Thursday, February 16, 2017

Create a globally partitioned hash index


Partitioned table layout:

CREATE TABLE SCOTT.ARCHIVED_NOTES
(
  NOTE_ID                     NUMBER(10)        NOT NULL,
  PPT_ID                      NUMBER(10),
  REGISTRED_DATE              DATE              NOT NULL,
  EDITED_DATE                 DATE,
  ALTERED_BY                  VARCHAR2(30 BYTE) NOT NULL,
  ALTERED_DATE                DATE              NOT NULL,
  CONTENT_CODE                VARCHAR2(4 BYTE),
  NOTEFORMAT                  VARCHAR2(1 BYTE)
)
TABLESPACE USERS
PARTITION BY RANGE (NOTE_ID)
(  
  PARTITION ARCHIVED20 VALUES LESS THAN (21000000),  
  PARTITION ARCHIVED21 VALUES LESS THAN (22000000),
  PARTITION ARCHIVED22 VALUES LESS THAN (23000000)
)
;

Create a global hash-partitioned index. Notice the GLOBAL keyword:
CREATE INDEX ARCHIVED_NOTES_IDX2 ON ARCHIVED_NOTES (REGISTRED_DATE)
  GLOBAL PARTITION BY HASH (REGISTRED_DATE)
  PARTITIONS 3;


Verify its existence:
SELECT INDEX_NAME,PARTITION_NAME 
FROM USER_IND_PARTITIONS 
WHERE INDEX_NAME = (SELECT INDEX_NAME FROM USER_INDEXES WHERE TABLE_NAME='ARCHIVED_NOTES'); 

INDEX_NAME PARTITION_NAME
ARCHIVED_NOTES_IDX2 SYS_P441
ARCHIVED_NOTES_IDX2 SYS_P442
ARCHIVED_NOTES_IDX2 SYS_P443


Let's take a deeper look at the index properties:
SELECT C.TABLE_NAME,I.INDEX_NAME,I.PARTITIONING_TYPE, I.SUBPARTITIONING_TYPE, I.LOCALITY,I.ALIGNMENT,C.COLUMN_NAME,C.COLUMN_POSITION
FROM USER_PART_INDEXES I JOIN USER_IND_COLUMNS C
ON (I.INDEX_NAME = C.INDEX_NAME)
WHERE I.TABLE_NAME IN (SELECT TABLE_NAME FROM USER_TABLES WHERE PARTITIONED='YES')
ORDER BY 1,2,7;

TABLE_NAME INDEX_NAME PARTITIONING_TYPE SUBPARTITIONING_TYPE LOCALITY ALIGNMENT COLUMN_NAME COLUMN_POSITION
ARCHIVED_NOTES ARCHIVED_NOTES_IDX2 HASH NONE GLOBAL PREFIXED REGISTRED_DATE
1

For global partitioned indexes, the "alignment" column will always show the index as "prefixed". This is the only supported index type for global partitioned indexes.


Create a globally partitioned range index


CREATE TABLE SCOTT.ARCHIVED_NOTES
(
  NOTE_ID                     NUMBER(10)        NOT NULL,
  PPT_ID                      NUMBER(10),
  REGISTRED_DATE              DATE              NOT NULL,
  EDITED_DATE                 DATE,
  ALTERED_BY                  VARCHAR2(30 BYTE) NOT NULL,
  ALTERED_DATE                DATE              NOT NULL,
  CONTENT_CODE                VARCHAR2(4 BYTE),
  NOTEFORMAT                  VARCHAR2(1 BYTE)
)
TABLESPACE USERS
PARTITION BY RANGE (NOTE_ID)
(  
  PARTITION ARCHIVED20 VALUES LESS THAN (21000000),  
  PARTITION ARCHIVED21 VALUES LESS THAN (22000000),
  PARTITION ARCHIVED22 VALUES LESS THAN (23000000)
)
;

Create a global range partitioned index. Notice the GLOBAL keyword:
CREATE INDEX SCOTT.ARCHIVED_NOTES_IDX1 
ON SCOTT.ARCHIVED_NOTES(NOTE_ID)
GLOBAL PARTITION BY RANGE(NOTE_ID)(  
  PARTITION ARCHIVED20 VALUES LESS THAN (21000000),  
  PARTITION ARCHIVED21 VALUES LESS THAN (22000000),  
  PARTITION ARCHIVED22 VALUES LESS THAN (23000000),
  PARTITION ARCHIVED_OTHERS VALUES LESS THAN (MAXVALUE)
);

