How to find the size pf a partition in a partition table?I guess we need to query views like dba_tab_partitions but I am not very sure. will running dbms_stats.gather_table_stats('schema_name,'table_name,'partition_name')
View 3 RepliesI have a table that partitioned into six partitions. each partitions placed in different table space and every two table space placed it on a different hardisk
when I will do query select with the non-partition keys condition, how the search process ? whether the sequence (scan sequentially from partition 1 to partition 6) or partition in a hardisk is accessed at the same time with other partition in other hardisk. ( in the image, partition 1,4 accessed at the same time with partition 2,5 and 3,6)
i have table with range partition and list sub-partition..can i add one more list sub-partition if it is not possible , i have to drop first sub-partition.
View 6 Replies View RelatedCan I add range sub partition to a hash partition table. Example like this.
CREATE TABLE test
(
test_id VARCHAR2(10 ) ,
test_TYPE VARCHAR2(5) ,
CREATE_DATE date
)
partition by hash (test_id, test_type)
Partitions 3
SUBPARTITION BY RANGE (CREATE_DATE);
When Tried, I am getting syntax error as invalid option.
i want to create a new partition for version 2
existing table is as below
create table test
(
name varchar2(100),
version NUMBER(12)
)
[Code]....
I have two tables in which one is partitioned table with the following details.
CREATE TABLE "SCOTT"."TBL_MITTAL"
("ACCOUNT_NAME" VARCHAR2(50 BYTE),
"BILL_NO" VARCHAR2(50 BYTE),
"BILL_DATE" VARCHAR2(50 BYTE),
"CLI" VARCHAR2(50 BYTE),
"ANI" VARCHAR2(50 BYTE),
[code].....
When I am trying to insert record from tbl_mittal into tbl_temp table. I am facing "ORA-14400: inserted partition key does not map to any partition" error
SQL> insert into tbl_temp select * from tbl_mittal;
insert into tbl_temp select * from tbl_mittal
*
ERROR at line 1:
ORA-14400: inserted partition key does not map to any partition
AS tbl_mittal is having hugh number of records so I am providing only few rows from tbl_mittal table as test data.
ACCOUNT_NAMEBILL_NOBILL_DATECLIANICHARGE_START_DATEDURFROM_LOCATIONTO_LOCATIONINVOICE_IDCIRCLE
10000010357423128271095119301-Feb-111723006000931488182328-JAN-11 11.30.54.000000000 AM59.04CHANDIGARHJAIPUR271095119
10000011844187128348720198715-Jun-121409470011825531896615-MAY-12 09.10.36.000000000 AM28.03CHANDIGARHBANTWAL348720198
10000011844187128348720198715-Jun-121409470011825531927115-MAY-12 09.10.41.000000000 AM38.32CHANDIGARHBANTWAL348720198
10000011844187128348720198715-Jun-121409470011825531933015-MAY-12 09.10.46.000000000 AM28.81CHANDIGARHBANTWAL348720198
10000011844187128348720198715-Jun-121409470011825531930215-MAY-12 09.10.53.000000000 AM28.96CHANDIGARHBANTWAL348720198
[code].....
I also tried to upload the same data using sqlldr.
sqlldr log file contents is as follows:
Total logical records skipped: 0
Total logical records read: 1857532
Total logical records rejected: 801092
Total logical records discarded: 37
[code].....
So some sqlldr bad file contents is as follows.
100000118441871,283487201987,15-JUN-12,1723958000,9355115251,10-JUN-12 05.56.05.000000 PM,36.99,CHANDIGARH,AMBALA,348720198,,
100000118441871,283487201987,15-JUN-12,1723958000,7520533825,10-JUN-12 05.56.14.000000 PM,44.12,CHANDIGARH,AGRA,348720198,,
100000118441871,283487201987,15-JUN-12,1723958000,9356452151,10-JUN-12 05.56.17.000000 PM,116.83,CHANDIGARH,JALANDHAR,348720198,,
100000118441871,283487201987,15-JUN-12,1723958000,9331223048,10-JUN-12 05.56.21.000000 PM,28.33,CHANDIGARH,KOLKATA,348720198,,
100000118441871,283487201987,15-JUN-12,1723958000,7827927893,10-JUN-12 05.56.24.000000 PM,3384.33,CHANDIGARH,DELHI,348720198,,
[code].....
