It is easy to create a default GLOBAL Index that is defined as a Unique Index on a column (or composite of columns) containing unique values.
But what if you have a Partitioned Table and want to create a LOCAL (i.e. equi-partitioned with the table) Index as a Unique Index ? Are there any constraints ?
Let me demonstrate a Partitioned table listing Users by Region Code.
But what if you have a Partitioned Table and want to create a LOCAL (i.e. equi-partitioned with the table) Index as a Unique Index ? Are there any constraints ?
Let me demonstrate a Partitioned table listing Users by Region Code.
SQL> create table users
2 (region_code varchar2(3),
3 username varchar2(30),
4 account_status varchar2(32),
5 created date,
6 profile varchar2(128))
7 partition by range (region_code)
8 (partition a_m values less than ('N'),
9 partition n_r values less than ('S'),
10 partition s_z values less than (MAXVALUE))
11 /
Table created.
SQL>
SQL> insert into users
2 select substr(username,1,3), username, account_status, created, profile
3 from dba_users
4 /
39 rows created.
SQL>
SQL> commit;
Commit complete.
SQL> exec dbms_stats.gather_table_stats('','USERS');
PL/SQL procedure successfully completed.
SQL>
SQL> col partition_name format a30
SQL> select partition_name, num_rows
2 from user_tab_partitions
3 where table_name = 'USERS'
4 order by partition_position
5 /
PARTITION_NAME NUM_ROWS
------------------------------ ----------
A_M 18
N_R 9
S_Z 12
SQL>
I now verify that I can create a GLOBAL (non-partitioned) Unique Index on USERNAME.
SQL> create unique index users_username_u1 on users(username) global;
Index created.
SQL> drop index users_username_u1;
Index dropped.
SQL> create unique index users_username_u1 on users(username);
Index created.
SQL> drop index users_username_u1;
Index dropped.
SQL>
I now verify that I can create a Non-Unique LOCAL Index (being equi-partitioned, the index is partitioned by REGION_CODE). (Being equi-partitioned, the default behaviour is the Index Partition Names inherit from the Table Partition Names).
SQL> create index users_username_l1 on users(username) local;
Index created.
SQL> select partition_name, num_rows
2 from user_ind_partitions
3 where index_name = 'USERS_USERNAME_L1'
4 order by partition_position
5 /
PARTITION_NAME NUM_ROWS
------------------------------ ----------
A_M 18
N_R 9
S_Z 12
SQL>
SQL> drop index users_username_l1;
Index dropped.
SQL>
I've proven (with the GLOBAL Index) that USERNAME is Unique across the whole table. Can I create a Unique LOCAL Index on this column ?
SQL> create unique index users_username_u_l1 on users(username) local;
create unique index users_username_u_l1 on users(username) local
*
ERROR at line 1:
ORA-14039: partitioning columns must form a subset of key columns of a UNIQUE index
SQL>
Note the Error Message. The Partition Key must be a subset of the Unique Index columns. Let me try adding the Partition Key (in my example table, the Partition Key is a single column -- it could be a composite of multiple columns. In that case all the columns of the Partition Key must for a subset of the Unique Index).
SQL> create unique index users_rc_un_u_l1 on users(region_code, username) local;
Index created.
SQL> select partition_name, num_rows
2 from user_ind_partitions
3 where index_name = 'USERS_RC_UN_U_L1'
4 order by partition_position
5 /
PARTITION_NAME NUM_ROWS
------------------------------ ----------
A_M 18
N_R 9
S_Z 12
SQL> drop index users_rc_un_u_l1;
Index dropped.
SQL> create unique index users_un_rc_u_l1 on users(username, region_code) local;
Index created.
SQL> select partition_name, num_rows
2 from user_ind_partitions
3 where index_name = 'USERS_UN_RC_U_L1'
4 order by partition_position
5 /
PARTITION_NAME NUM_ROWS
------------------------------ ----------
A_M 18
N_R 9
S_Z 12
SQL>
It doesn't matter if I put the Partition Key (REGION_CODE) before or after the target column(s) (USERNAME) when I create the LOCAL Unique Index. What is necessary is that the Partition Key be a subset of the Unique Index definition.
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