Also note that Oracle required the MAXVALUES clause as the last partition in the index, to ensure that all rows in the table will be represented in the index. Without MAXVALUES, Oracle will throw the error:
ORA-14021: MAXVALUE must be specified for all columns

Verify its existence:
SELECT INDEX_NAME,PARTITION_NAME 
FROM USER_IND_PARTITIONS 
WHERE INDEX_NAME = (SELECT INDEX_NAME FROM USER_INDEXES WHERE TABLE_NAME='ARCHIVED_NOTES'); 

INDEX_NAME PARTITION_NAME
ARCHIVED_NOTES_IDX1 ARCHIVED20
ARCHIVED_NOTES_IDX1 ARCHIVED21
ARCHIVED_NOTES_IDX1 ARCHIVED22
ARCHIVED_NOTES_IDX1 ARCHIVED_OTHERS

You could also make the index UNIQUE:
CREATE UNIQUE INDEX SCOTT.ARCHIVED_NOTES_IDX1 ... 

A unique index like this could support a primary key constraint on the table, if desirable:
ALTER TABLE ARCHIVED_NOTES ADD CONSTRAINT ARCHIVED_NOTES_PK
PRIMARY KEY (NOTE_ID)
USING INDEX ARCHIVED_NOTES_IDX1;

Let's take a deeper look at the index Properties:
SELECT C.TABLE_NAME,I.INDEX_NAME,I.PARTITIONING_TYPE, I.SUBPARTITIONING_TYPE, I.LOCALITY,I.ALIGNMENT,C.COLUMN_NAME,C.COLUMN_POSITION
FROM USER_PART_INDEXES I JOIN USER_IND_COLUMNS C
ON (I.INDEX_NAME = C.INDEX_NAME)
WHERE I.TABLE_NAME IN (SELECT TABLE_NAME FROM USER_TABLES WHERE PARTITIONED='YES')
ORDER BY 1,2,7;

TABLE_NAME INDEX_NAME PARTITIONING_TYPE SUBPARTITIONING_TYPE LOCALITY ALIGNMENT COLUMN_NAME COLUMN_POSITION
ARCHIVED_NOTES ARCHIVED_NOTES_IDX1 RANGE NONE GLOBAL PREFIXED NOTE_ID
1


For globally partitioned indexes, the "alignment" column will always show the index as "prefixed". This is the only supported index type for global partitioned indexes.

An important point regarding globally partitioned indexes is pointed out in the Oracle Documentation:

"Normally, a global index is not equipartitioned with the underlying table."

and

"There is nothing to prevent an index from being equipartitioned with the underlying table, but Oracle does not take advantage of the equipartitioning when generating query plans or executing partition maintenance operations. So an index that is equipartitioned with the underlying table should be created as LOCAL."

In other words, while the example above would work, it may not be a good idea. It would make better sense for the global partitioned index to be created on another column:
CREATE INDEX SCOTT.ARCHIVED_NOTES_IDX1 
ON SCOTT.ARCHIVED_NOTES(REGISTRED_DATE)
GLOBAL PARTITION BY RANGE(REGISTRED_DATE)(  
  PARTITION NOTES_2012 VALUES LESS THAN (to_date('01.01.2013', 'dd.mm.yyyy')),  
  PARTITION NOTES_2013 VALUES LESS THAN (to_date('01.01.2014', 'dd.mm.yyyy')),  
  PARTITION NOTES_2014 VALUES LESS THAN (to_date('01.01.2015', 'dd.mm.yyyy'))
  ,PARTITION ARCHIVED_OTHERS VALUES LESS THAN (MAXVALUE)
);


Create a local non-prefixed partitioned index


The table below is partitioned by range, using the column NOTE_ID as the partition key:

CREATE TABLE SCOTT.ARCHIVED_NOTES
(
  NOTE_ID                     NUMBER(10)        NOT NULL,
  PPT_ID                      NUMBER(10),
  REGISTRED_DATE              DATE              NOT NULL,
  EDITED_DATE                 DATE,
  ALTERED_BY                  VARCHAR2(30 BYTE) NOT NULL,
  ALTERED_DATE                DATE              NOT NULL,
  CONTENT_CODE                VARCHAR2(4 BYTE),
  NOTEFORMAT                  VARCHAR2(1 BYTE)
)
TABLESPACE USERS
PARTITION BY RANGE (NOTE_ID)
(  
  PARTITION ARCHIVED20 VALUES LESS THAN (21000000),  
  PARTITION ARCHIVED21 VALUES LESS THAN (22000000),
  PARTITION ARCHIVED22 VALUES LESS THAN (23000000)
)
;

