Can the command ADD partition (in table )invalid procedures,packages. ??

Hi Folks.
I know that drop a partition make invalid public synonyms, packages, functions.
But.
Add partition, can make invalid some object?
I tested and didnt find objects invalid.
Tks

>
I tested and didnt find objects invalid.
>
Please confirm that your tests included these steps
1. created a partitioned table
2. create at least one of every possible type of object that might depend on that table
3. performed an ADD PARTITION that succeeded
4. confirmed that NONE of the objects from step #2 were invalidated
5. performed an ADD PARTITION that failed in various ways (local index partition can't be created, tablespace can't be extended, non-existent tablespace, etc)
6. confirmed that NONE of the objects from step #2 were invalidated.
See ALTER TABLE in the SQL Language Reference
http://docs.oracle.com/cd/B28359_01/server.111/b28286/statements_3001.htm
>
Note:
If you alter a table that is a master table for one or more materialized views, then Oracle Database marks the materialized views INVALID. Invalid materialized views cannot be used by query rewrite and cannot be refreshed. For information on revalidating a materialized view, see ALTER MATERIALIZED VIEW.
>
Did you test that the above MV is not invalidated?
Did you test that adding a partition did not invalidate any local or global indexes?

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Limit on the number of partitions a table can have

Hi,
It's been quite some time I'm thinking of a simple and the best solution to a performance issue in my application. Benefiting out of partitioning the table seems a simplest approach; it involves very minimal changes to my application code and in addition, it keeps the overall application logic 100% intact. However, at times, the number of partitions required can grow to as large as 500 thousand. I know of another application implementing as many as 50 thousand partitions for one of its tables. I'm wondering if Oracle recommends or poses any limit to a number of partitions the table can have. How does that limit, if it exists, vary if each of my table partition holds only a small amount of data; say, not more than 2 thousand records each of 1 KB size. What are the important considerations that one needs to keep in mind while creating the huge number of partitions for a table?
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ps: Consider Oracle releases 10g onwards.
Edited by: Aniruddh on Dec 30, 2009 9:46 AM
Edited by: Aniruddh on Dec 30, 2009 9:50 AM

Aniruddh,
>
What are the important considerations that one needs to keep in mind while creating the huge number of partitions for a table?
>
I doubt if you are using partitioning for the right causes. Please explain how is your data structured and why you need to create so many partitions..?
when creating partition, you need to consider..
a) why you need to partition?
b) how is your data structured
c) how your data is indexed.
d) What kind of queries would you firing on this table to get the data. Will they use the partition key ?
Thanks,
Rajesh.
Please mark this/any other answer as helpful or answered if it is so. If not, provide additional details/feedback.
Always try to provide create table and insert statements to help the forum members help you better.