I think that performance better partition table than non-partition table. How to assure partition table is better than non-partition table at SELECT operation?
I have compare a specific query EXPLAIN PLAN at partition table and non-partition table. both tables data is same. Is it true way or not?
Can i alter the table to create partition on non partition table, i have tried and could not create it. Do we have some other means to do it as this is the live table and cannot drop them else will lose the data.
View 1 Replies View RelatedI Know we can create dynamic partitions on table in oracle 11g. Is it possible to create normal partition and sub partition both dynamically.I have to create Normal partition range on date and sub partition list on Batch ID (varchar).
View 3 Replies View RelatedAt present we have a non partitioned table.
Can we apply redefinition and create range partition and hash sub partition on it?
I am trying to add partition to table without partition
with following code
ALTER TABLE ACC_LOC1_TAB
ADD PARTITION testpart BY RANGE (ALT_AUTHDT)
(PARTITION UPTO_2010 values less than (31-mar-2010),
PARTITION APR_JUN_10 VALUES less than (30-JUN-2010),
PARTITION JUL_SEP_10 VALUES less than (30-SEP-2010),
PARTITION OCT_DEC_10 VALUES less than (31-DEC-2010),
PARTITION JAN_MAR_11 VALUES less than (31-MAR-2011))
it will raise error ora-14020
I am using Oracle 11.2.0.1 Oracle Database.I have a table with 10 Million records and it's a Non Partitioned Table.
1) I would like to partition the table (with partition by range ) without creating new table . I should do it in the existing table itself (not sure DBMS_ REDEFINITION is the only option ) (or) can i use alter table ...?
2) Add one partition which will have data for the unspecified range.
I think that performance better partition table than non-partition table. How to assure partition table is better than non-partition table at SELECT operation?
I have compare a specific query EXPLAIN PLAN at partition table and non-partition table. both tables data is same. Is it true way or not?
I have two tables Activity and Activity1.
Activity Structure
ACTIVITY_TYPE CHAR (1) NOT NULL,
ACTIVITY_DATE DATE DEFAULT sysdate NOT NULL,
ACTIVITY_ON VARCHAR2 (30) NOT NULL,
REFERENCE_NO VARCHAR2 (19),
CHILD_REFERENCE_NO VARCHAR2 (19),
USER_ID VARCHAR2 (30) DEFAULT user NOT NULL,
TERMINAL VARCHAR2 (30) DEFAULT userenv ('TERMINAL') NOT NULL )
Activity1 Structure Which I have Done Partitioning When I insert data from Activity to Activity1 it gives that error ORA-14400: inserted partition key does not map to any partition what I am doing wrong
CREATE TABLE ACTIVITY1(
ACTIVITY_TYPE CHAR (1) NOT NULL,
ACTIVITY_DATE DATE DEFAULT sysdate NOT NULL,
ACTIVITY_ON VARCHAR2 (30) NOT NULL,
REFERENCE_NO VARCHAR2 (19),
[code]....
Insert Statement
insert into ACTIVITY1(ACTIVITY_TYPE,
ACTIVITY_DATE,
ACTIVITY_ON,
REFERENCE_NO,
CHILD_REFERENCE_NO,
[code]....
I have partition based table one the basis of year month. And we have 8 local indexes on this table. Every month we have to create a new partition and load data into this partition and the volume of the data is around 14million and the load process is taking long time due to indexes. Is it possible to drop the indexes from particular partition?
