Element Object version Number

Question

Where can I get the object version number of an element. I cant seem to find it in any of the element tables
Thanks

Answer 1

If I correctly understood you:
SQL> INSERT INTO checkpoint VALUES
  2  ( 1
  3  , person_typ('Ignat'));
1 row created.
SQL> INSERT INTO checkpoint VALUES
  2  ( 2
  3  , employee_typ ( 'Scott', 111 ) );
1 row created.
SQL> INSERT INTO checkpoint VALUES
  2  ( 3
  3  , visitor_typ ( 'Adams', 333 ) );
1 row created.
SQL> column person format a30
SQL> select  id,
  2          person,
  3          sys_typeid(person)
  4    from  checkpoint
  5  /
        ID PERSON(NAME)                   SYS_TYPEID(PERSON)
         1 PERSON_TYP('Ignat')            01
         2 EMPLOYEE_TYP('Scott', 111)     02
         3 VISITOR_TYP('Adams', 333)      03
SQL> As you can see, function sys_typeid gives you type "version" number. Now, how to trace that "version" number to type name? Unfortunately, data dictionary views do not provide that. This is another example of DD inefficiency. If you check SYS.TYPE$ table:
SQL> desc sys.type$
Name                                                                     Null?    Type
TOID                                                                     NOT NULL RAW(16)
VERSION#                                                                 NOT NULL NUMBER
VERSION                                                                  NOT NULL VARCHAR2(30)
TVOID                                                                    NOT NULL RAW(16)
TYPECODE                                                                 NOT NULL NUMBER
PROPERTIES                                                               NOT NULL NUMBER
ATTRIBUTES                                                                        NUMBER
METHODS                                                                           NUMBER
HIDDENMETHODS                                                                     NUMBER
SUPERTYPES                                                                        NUMBER
SUBTYPES                                                                          NUMBER
EXTERNTYPE                                                                        NUMBER
EXTERNNAME                                                                        VARCHAR2(4000)
HELPERCLASSNAME                                                                   VARCHAR2(4000)
LOCAL_ATTRS                                                                       NUMBER
LOCAL_METHODS                                                                     NUMBER
TYPEID                                                                            RAW(16)
ROOTTOID                                                                          RAW(16)
SPARE1                                                                            NUMBER
SPARE2                                                                            NUMBER
SPARE3                                                                            NUMBER
SUPERTOID                                                                         RAW(16)
HASHCODE                                                                          RAW(17)
SQL> you'll notice ROOTTOID which is, unfortunately, not mapped into DD XXX_TYPE views. What you can do is something like:
SQL> CREATE OR REPLACE
  2    TYPE TRAINEE_TYP UNDER EMPLOYEE_TYP(trainee_id NUMBER)
  3      NOT FINAL
  4  /
Type created.
SQL> set linesize 132
SQL> column owner format a5
SQL> column root_type_owner format a15
SQL> column supertype_owner format a15
SQL> column type_name format a15
SQL> column root_type_name format a15
SQL> column supertype_name format a15
SQL> column typeid format a6
SQL> select  tc.owner,
  2          tc.type_name,
  3          tc.typeid,
  4          roottoid,
  5          tc.supertype_owner,
  6          tc.supertype_name,
  7          tp.owner root_type_owner,
  8          tp.type_name root_type_name
  9    from  dba_types tc,
10          sys.type$ t,
11          dba_types tp
12    where t.toid = tc.type_oid
13      and tp.type_oid = nvl(t.roottoid,t.toid)
14      and tc.owner = 'SCOTT'
15      and tc.type_name in ('PERSON_TYP','EMPLOYEE_TYP','VISITOR_TYP','TRAINEE_TYP')
16    order by tc.typeid
17  /
OWNER TYPE_NAME       TYPEID ROOTTOID                         SUPERTYPE_OWNER SUPERTYPE_NAME  ROOT_TYPE_OWNER ROOT_TYPE_NAME
SCOTT PERSON_TYP      01                                                                      SCOTT           PERSON_TYP
SCOTT EMPLOYEE_TYP    02     33A90A8C974E4BF4B369D443E27A83B9 SCOTT           PERSON_TYP      SCOTT           PERSON_TYP
SCOTT VISITOR_TYP     03     33A90A8C974E4BF4B369D443E27A83B9 SCOTT           PERSON_TYP      SCOTT           PERSON_TYP
SCOTT TRAINEE_TYP     04     33A90A8C974E4BF4B369D443E27A83B9 SCOTT           EMPLOYEE_TYP    SCOTT           PERSON_TYP
SQL> SY.