Partitioning -- 3c : Unique Index[es] on a Partitioned Table

Let's explore what sort of Unique Indexes you can create on a Partitioned Table.

There are three types of partitioning for Indexes :

a  Global (Non-Partitioned)

b  Global Partitioned

c  Local Partitioned

Can a Unique Index be created using either type ?

Let me start with another table, SALES_DATA_2  which has the same structure and Partition Key as SALES_DATA, except that it doesn't have the Primary Key definition that builds the Unique Index.

SQL> l
  1  CREATE TABLE SALES_DATA_2
  2 ( SALE_ID NUMBER,
  3   SALE_DATE DATE,
  4   INVOICE_NUMBER VARCHAR2(21),
  5   CUSTOMER_ID NUMBER,
  6   PRODUCT_ID NUMBER,
  7   SALE_VALUE NUMBER
  8 )
  9    TABLESPACE HEMANT
 10    PARTITION BY RANGE (SALE_DATE)
 11   (PARTITION P_2018  VALUES LESS THAN (TO_DATE(' 2019-01-01','YYYY-MM-DD'))
 12    TABLESPACE TBS_YEAR_2018 ,
 13   PARTITION P_2019 VALUES LESS THAN (TO_DATE(' 2020-01-01','YYYY-MM-DD'))
 14    TABLESPACE TBS_YEAR_2019 ,
 15   PARTITION P_2020 VALUES LESS THAN (TO_DATE(' 2021-01-01','YYYY-MM-DD'))
 16    TABLESPACE TBS_YEAR_2020 ,
 17   PARTITION P_MAXVALUE VALUES LESS THAN (MAXVALUE)
 18*   TABLESPACE HEMANT )
SQL> /

Table created.

SQL> 

Next, I try a Global (Non-Partitioned) Unique Index on SALE_ID.  Note that the "GLOBAL" Keyword is optional if it is Non-Partitioned.

SQL> create unique index sales_2_uk
  2  on sales_data_2 (sale_id) global
  3  tablespace hemant
  4  /

Index created.

SQL>
SQL> select partitioned, status
  2  from user_indexes
  3  where index_name = upper('sales_2_uk')
  4  /

PAR STATUS
--- --------
NO  VALID

SQL> drop index sales_2_uk; 

Index dropped.

SQL> 

Effectively, this Global Index is the same as the Primary Key index on SALES_DATA that I built earlier.

Next, I try a Unique Global Partitioned Index on the same column.

SQL> create unique index sales_2_uk
  2  on sales_data_2 (sale_id) global
  3  partition by range (sale_id)
  4  (partition p_1mill values less than (1000001) tablespace new_indexes,
  5   partition p_2mill values less than (2000001) tablespace new_indexes,
  6   partition p_3mill values less than (3000001) tablespace new_indexes,
  7   partition p_maxval values less than (maxvalue) tablespace new_indexes)
  8  /

Index created.

SQL>
SQL> select uniqueness, partitioned, status
  2  from user_indexes
  3  where index_name = upper('sales_2_uk')
  4  /

UNIQUENES PAR STATUS
--------- --- --------
UNIQUE    YES N/A

SQL>
SQL> l
  1  select column_position, column_name
  2  from user_part_key_columns
  3  where name = upper('sales_2_uk')
  4* order by column_position
SQL> /

COLUMN_POSITION COLUMN_NAME
--------------- ----------------
              1 SALE_ID

SQL> 
SQL> select partition_name, status
  2  from user_ind_partitions
  3  where index_name = upper('sales_2_uk')
  4  order by partition_position
  5  /

PARTITION_NAME                 STATUS
------------------------------ --------
P_1MILL                        USABLE
P_2MILL                        USABLE
P_3MILL                        USABLE
P_MAXVAL                       USABLE

SQL> 

So, that is a valid Unique Global Partitioned Index.

The next attempt is a Unique Local Partitioned Index -- i.e. partitioned by the same key as the Table.

SQL> create unique index sales_2_uk
  2  on sales_data_2 (sale_id) local
  3  /
on sales_data_2 (sale_id) local
   *
ERROR at line 2:
ORA-14039: partitioning columns must form a subset of key columns of a UNIQUE
index


SQL> !oerr ora 14039              
14039, 00000, "partitioning columns must form a subset of key columns of a UNIQUE index"
// *Cause:  User attempted to create a UNIQUE partitioned index whose
//          partitioning columns do not form a subset of its key columns
//          which is illegal
// *Action: If the user, indeed, desired to create an index whose
//          partitioning columns do not form a subset of its key columns,
//          it must be created as non-UNIQUE; otherwise, correct the
//          list of key and/or partitioning columns to ensure that the index'
//          partitioning columns form a subset of its key columns

SQL>
SQL> create unique index sales_2_uk
  2  on sales_data_2 (sale_id, sale_date) local
  3  /

Index created.

SQL> 
SQL> select uniqueness, partitioned, status
  2  from user_indexes
  3  where index_name = upper('sales_2_uk')
  4  /

UNIQUENES PAR STATUS
--------- --- --------
UNIQUE    YES N/A

SQL> select column_position, column_name
  2  from user_part_key_columns
  3  where name = upper('sales_2_uk')
  4  order by column_position
  5  /

COLUMN_POSITION COLUMN_NAME
--------------- ----------------
              1 SALE_DATE

SQL> select column_position, column_name 
  2  from user_ind_columns
  3  where index_name = upper('sales_2_uk')
  4  order by column_position
  5  /

COLUMN_POSITION COLUMN_NAME
--------------- ----------------
              1 SALE_ID
              2 SALE_DATE

SQL> 
SQL> select partition_name, tablespace_name, status
  2  from user_ind_partitions
  3  where index_name = upper('sales_2_uk')
  4  order by partition_position
  5  /

PARTITION_NAME                 TABLESPACE_NAME                STATUS
------------------------------ ------------------------------ --------
P_2018                         TBS_YEAR_2018                  USABLE
P_2019                         TBS_YEAR_2019                  USABLE
P_2020                         TBS_YEAR_2020                  USABLE
P_MAXVALUE                     HEMANT                         USABLE

SQL> 

So, a Unique Local Partitioned Index must include the Table Partition Key as a subset of the Index Key columns.  This is something you must consider when Partitioning the Table and Index both.
(Also, note how USER_PART_KEY_COLUMNS doesn't show SALE_ID as a Partition Key.  This is in 11.2.0.4)