Hi Experts,
  

  ALTER TABLE EMP ADD STATUS INTEGER DEFAULT 1 NOT NULL;
  UPDATE EMP SET STATUS=2 WHERE EMPNO > 7700
  SELECT COUNT(*) OVER() TOTALRECORDS,EMP.DEPTNO,EMP.EMPNO FROM EMP,DEPT WHERE EMP.DEPTNO=EMP.DEPTNO

  The Select query will return the total number of the EMP, DEPT tables (the two status with 1 and 2).
  But I need to get the number of 1 and 2 separately.
  
  Is this possible with changes in an existing SQL query?
  
  Thank you
  Dharan V

  ALTER TABLE EMP ADD STATUS INTEGER DEFAULT 1 NOT NULL;
  UPDATE EMP SET STATUS=2 WHERE EMPNO > 7700
  SELECT COUNT(*) OVER() TOTALRECORDS,EMP.DEPTNO,EMP.EMPNO FROM EMP,DEPT WHERE EMP.DEPTNO=EMP.DEPTNO
  

  change this option like this

  SELECT COUNT(decode(status,1,1,null)) OVER() RECORDS_1,
         COUNT(decode(status,2,2,null)) over() RECORDS_2,
         EMP.DEPTNO,EMP.EMPNO
    FROM EMP,DEPT WHERE EMP.DEPTNO=EMP.DEPTNO
  

• How to know the status of the process flow using a SQL query

  Hello
  I want to know the final status of a process flow after execution by using a sql query?
  
  The available entries are:
  -Item_Key
  -Process_Flow_Package_Name
  -Process_Flow_Name
  
  I ran the process flow using a WF_Engine.LaunchProcess ()procedure, but I was not able to know the status of the process flow.
  
  
  Is it possible to know the status of the process flow?
  
  using a procedure or an sql query?
  
  Thanks in advance,
  SriGP.

  Once the process is completed, or not, you can see the status by calling:

  UAS (your_OWF_shema) .wf_engine. ItemStatus()

• Generate the Trace file to a sql query

  Hi all
  
  I want to generate a trace for a sql query file so that I can generate a .out file corresponding I need to check the performance of an application before using it.
  Anyone can guide me please how to do this.
  I know how do to generate a trace for a concurrent program, but right now, I want to track for a simple sql query.
  
  Kind regards
  Ankur

  Hello
  Agreed. but I thought that if OP do not have access to metalink then?

  In any case I not substitute me your answer.

  Oops. Sorry I did not read the lines below.

  You can turn simple trace for this particular session.
  Oracle will generate trace files (.trc) for each session where the value of SQL_TRACE = TRUE and write them to the USER_DUMP_DEST destination. That you can use tkprof to read the generated trace file.

  Kind regards
  Taj

  Published by: Mohammed Taj on July 14, 2009 10:11

• Access to the CONSTANT packet in a SQL query

  Hello
  Can someone explain why I cannot access the package constant in a SQL, but you can use dbms_output.put_line:
  
  SQL > create package scott.xx_test_const_pkg
  2 as
  3 a constant varchar2 (1): = 'A ';
  4 end;
  5.
  
  Package created.
  
  SQL > start
  2 dbms_output.put_line (scott.xx_test_const_pkg.a);
  3 end;
  4.
  A <-THIS WORKS
  
  PL/SQL procedure successfully completed.
  
  SQL > select * from scott.emp where ename = scott.xx_test_const_pkg.a;
  Select * from scott.emp where ename = scott.xx_test_const_pkg.a <-THIS DOES NOT WORK
  
  *
  ERROR on line 1:
  ORA-06553: PLS-221: 'A' is not a procedure or is not defined
  
  
  Thank you

  Impossible to refer to a pl/sql package constant in a sql statement.
  Create a function that returns the value of constant, and then refer to the function in a sql statement.

• How can I show all the results returned by a sql query?

  Hi guys,.
  
