11.2.0.1...How do I create an index on a view or any workaround that my view won't get duplicates?
SQL> create unique index indx01 on db_backup_details_vw(id);
create unique index indx01 on db_backup_details_vw(id)
*
ERROR at line 1:
ORA-01702: a view is not appropriate here
I have following records in table (T)
Idtype_cdNbrDt
1A1001/1/2013
2A101Null
3B100Null
4C100Null
5A100Null
Based on the type-cd, I need to create Unique index on Nbr & Dt columns
In above record set, I want to create Unique index on type_cd=A and not on B&C (Here i expect duplictes for a combination of Nbr & Dt)
Can we create Unique index on subset of record?
How to create index on materialized view which is refreshed daily basis.
View 3 Replies View RelatedWe have been recommended to store data in CLOD data type.
Sample data: 1:2:2000000:20000:4455:000099:444:099999:....etc it will grow to a large number.
We want to create a Unique index, for functional reason. Is it advised to create a unique index on a CLOB datatype?
Using Oracle 11g, below is the table, partitions, unique and non-unique local index:
CREATE TABLE DOCA( DOCA_ID NUMBER NOT NULL , DOCA_BKG_PAX_ID NUMBER NULL , ROW_PURGE_DATE DATE NULL ,)PARTITION BY RANGE(ROW_PURGE_DATE)INTERVAL(NUMTOYMINTERVAL(1, 'MONTH'))( PARTITION P2007 VALUES LESS THAN (TO_DATE('01/01/2008', 'dd/mm/yyyy')), PARTITION P200801 VALUES LESS THAN (TO_DATE('01/02/2008', 'dd/mm/yyyy')),) TABLESPACE T0; ALTER TABLE DOCA ENABLE ROW MOVEMENT;
CREATE UNIQUE INDEX XPKDOCA ON DOCA( DOCA_ID ASC, ROW_PURGE_DATE ASC)LOCALREVERSE TABLESPACE I0; ALTER TABLE DOCA ADD CONSTRAINT XPKDOCA PRIMARY KEY (DOCA_ID); CREATE INDEX XFKDOCA_DOCA_BKG_PAX_ID ON DOCA( DOCA_BKG_PAX_ID ASC)LOCALREVERSETABLESPACE I0;
I would like to know the difference between the performance of the unique and non-unique local indexes?.
I have a running application where i have a table with pk defined on it and the pk clumn is inde in asc order by default. Can i alter the index in desc order as i always need to see the data in desc order.
View 7 Replies View RelatedI have created one unique index on one column of my table. Now i would like to add one more column in the same index without dropping the index.
SQL > CREATE TABLE DEBUG_TABLE
2 (
3 SLNO NUMBER,
4 MESSAGE VARCHAR2(4000 BYTE),
5 CREATED_DATE DATE DEFAULT SYSDATE,
6 CREATED_TIME TIMESTAMP(6) DEFAULT SYSDATE
7 );
Table created.
SQL > CREATE UNIQUE INDEX index_debug1 ON debug_table (SLNO);
Index created.
SQL > ALTER INDEX index_debug1 ADD COLUMN MESSAGE;
ALTER INDEX index_debug1 ADD COLUMN MESSAGE
*
ERROR at line 1:
ORA-02243: invalid ALTER INDEX or ALTER MATERIALIZED VIEW option
SQL >
I've got a table which has no unique identifier for a row. I have a field for an ID value but somhow there are duplicate values in this field.
Now I want to create an unique index over this field but therefore I have to remove the duplicate values.
So what I need is an update statement, maybe in a for each loop, which updates the ID field in each row starting with 1 and increasing the value.
In Informix there is a rowid value which is automatically set by Informix. But I think there´s no correspondent function/value in Oracle....
I have a table which sees a lot of use for queries
CREATE TABLE CASE_STAGE
(
ID NUMBER(9) NOT NULL,
STAGE_ID NUMBER(9) NOT NULL,
CASE_PHASE_ID NUMBER(9) NOT NULL,
"CURRENT" NUMBER(1) NOT NULL,
--and other columns
)
ID is a primary key
CASE_PHASE_ID is a foreign key
"CURRENT" should only ever have values of 0 or 1. When it has a value of 1 it is unique for that CASE_PHASE_ID
What I have tried that doesn't work is
create unique index case_stage_F_IDX1 on case_stage("CURRENT", case_Phase_id) which gives me
ORA-01452: cannot CREATE UNIQUE INDEX; duplicate keys found
What is the correct syntax, something like ("CURRENT"=1,case_phase_id) seems right but fails with an error about a missing bracket. Do I need to use a CASE statement here?