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Hello Friends ,
Hope you are all fine and doing great.
By the way - I have a quick question on oracle 10g Partitioning of tables...
I have a table with partition as stated below ..
CREATE TABLE X_ACC_ASSETS_GPC_AGG
X_ACC_ASSETS_GPC_AGG_RK NUMBER(10) NOT NULL,
X_AS_OF_DT DATE NOT NULL,
ACCOUNT_RK NUMBER(10) NOT NULL,
X_UNIV_ACCOUNT_ID NUMBER(10),
ACCOUNT_ID VARCHAR2(10 BYTE),
X_ASSET_TYPE_CD VARCHAR2(6 BYTE),
X_AUC_AMT NUMBER(18,5),
X_FIRM_AMT NUMBER(18,5),
X_ADJ_SRCE_AMT NUMBER(18,5),
PROCESSED_DTTM DATE
)PARTITION BY RANGE (X_AS_OF_DT)
PARTITION P200712 VALUES LESS THAN (TO_DATE(' 2008-01-01 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN'))
NOLOGGING
NOCOMPRESS
TABLESPACE KAW_DATA
PCTFREE 10
INITRANS 1
MAXTRANS 255
STORAGE (
INITIAL 64K
MINEXTENTS 1
MAXEXTENTS 2147483645
BUFFER_POOL DEFAULT
PARTITION P201112
NOLOGGING
NOCOMPRESS
TABLESPACE KAW_DATA
PCTFREE 10
INITRANS 2
MAXTRANS 255
STORAGE (
INITIAL 64K
MINEXTENTS 1
MAXEXTENTS 2147483645
BUFFER_POOL DEFAULT
PARTITION PMAX
NOLOGGING
NOCOMPRESS
TABLESPACE KAW_DATA
PCTFREE 10
INITRANS 2
MAXTRANS 255
STORAGE (
INITIAL 64K
MINEXTENTS 1
MAXEXTENTS 2147483645
BUFFER_POOL DEFAULT
)NOPARALLEL;
My question is :
1) How to add a partition say P201201 ..
Since I have already stated PARTITION PMAX can I still add partition?
2) Can I add partition even though the table has data ?
==========================
I tried to add partition ..say
ALTER TABLE X_ACC ADD PARTITION P201201 VALUES LESS THAN
(TO_DATE(' 2012-01-01 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN'))
NOLOGGING
NOCOMPRESS
TABLESPACE KAW_DATA
PCTFREE 10
INITRANS 1
MAXTRANS 255
STORAGE (
INITIAL 64K
MINEXTENTS 1
MAXEXTENTS 2147483645
BUFFER_POOL DEFAULT
but getting the error ..
"partition bound must collate higher than that of the last partition"
Thanks/kumar

You have partitions
200712
200812
200912
201012
201112
PMAX.
so your condition for partition p201112 is '201012-01-01' to '201112-01-01' i.e the values between these two dates will be in partition p201112. and then everything else will be in PMAX. So now you are splitting your pmax.
alter table X_acc split partition pmax at ( to_date('2012-01-01 00:00:00','YYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN') )
into (partition p201201, partition pmax); The condition you have to change. I just showed as an example. It could be the date range between two dates or it could be greater than a date value. It is upto how you want this new partition.
how to add partition between two partition
Edited by: pransuj on Mar 25, 2011 11:13 AM
Edited by: pransuj on Mar 25, 2011 11:14 AM

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i should want to know what are the benefit to partition a table, because I follow the step on:
http://docs.oracle.com/cd/B19306_01/server.102/b14231/tables.htm#i1006754
the example 1.
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If there is a solution of course.
thanks
Francesco