Create a global partitioned index:
CREATE INDEX SCOTT.REGISTRED_DATE_SK1 ON SCOTT.ARCHIVED_NOTES
(REGISTRED_DATE)
LOCAL (  
  PARTITION ARCHIVED20,  
  PARTITION ARCHIVED21,  
  PARTITION ARCHIVED22
)

Confirm its existence:
SELECT INDEX_NAME,PARTITIONED
FROM USER_INDEXES
WHERE TABLE_NAME='ARCHIVED_NOTES';

INDEX_NAME PARTITIONED
REGISTRED_DATE_SK1 YES


Confirm that the expected index partitions have been created:
SELECT INDEX_NAME,PARTITION_NAME 
FROM USER_IND_PARTITIONS 
WHERE INDEX_NAME = (SELECT INDEX_NAME FROM USER_INDEXES WHERE TABLE_NAME='ARCHIVED_NOTES'); 

INDEX_NAME PARTITION_NAME
REGISTRED_DATE_SK1 ARCHIVED20
REGISTRED_DATE_SK1 ARCHIVED21
REGISTRED_DATE_SK1 ARCHIVED22



Add a new partition:
ALTER TABLE SCOTT.ARCHIVED_NOTES
 ADD 
  PARTITION ARCHIVED23 VALUES LESS THAN (24000000);

If you confirm the index partitions again, you'll see that a new index partition has been added:

INDEX_NAME PARTITION_NAME
REGISTRED_DATE_SK1 ARCHIVED23

Let's take a deeper look at the properties of this index:
SELECT C.TABLE_NAME,I.INDEX_NAME,I.PARTITIONING_TYPE, I.SUBPARTITIONING_TYPE, I.LOCALITY,I.ALIGNMENT,C.COLUMN_NAME,C.COLUMN_POSITION
FROM USER_PART_INDEXES I JOIN USER_IND_COLUMNS C
ON (I.INDEX_NAME = C.INDEX_NAME)
WHERE I.TABLE_NAME IN (SELECT TABLE_NAME FROM USER_TABLES WHERE PARTITIONED='YES')
ORDER BY 1,2,7;

TABLE_NAME INDEX_NAME PARTITIONING_TYPE SUBPARTITIONING_TYPE LOCALITY ALIGNMENT COLUMN_NAME COLUMN_POSITION
ARCHIVED_NOTES REGISTRED_DATE_SK1 RANGE NONE LOCAL NON_PREFIXED REGISTRED_DATE
1



The "alignment" column of the above output reveals that the index is of type "non_prefixed", which means that the partitioning key is not the first column in the index.



Wednesday, February 15, 2017

Create a local prefixed partitioned index


The table below is partitioned by range, using the column NOTE_ID as the partition key:

CREATE TABLE SCOTT.ARCHIVED_NOTES
(
  NOTE_ID                     NUMBER(10)        NOT NULL,
  PPT_ID                      NUMBER(10),
  REGISTRED_DATE              DATE              NOT NULL,
  EDITED_DATE                 DATE,
  ALTERED_BY                  VARCHAR2(30 BYTE) NOT NULL,
  ALTERED_DATE                DATE              NOT NULL,
  CONTENT_CODE                VARCHAR2(4 BYTE),
  NOTEFORMAT                  VARCHAR2(1 BYTE)
)
TABLESPACE USERS
PARTITION BY RANGE (NOTE_ID)
(  
  PARTITION ARCHIVED20 VALUES LESS THAN (21000000),  
  PARTITION ARCHIVED21 VALUES LESS THAN (22000000),
  PARTITION ARCHIVED22 VALUES LESS THAN (23000000)
)
;

Create a local index:
CREATE UNIQUE INDEX SCOTT.ARCHIVED_NOTES_PK ON SCOTT.ARCHIVED_NOTES
(NOTE_ID)
LOCAL (  
  PARTITION ARCHIVED20,  
  PARTITION ARCHIVED21,  
  PARTITION ARCHIVED22
);

Confirm its existence:

INDEX_NAME PARTITIONED
ARCHIVED_NOTES_PK YES

The index we just created can be used to support a unique or a primary key constraint:
ALTER TABLE ARCHIVED_NOTES ADD CONSTRAINT ARCHIVED_NOTES_PK
PRIMARY KEY (NOTE_ID)
USING INDEX ARCHIVED_NOTES_PK;