I have a table with a non-unique index consisting of three columns. The first column is not null while the remaining two are nullable. Queries using this index will chiefly be made in two ways.
1. Column one and two having values. Column three is null.
2. Column one and thre having values. Column two is null.
In both cases I expect range scan will be used since it's non-unique. In the first case the scan will be on values in column one and two. But what happens in case two. Will the range be on colum one, column two(being null) and cxolumn three? Or will it be on just column one since the second column is null? I have done some testing. I can see , using EXPLAN PLAN, that range scan is used in both cases. how the index is used?
Is there any other drawbacks with an index like this?
I'm having problem with my database, which contains more than 1 rows with a same value on a field that has uniqueness contraint.
Here is the log from sqlplus. When I select on RI field, it shows 2 rows. But when I select on SCNUM field, it shows only 1 row. This SCNUM has an unique index on it.
And it is still in VALID state
SQL> set autotrace on
SQL> select ri, scnum from scratch1_p where ri in (536964983, 536955574);
select ri from scratch1_p where scnum='444393975';
RI SCNUM
---------- ----------
536955574 444393975
[code].....
Is there any hint to force "Index Unique Scan" over "Index Range Scan".
My query is generating different plans with the above two, and very slow when it uses "Index Range Scan".
What should be the order of columns while creating b-tree non-unique index.
Low cardinality first
or
Hight cardinality first
We are getting an error as below when trying to load data into a table.
INSERT /*+ APPEND parallel(IA_SBSCR_DED_MAX,4) */ INTO EDW.IA_SBSCR_DED_MAX
*
ERROR at line 1:
ORA-12801: error signaled in parallel query server P000
ORA-26026: unique index EDW.XPKIA_SBSCR_DED_MAX_A initially in unusable state
The index has been rebuilt but we still have this issue.
Can we create non-cluster index on a clustered index?
View 5 Replies View RelatedA Master form in a tabular format. Id is a display item. I want to auto generate Id with following query.
select nvl(max(issue_id),0)+1 into :issue_mt.issue_id from issue_mt;
It works:
1)When each record is saved before creating new record.
It doesn't work:
2)If multiple records are entered with out saving each record.
What to do to generate unique Id in both cases?
i want to insert unique number with out gaps in a col of a table every time my query runs . it shouel generate numbers from 1 to n. I can do it using sequence , but sequence need reseting and other user can reset it also and it creats gaps too.
Any function/technique which can give me session or transaction specific unique number on each call to sp.
i want to inserting these numbers in a table and insertion is going on in a loop.
is it possible to create primary key on view and use this view for creating foreign key .
View 3 Replies View RelatedI have a table that has three columns.
UserId - DateStamp - Action
An action contains user activities such as "logon". A user may logon multiple times in a day.I'm trying to create a query that displays the number of unique logons for each user per day.
By doing the following:
select userid, to_char(creationtimestamp,'YYYY-MM-DD'), count(*)
from useraction
where action = 'logon'
group by to_char(creationtimestamp,'YYYY-MM-DD'), userid
order by to_char(creationtimestamp,'YYYY-MM-DD') desc;
I can get the following results
userid - date - count
1 - 2/21/2010 - 8
1 - 2/22/2010 - 3
2 - 2/22/2010 - 2
1 - 2/23/2010 - 1
4 - 2/23/2010 - 6
6 - 2/23/2010 - 1
What I'm trying to get is
2/21/2010 - 1
2/22/2010 - 2
2/22/2010 - 3
When you create a MAV, you automatically get a hidden column and an index. Here's an example,drop user jon cascade;
grant dba to jon identified by jon;
conn jon/jon
create table emp as select * from scott.emp;
create materialized view mv1 enable query rewrite as
select deptno,sum(sal) from emp group by deptno;
select object_name,object_type from user_objects;
select index_name,column_name from user_ind_columns where table_name='MV1';
select column_name,hidden_column from user_tab_cols where table_name='MV1';
select deptno,"SUM(SAL)",sys_nc00003$ from mv1;
I have a view on base tables holding historical data for previous 60 months(one table per month) with union all operators.create index on those base tables will improve performance or creating a primary key with disabled novalidate will improve for retrieving data?