  I need your help.
  Let's say I have a table: TableA. Fields in TableA are aleg, anon, monkeys. The following sentence can return, in general, several rows: select anon from TableA where aleg = somevalue. I would like to show the result of the anon column but no luck. If I try to display the results in a TextArea and the origin is a sql query shows that the first value of the line. I tried to see the like: display text (based in PLSQL) and a block anonymous plsql and coding
  DECLARE
  v_anon TableA.anon%TYPE;
  CURSOR v_cur IS
  Select anon from TableA where aleg = somevalue;
  BEGIN
  OPEN v_cur.
  LOOP
  SEEK v_cur INTO v_anon;
  EXIT WHEN v_cur % NOTFOUND;
  : FIELD_IN_FORM: = v_anon;
  END LOOP;
  CLOSE V_cur;
  END;
  but in this case, he showed no results.
  So the first question is what type of field can I use to show the result. And the second is what can I do to be able to show all the results returned by the query (if more than one line).
  
  concerning

  GMC

  We are all happy that you found a solution, but you must sign your messages and change your ID forum. It is always good to know the real name of the person you are communicating with. In addition, if you plan to get answer to your questions in the future, you must mark responses as useful or correct if they helped you solve your problems.

  Denes Kubicek
  -------------------------------------------------------------------
  http://deneskubicek.blogspot.com/
  http://www.Opal-consulting.de/training
  http://Apex.Oracle.com/pls/OTN/f?p=31517:1
  -------------------------------------------------------------------

• Need help on the design of a PL/SQl query?

  Hi all
  
  I question that brand a customer details on some factors and gives it a rank
  
  I have a table called agents which keeps officers details.
  
  Here's how to get the rank:
  Select *.
  BeO
  Select client_name, agent_number
  customer_number,
  customer_score,
  agent_score,
  customer_score + agent_score total_score,
  Rank() over (PARTITION BY name ORDER BY (customer_score + agent_score) DESC) as Ranker
  the customer, agents)
  where ranker = 1
  
  The agents table has agent_number and agent_Name...
  
  By the above query, I check various score against every available agent and assign customers to agents with the highest score or rank is 1
  
  But my problem is who to go through the process... I mean after the intial customers are assinged with agents
  
  It should work again for subscribers and for the officers who are free (and rank for them is 1) should be checked for the allocation of
  
  Once all the agents are assigned, but more customers are available then agents even thoug they are already assigned to that they can now be assigned because no agent is free...
  
  Hope my quetsion is clear... Sorry for my English... starting from PL/SQl

  Solomon Yakobson says:
  If the customer number is a string, change of type ranking_obj and you should be all together regarding the display of the zeros.

  In fact, it's a bit more than that. And I found an error in my code. List of agents must be removed before running SQL, not after:

  drop type ranking_obj_tbl
  /
  create or replace
    type NumList
      as table of number;
  /
  create or replace
    type StrList
      as table of varchar2(4000);
  /
  create or replace
    type ranking_obj
      as object(
                customer_name  varchar2(20),
                customer_num   varchar2(20),
                customer_score number,
                agent_name     varchar2(20),
                agent_number   number,
                agent_score    number,
                total_score    number
               )
  /
  create or replace
    type ranking_obj_tbl
      as table of ranking_obj
  /
  create or replace
    function ranking_function
      return ranking_obj_tbl
      pipelined
      is
          v_agent_number_list NumList := NumList();
          v_customer_number_list StrList := StrList();
          v_customer_cnt number;
          v_agent_cnt number;
          v_ranking_obj ranking_obj;
      begin
          select  count(distinct agent_number)
            into  v_agent_cnt
            from  agent;
          select  count(*)
            into  v_customer_cnt
            from  customer;
          for v_i in 1..v_customer_cnt loop
            if v_agent_number_list.count = v_agent_cnt
              then v_agent_number_list.delete;
            end if;
            select  ranking_obj(
                                customer_name,
                                customer_num,
                                customer_score,
                                agent_name,
                                agent_number,
                                agent_score,
                                total_score
                               )
              into  v_ranking_obj
              from  (
                     select  customer_name,
                             c.customer_num,
                             customer_score,
                             agent_name,
                             agent_number,
                             agent_score,
                             customer_score + agent_score total_score,
                             row_number() over(order by customer_score + agent_score desc) rn
                       from  customer c,
                             agent a
                       where c.customer_num = a.customer_num
                         and c.customer_num not member of v_customer_number_list
                         and a.agent_number not member of v_agent_number_list
                    )
              where rn = 1;
            v_agent_number_list.extend;
            v_agent_number_list(v_agent_number_list.count) := v_ranking_obj.agent_number;
            v_customer_number_list.extend;
            v_customer_number_list(v_customer_number_list.count) := v_ranking_obj.customer_num;
            pipe row(v_ranking_obj);
          end loop;
          return;
  end;
  /
  select  *
    from  customer
  /
  select  *
    from  agent
  /
  select  *
    from  table(ranking_function)
  / 
  

  For example:

  SQL> drop type ranking_obj_tbl
    2  /
  
  Type dropped.
  
