I have one table and I created partition month wise. I also creates index on that table. How can I check whether parition will be used when I query that table? In explain plan I can see PARTITION RANGE as ALL. Is it using the partition?
View 2 RepliesI want to make partition of table which is already created and has huge amount of Data I have done indexing already
my structure are as follows
EMPLOYEE CHAR(10),
MONTH_YEAR NUMBER (6),
TYPE CHAR (1),
SALARY_CODE CHAR (5),
AMOUNT NUMBER (16,2)
Indexing on Employee,Month,Salary_Code
How can I do partitioning on this table.
I have table A with size 120 Million and two more tables are of size 2 Million on Table B and less than 1 Million size on table C.I had created Partition and Parallel degree 4 on the table A. Created table B with Parallel degree 2 and Created table C with NOPARALLEL.
My query is using above tables with joins and inserting into table D using HINT /*+ APPEND NOLOGGING*/I had executed the explain on the above criteria the cost is showing 20767.
Later created Tables A,B and C with NOPARALLEL. Applied HINT on Table D /*+ APPEND NOLOGGING*/ and als applied HINT /*+ PARALLEL(A, 4) PARALLEL(B C, 2) on select query which uses to insert into Table D.
My question which is best practice on PARALLEL degree creation at table level or query level:
a) Creating table with Paralle (degree 4)
b) Applying HINT /*+ PARALLEL (TABLE A , 4) */ at query level
Is their any query to find cpu usage & memory usage of all the queries currently running on DB?
View 5 Replies View RelatedIs there a way to find when was a database role created and who has created?
View 5 Replies View Relatedtell me if a REFRESH GROUP is automatically created when a materialized view is created?
View 3 Replies View Relatedi 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 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 RelatedI 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)
At present we have a non partitioned table.
Can we apply redefinition and create range partition and hash sub partition on it?
Can 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 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 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 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 RelatedI 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].....
needed 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]...........
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 Replies View RelatedThere is an 'emp' table with a column name as 'mgr' with datatype 'number'. following is the detailed description of the table:
SQL> desc emp;
Name Null? Type
----------------------------------------- -------- ---------------------------
EMPNO NOT NULL NUMBER(4)
ENAME VARCHAR2(10)
JOB VARCHAR2(9)
MGR NUMBER(4)
HIREDATE DATE
SAL NUMBER(7,2)
COMM NUMBER(7,2)
DEPTNO NUMBER(2)
Now when I run the 'select mgr from emp e;' query the output which I get is:
7902
7698
7698
7839
7698
7839
7839
7566
7698
7788
7698
7566
7782
Note: One value in between here is null, the required to me is that I want to print a character value 'President' in place of null .
when to use || NULL oerator or what is the meaning of this statement when it is used with the column name.
and A.RtmtCd = B.RtmtCd|| NULL
or
A.RtmtCd||NULL IS NOT NULL
I have one procedure which is executing daily two times(i.e morning run and evening run) . In morning run it's executing around 150 mins and evening run executing around 25 mins.
It's happening for all procedures executing in longer run in morning. There is NO LOCKS, I have checked LOCKS while executing morning run. I suspects CPU usage in morning run.
How to check CPU usage ? AND also each session how much utilizing CPU. I'm using oracle 8i database, I know it's older version but my company is using oracle 8i. So i have to look in this database version only.
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 a table "tl" which is partitioned--say 30 partition and for each partition there is a seperate view like as follows
create view view_t130 as select * from tl partition (p30);
create view view_t129 as select * from t1 partition (p29);
.
.
.
create view view_t101 as select * from t1 partition (p01);
my question is how to use hints on this table if your are using view to access the data from internal table.
Normal structure is if i don't wrong:-
index( <<view name|view alia name>> <<table name|table alias name>> name of index)
Consider my case
select * from view_t130 where <index_column> --not picking up index
i want to give expicit index hint.so i used the same structure that i specified above but it didn't work.
select /*+ index( view_t130 t1 <index_name) */ * from view_t130 where <index_name>
how to give explicit index hint..but one constraint is i cannot give any alias names for internal tables because those(view structure) are generated by predefined scripts..so it's not possible to change it.
I had a question on BFILENAME() :
first,
create or replace directory UTL_FILE_DIR as 'k:lob_test';
GRANT READ,WRITE ON DIRECTORY UTL_FILE_DIR TO user1;
and then ,I login use user : user1
create or replace package body p_utl_mail_attach_raw as
procedure send_single_user_attach_raw(........)
fil BFILE;
filenm VARCHAR2(50) := 'aaa.jpg'
begin
fil := BFILENAME('UTL_FILE_DIR', filenm);
end ;
this result it show :
UTL_FILE_DIR/aaa.jpg, exists=F, length=0, open=F
I'd like to have a conditional that only gets the next value from a sequence is the current (or supplied) value is null.
Here are some trivial examples:
CREATE SEQUENCE ts
MINVALUE 1
START WITH 1
INCREMENT BY 1;
SELECT ts.NEXTVAL FROM DUAL;
SELECT COALESCE(ts.CURRVAL, ts.NEXTVAL) FROM DUAL;
SELECT NVL(ts.CURRVAL, ts.NEXTVAL) FROM DUAL;
SELECT CASE
WHEN ts.CURRVAL IS NOT NULL THEN ts.CURRVAL
ELSE ts.NEXTVAL
END
FROM DUAL;
Ideally multiple executions of any of these (or a better one of your design) should return the same value from the sequence, but mine do not.
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 RelatedI 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 can get "CPU used by this session' value from v$sesstat. That is the value in centiseconds used by each session. How to convert the value into a percentage of total CPU resource ?
Or there are some other ways to gain the value. i.e. Percentage of Total CPU usage for each Oracle session.
I am building a reporting table using the count analytic function in order to count up several different attributes in one statement.What I find is that this method quickly eats up my TEMP space. This is 10gR2. I have attempted to use MANUAL workarea policy with as large ofsort_area_size as possible (2G) but that does not seem to have any effect on performance or TEMP usage. The RAW table is about 12G with 75 million rows. I am not that concerned about execution time, but rather TEMP usage.
--INSERT into <object>...
select distinct
file_sid,filename,control_numb,processing_date,file_class,
vendor_id,vendor_desc,
c_status_id,c_status_desc,
[code]...
I am not seeing any increase in onepass or multipass executions on the PGA during execution of this statement using...
SELECT CASE WHEN low_optimal_size < 1024*1024
THEN to_char(low_optimal_size/1024,'999999') ||
'kb <= PGA < ' ||
(high_optimal_size+1)/1024|| 'kb'
ELSE to_char(low_optimal_size/1024/1024,'999999') ||
[code]...
I'd like to get a better explaination of how analytics use the instance resources and TEMP space. For example if I add
a count with a different window (such as the last two columns commented in the above query) I blow out my temp space (70G).
Is the critcal factor the use of distinct? or multiple windows? or something else?