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who is oracle repository owner?

Posted on 2004-04-02
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Last Modified: 2008-03-17
Hi,
I want to know how to check to know who is the repository owner?

Thank you very much!
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Question by:fengq
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5 Comments
 
LVL 8

Expert Comment

by:baonguyen1
ID: 10744161
Do you mean the OEM repository ?


To do this:    
1.  Connect as a DBA .  
2.  Query the owner.SMP_REP_VERSION or the owner.SMP_VDS_REPOS_VERSION table,  owner is the  actual schema returned from the first query.  

Hope this helps
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Accepted Solution

by:
baonguyen1 earned 50 total points
ID: 10744188
I found this script to automate the above steps from Oracle.  Here are the instructions for running the script:
 
  1. Save the SQL script listed below to a file.
 
  2. Connect as a DBA user in SQL*Plus or SQL*Plus Worksheet.
 
  3. Run the following SQL command:  
 
       @file_name  
 
     where file_name is the full path and file name of the file saved  
     in Step #1.
 
 
Contents of the script file:
----------------------------
 
SET SERVEROUTPUT ON
 
        DECLARE  
        cursor1 integer;
        v_count number(1);
        v_schema dba_tables.owner%TYPE;
        v_version varchar2(10);
        v_component varchar2(20);
        v_i_name varchar2(10);
        v_h_name varchar2(30);
        stmt varchar2(200);
        rows_processed integer;
 
        CURSOR schema_array IS
        SELECT owner  
        FROM dba_tables WHERE table_name = 'SMP_REP_VERSION';
 
        CURSOR schema_array_v2 IS
        SELECT owner  
        FROM dba_tables WHERE table_name = 'SMP_VDS_REPOS_VERSION';
 
        BEGIN
        DBMS_OUTPUT.PUT_LINE ('.');
        DBMS_OUTPUT.PUT_LINE ('OEM REPOSITORY LOCATIONS');
 
        select instance_name,host_name into v_i_name, v_h_name from
        v$instance;
        DBMS_OUTPUT.PUT_LINE ('Instance: '||v_i_name||' on host: '||v_h_name);
 
        OPEN schema_array;
        OPEN schema_array_v2;
 
        cursor1:=dbms_sql.open_cursor;
 
        v_count := 0;
 
        LOOP -- this loop steps through each valid schema.
        FETCH schema_array INTO v_schema;
        EXIT WHEN schema_array%notfound;
        v_count := v_count + 1;
        dbms_sql.parse(cursor1,'select c_current_version, c_component from
        '||v_schema||'.smp_rep_version', dbms_sql.native);
        dbms_sql.define_column(cursor1, 1, v_version, 10);
        dbms_sql.define_column(cursor1, 2, v_component, 20);
 
        rows_processed:=dbms_sql.execute ( cursor1 );
 
        loop -- to step through cursor1 to find console version.
        if dbms_sql.fetch_rows(cursor1) >0 then
        dbms_sql.column_value (cursor1, 1, v_version);
        dbms_sql.column_value (cursor1, 2, v_component);
        if v_component = 'CONSOLE' then
        dbms_output.put_line ('Schema '||rpad(v_schema,15)||' has a repository
        version '||v_version);
        exit;
 
        end if;
        else
        exit;
        end if;
        end loop;
 
        END LOOP;
 
        LOOP -- this loop steps through each valid V2 schema.
        FETCH schema_array_v2 INTO v_schema;
        EXIT WHEN schema_array_v2%notfound;
 
        v_count := v_count + 1;
        dbms_output.put_line ( 'Schema '||rpad(v_schema,15)||' has a repository
        version 2.x' );
        end loop;
 
        dbms_sql.close_cursor (cursor1);
        close schema_array;
        close schema_array_v2;
        if v_count = 0 then
        dbms_output.put_line ( 'There are NO OEM repositories in this instance.');
        end if;
        end;
        /
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LVL 1

Author Comment

by:fengq
ID: 10744638
Thanks, but I am actually looking for the answer for Oracle Designer Repository...
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LVL 3

Expert Comment

by:dbms_chu
ID: 10751068
It looks like the Oracle Designer Repository installation is similar to the OEM repository installation in that the DBA/installer can define the repos_owner at install time.  However, according to the 9i SCM Repository Installation Guide, the default repos_owner is REPOS_MANAGER.

http://otn.oracle.com/documentation/designer.html


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LVL 22

Expert Comment

by:Helena Marková
ID: 10755303
In RON you can see it in a property pallette.
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