How to find the users connected to the system

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  The test user connection is requested from their system, but he could not succeed because they do not have the password.
  
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  --------------------------------------------------------------------------------
  OS_PROCESS       TO_CHAR(TIMESTAMP,'
  ---------------- -------------------
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  I need your help, and I know that, "Yes, we can change." Change this thread to a question answered.
  
  So: How to find the child level for each table in a relational model?
  
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  .
       O /* This is a child who makes N references to each of the follow N parent tables (here: three), and so on. */
      /↑\ Fks
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     ↑ Fk
    "º"

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  I started to think about the following:
  
  First of all, I need to know how to take the kids:

  select distinct child.table_name child
    from all_cons_columns father
    join all_cons_columns child
   using (owner, position)
    join (select child.owner,
                 child.constraint_name fk,
                 child.table_name child,
                 child.r_constraint_name pk,
                 father.table_name father
            from all_constraints father, all_constraints child
           where child.r_owner = father.owner
             and child.r_constraint_name = father.constraint_name
             and father.constraint_type in ('P', 'U')
             and child.constraint_type = 'R'
             and child.owner = 'OWNER') aux
   using (owner)
   where child.constraint_name = aux.fk
     and child.table_name = aux.child
     and father.constraint_name = aux.pk
     and father.table_name = aux.father;

  Thought...
  We will share!
  
  Thanks in advance,
  Philips
  
  Published by: BluShadow on April 1st, 2011 15:08
  formatting of code and hierarchy for readbility

  Have you looked to see if there is a cycle in the graph of dependence? Is there a table that has a foreign key to B and B has a back of A foreign key?

  SQL> create table my_emp (
    2    emp_id number primary key,
    3    emp_name varchar2(10),
    4    manager_id number
    5  );
  
  Table created.
  
  SQL> ed
  Wrote file afiedt.buf
  
    1  create table my_mgr (
    2    manager_id number primary key,
    3    employee_id number references my_emp( emp_id ),
    4    purchasing_authority number
    5* )
  SQL> /
  
  Table created.
  
  SQL> alter table my_emp
    2    add constraint fk_emp_mgr foreign key( manager_id )
    3         references my_mgr( manager_id );
  
  Table altered.
  
  SQL> ed
  Wrote file afiedt.buf
  
    1   select level lvl,
    2          child_table_name,
    3          sys_connect_by_path( child_table_name, '/' ) path
    4     from (select parent.table_name      parent_table_name,
    5                  parent.constraint_name parent_constraint_name,
    6                  child.table_name        child_table_name,
    7                  child.constraint_name   child_constraint_name
    8             from user_constraints parent,
    9                  user_constraints child
   10            where child.constraint_type = 'R'
   11              and parent.constraint_type = 'P'
   12              and child.r_constraint_name = parent.constraint_name
   13           union all
   14           select null,
   15                  null,
   16                  table_name,
   17                  constraint_name
   18             from user_constraints
   19            where constraint_type = 'P')
   20    start with child_table_name = 'MY_EMP'
   21*  connect by prior child_table_name = parent_table_name
  SQL> /
  ERROR:
  ORA-01436: CONNECT BY loop in user data
  

  If you have a cycle, you have some problems.

  (1) it is a NOCYCLE keyword does not cause the error, but that probably requires an Oracle version which is not so far off support. I don't think it was available at the time 9.2 but I don't have anything old enough to test on

  SQL> ed
  Wrote file afiedt.buf
  
    1   select level lvl,
    2          child_table_name,
    3          sys_connect_by_path( child_table_name, '/' ) path
    4     from (select parent.table_name      parent_table_name,
    5                  parent.constraint_name parent_constraint_name,
    6                  child.table_name        child_table_name,
    7                  child.constraint_name   child_constraint_name
    8             from user_constraints parent,
    9                  user_constraints child
   10            where child.constraint_type = 'R'
   11              and parent.constraint_type = 'P'
   12              and child.r_constraint_name = parent.constraint_name
   13           union all
   14           select null,
   15                  null,
   16                  table_name,
   17                  constraint_name
   18             from user_constraints
   19            where constraint_type = 'P')
   20    start with child_table_name = 'MY_EMP'
   21*  connect by nocycle prior child_table_name = parent_table_name
  SQL> /
  
         LVL CHILD_TABLE_NAME               PATH
  ---------- ------------------------------ --------------------
           1 MY_EMP                         /MY_EMP
           2 MY_MGR                         /MY_EMP/MY_MGR
           1 MY_EMP                         /MY_EMP
           2 MY_MGR                         /MY_EMP/MY_MGR
  

  (2) If you try to write on a table and all of its constraints in a file and do it in a valid order, the entire solution is probably wrong. It is impossible, for example, to generate the DDL for MY_EMP and MY_DEPT such as all instructions for a table come first, and all the instructions for the other are generated second. So even if NOCYCLE to avoid the error, you would end up with an invalid DDL script. If that's the problem, I would rethink the approach.

  -Generate the DDL for all tables without constraint
  -Can generate the DDL for all primary key constraints
  -Can generate the DDL for all unique key constraints
  -Can generate the DDL for all foreign key constraints

  This is not solidarity all the DOF for a given in the file object. But the SQL will be radically simpler writing - there will be no need to even look at the dependency graph.

  Justin

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