View 8 Replies View Relatedneeded in stored procedure to achieve this...how to get this..(stored procedure)
table:
studentid,sname,partitonid
901,x,null
902,y,null
903,z,null
904,p,null
905,q,null
906,a,null
907,b,null
908,d,null
909,f,null
910,m,null
For the above data set i need to divide into 5 partittions and need to updated the partitonid with the partition number for each partition set,like the below result set
studentid,sname,partitonid
901,x,p1
902,y,p1
903,z,p2
904,p,p2
905,q,p3
906,a,p3
907,b,p4
908,d,p4
909,f,p5
910,m,p5
write ALTER STATEMENT for adding new partition P1 and SUBPARTITION P1_201001 and P1_201002.
TABLE
=======
CREATE TABLE TEST
(
"REPORT_ID" NUMBER,
"MONTH_ID" NUMBER,
"GROUP_ID" NUMBER,
"AGE_GROUP_ID" NUMBER,
[code]...........
I have a partitioned table in which at the first level it is been partitioned based on date column(C1) and within in each partition it is again sub-partitioned based on a column(C1) which is a numeric value now at first level(C1 column level) there are 6 partitions and in the last partition i want to add another partition how can i do that.
View 8 Replies View Relatedhow to partition and index my table for a special problem.
The table:
CREATE TABLE TEST (
ID NUMBER PRIMARY KEY,
U_VALUE NUMBER NOT NULL, -- Ranges from 0 - 30.000.000
O_VALUE NUMBER NOT NULL, -- Ranges from U_VALUE - 30.000.000
CREATE_TS TIMESTAMP DEFAULT SYSTIMESTAMP NOT NULL,
UPDATE_TS TIMESTAMP NOT NULL,
ITEM_TYPE NUMBER NOT NULL --<< Only 4 different values >>
);
As you can see, U_VALUE is ALWAYS lower than O_VALUE.I need to have the CREATE_TS in either main- or subpartition do drop the partitions after some time so i don,t have to use DELETE statements. The table has 360 millions rows.
The application has only 8 query which will always use a WHERE clause like this:
SELECT * FROM TEST
WHERE U_VALUE <= :1 AND O_VALUE => :2 AND ITEM_TYPE = :3
1. Is there any good technique how to create a good index for the queries (application will execute 120 queries per second)?
2. how to partition this table?
I am trying to delete the partition, but I am getting error "ORA-01426: numeric overflow"
The original partitioning was:
TABLE_NAME PARTITION_NAME NUM_ROWS
F_TFP_CP_MONTH P_201108 0
F_TFP_CP_MONTH P_201201 (NULL)
F_TFP_CP_MONTH P_201202 (NULL)
F_TFP_CP_MONTH P_99999999 (NULL)
F_TFP_CP_MONTH P_201106 1159130358
F_TFP_CP_MONTH P_201105 0
F_TFP_CP_MONTH P_201104 1212566971
F_TFP_CP_MONTH P_201103 1002557990
F_TFP_CP_MONTH P_201102 0
F_TFP_CP_MONTH P_201101 0
F_TFP_CP_MONTH P_201012 0
F_TFP_CP_MONTH P_201011 0
F_TFP_CP_MONTH P_201010 0
F_TFP_CP_MONTH P_201009 0
F_TFP_CP_MONTH P_201112 (NULL)
F_TFP_CP_MONTH P_201111 (NULL)
F_TFP_CP_MONTH P_201110 (NULL)
F_TFP_CP_MONTH P_201109 (NULL)
F_TFP_CP_MONTH P_201107 1627218307
All partitions were dropped, but three; these three returned the same error when trying to drop them:
SQL> alter table dw.F_TFP_CP_MONTH drop partition P_201112;
alter table dw.F_TFP_CP_MONTH drop partition P_201112
*
ERROR at line 1:
ORA-01426: numeric overflow
SQL> alter table dw.F_TFP_CP_MONTH drop partition P_201111;
alter table dw.F_TFP_CP_MONTH drop partition P_201111
*
ERROR at line 1:
ORA-01426: numeric overflow
SQL> alter table dw.F_TFP_CP_MONTH drop partition P_201110;
alter table dw.F_TFP_CP_MONTH drop partition P_201110
*
ERROR at line 1:
ORA-01426: numeric overflow
So, the situation now is that the table only has these three partitions, and we are not able to empty the table, so that we can later purge it and recover the
space.
free that space and empty the contents of the table?