  SQL> create or replace
    2    type NumList
    3      as table of number;
    4  / 
  
  Type created.
  
  SQL> create or replace
    2    type StrList
    3      as table of varchar2(4000);
    4  / 
  
  Type created.
  
  SQL> create or replace
    2    type ranking_obj
    3      as object(
    4                customer_name  varchar2(20),
    5                customer_num   varchar2(20),
    6                customer_score number,
    7                agent_name     varchar2(20),
    8                agent_number   number,
    9                agent_score    number,
   10                total_score    number
   11               )
   12  / 
  
  Type created.
  
  SQL> create or replace
    2    type ranking_obj_tbl
    3      as table of ranking_obj
    4  / 
  
  Type created.
  
  SQL> create or replace
    2    function ranking_function
    3      return ranking_obj_tbl
    4      pipelined
    5      is
    6          v_agent_number_list NumList := NumList();
    7          v_customer_number_list StrList := StrList();
    8          v_customer_cnt number;
    9          v_agent_cnt number;
   10          v_ranking_obj ranking_obj;
   11      begin
   12          select  count(distinct agent_number)
   13            into  v_agent_cnt
   14            from  agent;
   15          select  count(*)
   16            into  v_customer_cnt
   17            from  customer;
   18          for v_i in 1..v_customer_cnt loop
   19            if v_agent_number_list.count = v_agent_cnt
   20              then v_agent_number_list.delete;
   21            end if;
   22            select  ranking_obj(
   23                                customer_name,
   24                                customer_num,
   25                                customer_score,
   26                                agent_name,
   27                                agent_number,
   28                                agent_score,
   29                                total_score
   30                               )
   31              into  v_ranking_obj
   32              from  (
   33                     select  customer_name,
   34                             c.customer_num,
   35                             customer_score,
   36                             agent_name,
   37                             agent_number,
   38                             agent_score,
   39                             customer_score + agent_score total_score,
   40                             row_number() over(order by customer_score + agent_score desc) rn
   41                       from  customer c,
   42                             agent a
   43                       where c.customer_num = a.customer_num
   44                         and c.customer_num not member of v_customer_number_list
   45                         and a.agent_number not member of v_agent_number_list
   46                    )
   47              where rn = 1;
   48            v_agent_number_list.extend;
   49            v_agent_number_list(v_agent_number_list.count) := v_ranking_obj.agent_number;
   50            v_customer_number_list.extend;
   51            v_customer_number_list(v_customer_number_list.count) := v_ranking_obj.customer_num;
   52            pipe row(v_ranking_obj);
   53          end loop;
   54          return;
   55  end;
   56  / 
  
  Function created.
  
  SQL> select  *
    2    from  customer
    3  / 
  
  CUSTOMER_NAME        CUSTOMER_NUM         CUSTOMER_SCORE
  -------------------- -------------------- --------------
  Customer1            00945982                         25
  Customer2            00998643                         25
  Customer3            01021982                         25
  Customer4            01033503                         25
  Customer5            10945982                         25
  Customer6            10998643                         25
  Customer7            11021982                         25
  Customer8            11033503                         25
  
  8 rows selected.
  