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The table partioned has only an LOCAL index more, than do not have a partitioned table:
For the table partioned:
SQL> EXPLAIN PLAN SET STATEMENT_ID='PART'
2 FOR
3 select * from EVENTI_PM e where e.key_id_evento='50000034' and dt_inserimento = (select min(dt_inserimento) from eventi_pm e2
where e.key_id_evento=e2.key_id_evento);
Spiegato.See the plan:
SQL> SELECT
2 CARDINALITY,
3 BYTES,
4 COST,
5 POSITION,
6 PARTITION_START,
7 PARTITION_STOP,
8 PARTITION_ID,
9 lpad(' ',level-1)||OPERATION||' '||OPTIONS||' '||object_name as Plan
10    FROM PLAN_TABLE a
11 CONNECT BY prior id = parent_id
12          AND prior statement_id = statement_id
13    START WITH id = 0
14          AND statement_id = 'PART'
15    ORDER BY id;
CARDINALITY BYTES COST POSITION PARTITION_START PARTITION_STOP PARTITION_ID PLAN
          1   144    5        5                                              SELECT STATEMENT
          1   144    5        1                                               NESTED LOOPS
          1    26    3        1                                                VIEW VW_SQ_1
          1    17    3        1                                                 HASH GROUP BY
          2    34    3        1                                                  FIRST ROW
          2    34    3        1                                                   INDEX RANGE SCAN (MIN/MAX) EVENTO_PK
          1   118    2        2 KEY             KEY                        6   PARTITION HASH ITERATOR
          1   118    2        1 KEY             KEY                        6    TABLE ACCESS BY LOCAL INDEX ROWID EVENTI_P
                                                                             M
          1          1        1 KEY             KEY                        6     INDEX UNIQUE SCAN INDICE_LOC
Selezionate 9 righe.in which the cost is 5.
the same thing for a non-partioned table:
SQL> EXPLAIN PLAN SET STATEMENT_ID='NON_PART'
2 FOR
3 select * from EVENTI_PM2 e where e.key_id_evento='50000034' and dt_inserimento = (select min(dt_inserimento) from eventi_pm e
2 where e.key_id_evento=e2.key_id_evento);
Spiegato.see the plan
SQL> SELECT
2 CARDINALITY,
3 BYTES,
4 COST,
5 POSITION,
6 PARTITION_START,
7 PARTITION_STOP,
8 PARTITION_ID,
9 lpad(' ',level-1)||OPERATION||' '||OPTIONS||' '||object_name as Plan
10    FROM PLAN_TABLE a
11 CONNECT BY prior id = parent_id
12          AND prior statement_id = statement_id
13    START WITH id = 0
14          AND statement_id = 'NON_PART'
15    ORDER BY id;
CARDINALITY BYTES COST POSITION PARTITION_START PARTITION_STOP PARTITION_ID PLAN
          1   144    5        5                                              SELECT STATEMENT
          1   144    5        1                                               NESTED LOOPS
          1    26    3        1                                                VIEW VW_SQ_1
          1    17    3        1                                                 HASH GROUP BY
          2    34    3        1                                                  FIRST ROW
          2    34    3        1                                                   INDEX RANGE SCAN (MIN/MAX) EVENTO_PK
          1   118    2        2                                                TABLE ACCESS BY INDEX ROWID EVENTI_PM2
          1          1        1                                                 INDEX RANGE SCAN EVENTO_PK2
Selezionate 8 righe.but the cost is the same, why?Even if only one partition was used. How i can demostrate to my self that the query in the partitioned table is better?
I'm going to get crazy, partioned a table is not useful at all.Please help me!!!!!!!!!!!!!!!!!!!