The view has around 8 million data and used as a fact table with 4 dimension tables.A DTS package from MSSql side refreshes OLAP cube by retrieving data from these tables in oracle.
i want to create a bitmap index,but getting the error shown below..i created primary key on column dname of dept and unique constraint on empno of emp.
SQL> create bitmap index bindx on dept d(d.dname) from emp e,dept d where e.deptno=d.deptno;
create bitmap index bindx on dept d(d.dname) from emp e,dept d where e.deptno=d.deptno
*
ERROR at line 1:
ORA-25954: missing primary key or unique constraint on dimension
Can we create view in another view
View 10 Replies View Relateddifference between the ways we create Reverse Key Index on the table.The two ways are as below:
1) CREATE INDEX <Index_Name> ON <Table_Name>(<Column_Name) REVERSE ;
2) CREATE INDEX <Index_Name> ON<Table_Name>(REVERSE(<Column_Name>)) ;
Which one is the appropraite one.
Are both going to act in the same way.
I created index in both ways one by one on the table.But when I fire the select statement against the same table, The explain plan doesn't show any cost and it shows the full table scan in both ways.The select statement used is as below...I want to compare the column with string RBO(i.e('RBO%') at the start.
SELECT *
FROM <Table_Name>
WHERE REVERSE (<Column_Name>) LIKE '%OBR' ;
The select statement is giving me the correct result. Only the explain plan is showing Full Table Scan.can we use this reverse with IN operator also in the same way by reversing the values.
I use Oracle 10.0.2.0.1.If I create a table with constraint key; after that I create an unique index key, I got an error. Does it mean when I create a table with constraint the unique index are automatically created and I could not create index key as I did as below?
create table test_const(ename varchar2(50) not null,
key_num number not null,
descr varchar2(100),
constraint constraint_test_const unique (ename, key_num));
create unique index test_const_idx on test_const
(
"ENAME","KEY_NUM"
)
tablespace tmp_data;
Error report: SQL Error: ORA-01408: such column list already indexed 01408. 00000 - "such column list already indexed" added [pre] tags by Sriram
I am going through this scenario:
* 35 | ID TABLE ACCESS BY INDEX ROW | S_ORG_EXT | 3064K| 2472M| | 1 (0)| 00:00:01 |
| 36 | INDEX FULL SCAN | S_ORG_EXT_U1 | 14 | | | 1 (0)| 00:00:01 |
Predicate Information (identified by operation id):
---------------------------------------------------
35 - filter("T2"."ACCNT_FLG"<>'N' AND ("T2"."INT_ORG_FLG"<>'Y' OR "T2"."PRTNR_FLG"<>'N'))
This unselective index scan on step 36 of the explain is returning 14 rows but optimizer is selecting 3064 K rows from the table .
I tried creating combined index on all 3 columns mentioned in the predicates for 35th step , but that is not utilized .
how to index this whole expression ::--
(ACCNT_FLG<>'N' AND (INT_ORG_FLG<>'Y' OR PRTNR_FLG<>'N'))
Something like CREATE INDEX XYZ on table((ACCNT_FLG<>'N' AND (INT_ORG_FLG<>'Y' OR PRTNR_FLG<>'N')) compute statistics ;
how to create a primary key with out creating an index?
View 10 Replies View Relatedwe create index on multiple tables in oracle
View 12 Replies View RelatedIs it a possible to create table using clause below together with index ?
create table the_table
as
select col1, col2 from table2
I got procedure which create a table in the schema B. The procedure is called from schema A. But when I write into procedure query for create index
then I got a error:
ORA-01031: insufficient privileges when
...executing
Therefore I think about to create table together with index.
begin
B.proc.cre_table;
end;
I have an IOT table with partitioned on list. I have p1,p2 and p3 partitions. Now I would like to create a bitmap index on partition key.
I gave ALTER TABLE .. MOVE MAPPING TABLE;
But getting the below error,
ORA-28660: Partitioned Index-Organized table may not be Moved as a whole.