Confirm that the expected index partitions have been created:
SELECT INDEX_NAME,PARTITION_NAME 
FROM USER_IND_PARTITIONS 
WHERE INDEX_NAME = (SELECT INDEX_NAME FROM USER_INDEXES WHERE TABLE_NAME='ARCHIVED_NOTES'); 

INDEX_NAME PARTITION_NAME
ARCHIVED_NOTES_PK ARCHIVED20
ARCHIVED_NOTES_PK ARCHIVED21
ARCHIVED_NOTES_PK ARCHIVED22

Add a new partition:
ALTER TABLE SCOTT.ARCHIVED_NOTES
 ADD 
  PARTITION ARCHIVED23 VALUES LESS THAN (23000000);

If you confirm the index partitions again, you'll see that a new index partition has been added:

INDEX_NAME PARTITION_NAME
ARCHIVED_NOTES_PK ARCHIVED23

Let's take a deeper look at the properties of this index:
SELECT C.TABLE_NAME,I.INDEX_NAME,I.PARTITIONING_TYPE, I.SUBPARTITIONING_TYPE, I.LOCALITY,I.ALIGNMENT,C.COLUMN_NAME,C.COLUMN_POSITION
FROM USER_PART_INDEXES I JOIN USER_IND_COLUMNS C
ON (I.INDEX_NAME = C.INDEX_NAME)
WHERE I.TABLE_NAME IN (SELECT TABLE_NAME FROM USER_TABLES WHERE PARTITIONED='YES')
ORDER BY 1,2,7;

TABLE_NAME INDEX_NAME PARTITIONING_TYPE SUBPARTITIONING_TYPE LOCALITY ALIGNMENT COLUMN_NAME COLUMN_POSITION
ARCHIVED_NOTES ARCHIVED_NOTES_PK RANGE NONE LOCAL PREFIXED NOTE_ID
1


The "alignment" column of the above output reveals that the index is of type "prefixed", which means that the partitioning key is the first column in the index.



How to lock table statstics in Oracle

Use dbms_stats.lock_table_stats

conn scott
exec dbms_stats.lock_table_stats('SCOTT','DEPT');

Verify the setting:

select table_name,object_type,stattype_locked 
from user_tab_statistics;

TABLE_NAME                     OBJECT_TYPE                          STATTYPE_LOCKED
------------------------------ ------------------------------------ ---------------
DEPT                           TABLE                                ALL
EMP                            TABLE
BONUS                          TABLE
SALGRADE                       TABLE
RECEIVED_DOCUMENTS             TABLE

Lock stats on a specific partition:
exec dbms_stats.lock_partition_stats('SCOTT','RECEIVED_DOCUMENTS','SETTELM');
Verify the setting:
select table_name,partition_name,subpartition_name,object_type,stattype_locked 
from user_tab_statistics 
where table_name='RECEIVED_DOCUMENTS';

TABLE_NAME                     PARTITION_NAME  SUBPARTITION_NAME OBJECT_TYPE          STATTYPE_LOCKED
------------------------------ --------------- ----------------- -------------------- ---------------
RECEIVED_DOCUMENTS                                               TABLE
RECEIVED_DOCUMENTS             APPLICANT                         PARTITION
RECEIVED_DOCUMENTS             APPLICATIONS                      PARTITION
RECEIVED_DOCUMENTS             SETTELM                           PARTITION            ALL

To reverse the process:
exec dbms_stats.unlock_table_stats('SCOTT','DEPT);
exec dbms_stats.unlock_partition_stats('SCOTT','RECEIVED_DOCUMENTS','SETTELM');

Wednesday, February 8, 2017

How to check if the "gather_database_stats" automatic maintenance jobs has succseeded

select id,operation,to_char(start_time,'dd.mm.yyyy hh24:mi:ss') "starttime",
                        to_char(end_Time,'dd.mm.yyyy hh24:mi:ss') "endtime",
                        to_dsinterval(to_char(end_time-start_Time,'dd.mm.yyyy hh24:mi:ss')) "elapsed",
                        status,job_name
from DBA_OPTSTAT_OPERATIONS
where operation = 'gather_database_stats (auto)'
order by start_time desc
fetch first 5 rows only;