  SQL> select  *
    2    from  agent
    3  / 
  
  AGENT_NUMBER AGENT_NAME           AGENT_SCORE CUSTOMER_NUM
  ------------ -------------------- ----------- --------------------
            43 Agent1                        50 00945982
            79 Agent2                        25 00945982
            15 Agent3                        25 00945982
            15 Agent3                        25 00998643
            79 Agent2                        25 00998643
            43 Agent1                       100 00998643
            15 Agent3                        25 01021982
            79 Agent2                        50 01021982
            43 Agent1                        50 01021982
            15 Agent3                        25 01033503
            43 Agent1                        50 01033503
  
  AGENT_NUMBER AGENT_NAME           AGENT_SCORE CUSTOMER_NUM
  ------------ -------------------- ----------- --------------------
            79 Agent2                        50 01033503
            43 Agent1                        50 10945982
            79 Agent2                        25 10945982
            15 Agent3                        25 10945982
            15 Agent3                        25 10998643
            79 Agent2                        25 10998643
            43 Agent1                       100 10998643
            15 Agent3                        25 11021982
            79 Agent2                        50 11021982
            43 Agent1                        50 11021982
            15 Agent3                        25 11033503
  
  AGENT_NUMBER AGENT_NAME           AGENT_SCORE CUSTOMER_NUM
  ------------ -------------------- ----------- --------------------
            43 Agent1                        50 11033503
            79 Agent2                        50 11033503
  
  24 rows selected.
  
  SQL> select  *
    2    from  table(ranking_function)
    3  /
  
  CUSTOMER_NAME        CUSTOMER_NUM         CUSTOMER_SCORE AGENT_NAME           AGENT_NUMBER AGENT_SCORE TOTAL_SCORE
  -------------------- -------------------- -------------- -------------------- ------------ ----------- -----------
  Customer2            00998643                         25 Agent1                         43         100         125
  Customer3            01021982                         25 Agent2                         79          50          75
  Customer1            00945982                         25 Agent3                         15          25          50
  Customer6            10998643                         25 Agent1                         43         100         125
  Customer4            01033503                         25 Agent2                         79          50          75
  Customer5            10945982                         25 Agent3                         15          25          50
  Customer7            11021982                         25 Agent2                         79          50          75
  Customer8            11033503                         25 Agent1                         43          50          75
  
  8 rows selected.
  
  SQL>  
  

  SY.

  Published by: Solomon Yakobson, July 25, 2011 16:42

• Create the worksheet based on a Sql query

  Hello
  
  I use SQL Developer Version 1.5.5.
  I want to run 3 different sql queries and then to get the result in the worksheet 3 in a xls.
  Now, when I run the query in the results tab.
  Yes
  Right-click-> export data-> xls
  But the results come in 3 xls and not as as I want. Is it possible to do?
  
  Thank you

  You can export as text to the Clipboard and paste the results yourself in Excel.

  Have fun
  K.

• How to measure the performance of the sql query?

  Hi Experts,
  
  How to measure the cost of performance, efficiency and CPU of an sql query?
  
  What are all the measures available to a sql query?
  
  How to identify the optimal query writing?
  
  I use Oracle 9i...
  
  It'll be useful for me to write the effective query...
  
  Thanks and greetings

  PSRAM wrote:
  Could you tell me how to activate the PLUSTRACE role?

  First put on when you do a search on PLUSTRACE: http://forums.oracle.com/forums/search.jspa?threadID=&q=plustrace&objID=f75&dateRange=all&numResults=15&rankBy=10001

  Kind regards
  Rob.

• IOM - OID! provisioning of the OID groups-QUICK HELP REQUIRED users

  Hello
  
  I installed IOM connected to OID.
  
  I've been assigned certain tasks:
  
  (1) creation of access policy such that when a user is created in the IOM, it is put into service in two groups in OID... ie. in cn = users and cn = employees (where cn employees = is the group I created under cn = Groups, dc = ad, dc = company, dc = com)
  
  (2) creation of an access policy so that when a user is created in the IOM, it is put into service two additional groups in OID, say I created two custom in IOM and membership rules groups attached to them. Now when I create a user meets the rule of two members, it is attributed to these two IOM groups and placed in service in the cn = users, dc = ad, dc is company, dc = com cn = group1, cn = groups, dc = ad, dc is company, dc = com and cn = Group2, dc = ad, dc = company, dc = com.
  
  I also want to fill these OID groups in a child table and create their research in the form of process
  
  Please help me to realize and understand these concepts.
  
  The task of the OID Lookup Recon for group works well, lookup.oid.group is populated with values.
  How can these groups be filled with child form process table (table of OID user group).
  
  Edited by: Nelly Saluja on February 12, 2010 12:51 AM

  As mentioned in my other post, you can make these groups in the form of access policy, and all users assigned by this policy will get these groups. Any question to go back.