ADD PARTITION시 NEW LOCAL INDEX PARTITION에 관하여

제품 : ORACLE SERVER
작성날짜 : 2004-08-16
ADD PARTITION시 NEW LOCAL INDEX PARTITION에 관하여
==================================================
(※ Oracle Partitioning Option은 8~10g Standard Edition에서는 지원하지 않는다.)
개 요
=======
ALTER TABLE ADD PARTITION을 할 경우 table partition의 경우 table partition,
tablespace, physical storage attribute를 지정할 수 있으나,
local index partition의 경우 관련 syntax를 지원하지 않는다 (8.1.6 현재)
오라클은 자동으로 index partition, tablespace, physical storage attribute를 지정한다.
(ALTER TABLE SPLIT PARTITION도 마찬가지)
new index partition의 이름은 new table partition과 같은 이름이 할당되고,
tablespace, physical storage attribute는 local index에 지정된 default값이 사용되고,
local index에 지정된 default값이 없으면 user의 default tablespace가 사용된다.
그러므로 PARTITION 추가하여 사용하는 경우 index partition 관리에 주의가 요구된다.
다음은 ALTER TABLE ADD PARTITION한 이후에 index partition을 관리하는 예제이다.
예 제
=======
-- Create partition table --
CREATE TABLE tbl_emp (
empno NUMBER,
ename VARCHAR2(20),
job VARCHAR2(20)
PARTITION BY RANGE ( empno ) (
PARTITION pt_tbl_emp01 VALUES LESS THAN( 1000 ) TABLESPACE data_emp01,
PARTITION pt_tbl_emp02 VALUES LESS THAN( 2000 ) TABLESPACE data_emp02,
PARTITION pt_tbl_emp03 VALUES LESS THAN( 3000 ) TABLESPACE data_emp03
-- Create local index --
CREATE INDEX idx_emp ON tbl_emp( empno )
TABLESPACE users --> 여기서 local index default tablespace 지정
LOCAL (
PARTITION pt_idx_emp01 TABLESPACE indx_emp01,
PARTITION pt_idx_emp02 TABLESPACE indx_emp02,
PARTITION pt_idx_emp03 TABLESPACE indx_emp03
-- PK 설정 --
ALTER TABLE tbl_emp ADD CONSTRAINT pk_emp PRIMARY KEY ( empno );
-- Local Index를 PK index로 사용함을 확인 --
SELECT index_name FROM user_indexes WHERE table_name = 'TBL_EMP';
INDEX_NAME
IDX_EMP
DROP INDEX idx_emp
ERROR at line 1:
ORA-02429: cannot drop index used for enforcement of unique/primary key
-- Add partition --
ALTER TABLE tbl_emp
ADD PARTITION pt_tbl_emp04 VALUES LESS THAN( 4000 )
TABLESPACE data_emp04;
-- 추가된 Local index partition의 이름, Tablespace를 확인한다 --
SELECT partition_name, tablespace_name, status
FROM user_ind_partitions
ORDER BY 1;
PARTITION_NAME TABLESPACE_NAME STATUS
PT_IDX_EMP01 INDX_EMP01 USABLE
PT_IDX_EMP02 INDX_EMP02 USABLE
PT_IDX_EMP03 INDX_EMP03 USABLE
PT_TBL_EMP04 USERS USABLE
-- Tablespace는 local index의 default tablespace에 생긴다 --
SELECT index_name, def_tablespace_name
FROM user_part_indexes;
INDEX_NAME DEF_TABLESPACE_NAME
IDX_EMP USERS
--<< Local index partition name 변경 >>--
ALTER INDEX idx_emp
RENAME PARTITION pt_tbl_emp04 TO pt_idx_emp04;
--<< Local index tablespace 변경 >>--
ALTER INDEX idx_emp
REBUILD PARTITION pt_idx_emp04 TABLESPACE indx_emp04;
SELECT partition_name, tablespace_name, status
FROM user_ind_partitions
ORDER BY 1;
-- 변경된 Local index partition의 이름, Tablespace를 확인한다 --
PARTITION_NAME TABLESPACE_NAME STATUS
PT_IDX_EMP01 INDX_EMP01 USABLE
PT_IDX_EMP02 INDX_EMP02 USABLE
PT_IDX_EMP03 INDX_EMP03 USABLE
PT_IDX_EMP04 INDX_EMP04 USABLE
* 다음과 같이 local index의 default tablespace를 변경할 수도 있다.
alter index <indexname>
modify default attributes tablespace <new tablespace name>;

>
1. If i drop one partition of a table why all the local indexes(year_1998 ... year_2000) get dropped but year_idx remains in database? I couldn't find index in dba_part_indexes view but i could find only year_idx line in dba_indexes view.
>
After dropping a partition, the row in DBA_PART_INDEXES still exists, so I'm unsure why it does not appear on your system. The index however does not appear in the DBA_INDEXES view before or after the drop partition:
for example
SQL> select owner, index_name from dba_part_indexes where
2 index_name = 'YEAR_IDX';
OWNER                          INDEX_NAME
APPS                           YEAR_IDX
SQL> alter table all_facts drop partition year_1999;
Table altered.
SQL> select owner, index_name from dba_part_indexes where
2 index_name = 'YEAR_IDX';
OWNER                          INDEX_NAME
APPS                           YEAR_IDX
SQL> select owner, index_name from dba_indexes
2 where index_name = 'YEAR_INDEX';
no rows selected

Can the command ADD partition (in table )invalid procedures,packages. ??

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