Gave the following result:
ID OPERATION starttime endtime elapsed STATUS JOB_NAME
58778
gather_database_stats (auto) 07.02.2017 22:00:02 08.02.2017 02:00:15 +00 04:00:13.178197 TIMED OUT ORA$AT_OS_OPT_SY_1965
58585
gather_database_stats (auto) 06.02.2017 22:00:06 07.02.2017 02:00:24 +00 04:00:18.089479 TIMED OUT ORA$AT_OS_OPT_SY_1963
58273
gather_database_stats (auto) 05.02.2017 06:00:04 06.02.2017 02:00:26 +00 20:00:21.968028 TIMED OUT ORA$AT_OS_OPT_SY_1950
58088
gather_database_stats (auto) 04.02.2017 06:00:02 05.02.2017 02:00:32 +00 20:00:29.696085 TIMED OUT ORA$AT_OS_OPT_SY_1945
58038
gather_database_stats (auto) 03.02.2017 22:00:02 04.02.2017 02:00:17 +00 04:00:15.314408 TIMED OUT ORA$AT_OS_OPT_SY_1943

The output indicates that the Automatic maintenance job never get to finish before the maintenance Windows Closes.

You can examine another view, dba_opstat_operation_tasks, to see the details on what the job didn't have time to do before the window closed:

select  opid, target,target_type, 
        to_char(start_Time,'dd.mm.yyyy hh24:mi:ss') "starttime",
        to_char(end_time,'dd.mm.yyyy hh24:mi:ss') "endtime"
        ,status
from dba_optstat_operation_tasks 
where opid=58778
order by start_Time desc
;

Sample output:
OPID TARGET TARGET_TYPE starttime endtime STATUS
58778
SCOTT.TABLE1.SYS_SUBP25275 TABLE 08.02.2017 02:00:11 08.02.2017 02:00:11 TIMED OUT
58778
SCOTT.TABLE1.SYS_SUBP25267 TABLE 08.02.2017 02:00:11 08.02.2017 02:00:11 TIMED OUT

Notice how the starttime and endtime is identical; this is because the window Closes, and the job aborts. Oracle then records this timestamp as the start time and end time for the Objects it never had time to analyze.


The Predefined Maintenance Windows in Oracle 12c

Oracle documentation states:

"By default there are seven predefined maintenance windows, each one representing a day of the week. The weekend maintenance windows, SATURDAY_WINDOW and SUNDAY_WINDOW, are longer in duration than the weekday maintenance windows. The window group MAINTENANCE_WINDOW_GROUP consists of these seven windows."

Verify by executing:
SELECT WINDOW_NAME,REPEAT_INTERVAL,DURATION,to_char(LAST_START_DATE,'dd.mm.yyyy hh24:mi') "laststart",
    to_char(NEXT_START_DATE,'dd.mm.yyyy hh24:mi') "nextstart",ENABLED "window enabled?",ACTIVE "window open?"
FROM DBA_SCHEDULER_WINDOWS 
WHERE RESOURCE_PLAN='DEFAULT_MAINTENANCE_PLAN';

WINDOW_NAME REPEAT_INTERVAL DURATION laststart nextstart window enabled? window open?
SUNDAY_WINDOW freq=daily;byday=SUN;byhour=6;byminute=0; bysecond=0 +00 20:00:00.000000 05.02.2017 06:00 12.02.2017 06:00 TRUE FALSE
SATURDAY_WINDOW freq=daily;byday=SAT;byhour=6;byminute=0; bysecond=0 +00 20:00:00.000000 04.02.2017 06:00 11.02.2017 06:00 TRUE FALSE
FRIDAY_WINDOW freq=daily;byday=FRI;byhour=22;byminute=0; bysecond=0 +00 04:00:00.000000 03.02.2017 22:00 10.02.2017 22:00 TRUE FALSE
THURSDAY_WINDOW freq=daily;byday=THU;byhour=22;byminute=0; bysecond=0 +00 04:00:00.000000 02.02.2017 22:00 09.02.2017 22:00 TRUE FALSE
WEDNESDAY_WINDOW freq=daily;byday=WED;byhour=22;byminute=0; bysecond=0 +00 04:00:00.000000 01.02.2017 22:00 08.02.2017 22:00 TRUE FALSE
TUESDAY_WINDOW freq=daily;byday=TUE;byhour=22;byminute=0; bysecond=0 +00 04:00:00.000000 07.02.2017 22:00 14.02.2017 22:00 TRUE FALSE
MONDAY_WINDOW freq=daily;byday=MON;byhour=22;byminute=0; bysecond=0 +00 04:00:00.000000 06.02.2017 22:00 13.02.2017 22:00 TRUE FALSE

The weekend Windows starts at 6 in the morning and run for 20 hours.
The weekday Windows starts at 22:00 and runs until 02:00 in the morning.

An overview of some important data Dictionary views can be found here