I have the below questions on the locking mechanism in a partition table. Example I execute the below query,
delete from table_name subpartition (subpartition_name);
In that case,
1. If we perform above query, then which level of lock is happened to the table/partition & sub-partitions?
2. If we perform above query on same sub-partition of the same table with different where clause, what will happen?
3. If we perform above query for same table but different sub partition at same time, what will happen?
i want to ask about indexing in partition table. i have table that indexed by local index. when i want to select all data. I execute this query
select * from Book_Issue_Part where status='Pinjam';
but it does not select all data, only partly data have selected. is it a wrong query to select all data in indexing partition table? so what query should i execute to get all data.
can i able to partition the table based on the column which is in another table ??
For example Table X need to be partitioned based on the column in The Table Y . and Table X and table Y has some relation.
We have a transaction table and has 30 million rows. The table is not partitioned till date. We need to create partition on this table. We had an idea of moving this data to a temporary table and create partition[range]on the original table and move the data back.
View 5 Replies View RelatedCurrently we are deleting the partition daily. Table structure is Interval partitioning. After deleting partition how to analyze the partition table.
View 12 Replies View RelatedI am facing a problem in fetching / updating records from a customer details table having around 20 million records. The table contains around 30 fields with 'MOBILE_NO' as primary key. most of the queries are having 'mobile_no' in where clause .I am planning to hash partition that table using mobile_no column as there is no other column available which can be used for partition.
clarify whether creating hash partition on such key would increase performance of data extraction as I have read on net that hash partitioning is not effective for performance tuning.
How can we partition the existed table.
below scenario.
CREATE TABLE table_partition(sales number,year date,item char(4))
partition by range(year)
(
PARTITION p1 VALUES LESS THAN 1980,
PARTITION p2 VALUES LESS THAN 1982,
PARTITION p3 VALUES LESS THAN 1985
);
The above code will create table with partition.
That is the table is not existed before we are creating table with partition.
But my requirement is the table is already existed with 100000 rows.Now I want to range partition that table.
I’m facing an issue in my current project where we have to run batch jobs and interfaces on the same master tables. These tables are huge (millions) and we get very poor response time.
We thought of partitioning the tables, but the problem is our batch jobs queries are based on dates (some run on monthly data, some runs on yearly data..) but interfaces uses primary keys.
I’m not sure on how to partition the tables in this situation, is there any way I can partition the tables in such a way that both batches and interfaces will get benefit out of it?
Our database infrastructure has a 3 Node RAC (each with 8 CPUs) and oracle 10g R2. We have almost 60gig of RAM allocated to oracle.
PS: We are not thinking about Mviews now because client wants to explore the partition option first.
create or replace
Procedure ReadingsPurge
As
v_sql varchar2(500);
v_date date;
p_count NUMBER;
[Code]...
-- Code below drops partitions that are older than the NoOfDays Parameter
OPEN c1;
LOOP
FETCH c1 INTO v_partition_name, v_high_value;
EXIT WHEN c1%NOTFOUND;
[Code]....
Above code is compiling successfully.
After I added the lines makred in the red font, when I tried to execute the stored procedure, I got an error
Error starting at line 1 in command:
execute ReadingsPurge
Error report:
ORA-00933: SQL command not properly ended
ORA-06512: at "CDC_USER.READINGSPURGE", line 30
ORA-06512: at line 1
00933. 00000 - "SQL command not properly ended"
*Cause:
*Action:
I have a table where i have ser_id column, this column has got 200000 records and it get around 10 record every day, i want to apply partition on this table.
I am applying range partition of 5000, How can i create partition automatically for next 5000 records inserted into them
0-5000 p1
5001 -10000 p2
.
.
.
200000-205000
pn should be created automatically