update the indexed column

I got a name emp table that holds at least 30000 lines and I had created an index on emp_city

What happens if I update a value of emp_city? What effect will create it on the index or table?

the index will be updated automatically?

the index become invalid?

the line in the journal in the index will be deleted and inserted?

or updated in the journal will take place?

According to the name of the field, I understand that the distribution is pretty sparse for normal index on emp_city.

Will be better to create a bitmap index where you apply to this table with OLAP queries or not create an index on this field.

I don't care

If you have an index on emp_city or emp_city is a part of some regular index and you will upadate all value emp_city:

the index will be automatically updated. (Yes)

the index become invalid? (No, but I suggest that you run statistics on index or table wole)

the line in the journal in the index will be deleted and inserted? or updated in the journal will take place?

It's block of sheets (not rank of petals). Entire block will be replaced.

Best,

Ark

Tags: Database

Similar Questions

Identify the weekend and the days of the week a date and update the empty column

Hi, I look at previous posts that are similar to my question, but nothing helps.

I have a column called incident_date that is in this format and comes from the Nov_Final table

01-JUL-2009 20.10
04-NOV-2009 17.00
16-DEC-2009 16.00

In Nov_Final, I created a column called weekend

What I want to do is to perform an update, something like this:

Update Nov_Final

SET Weekend=(case when Incident_date ='WEEKEND' THEN 'Y' ELSE 'N' END)

Banner:
Oracle Database 11 g Release 11.2.0.2.0 - 64 bit Production
PL/SQL Release 11.2.0.2.0 - Production
"CORE 11.2.0.2.0 Production."
AMT for Linux: Version 11.2.0.2.0 - Production
NLSRTL Version 11.2.0.2.0 - Production

Weekend would be Saturday or Sunday. 
In short I need to find out if the incident_date falls on a weekend and if it does update the weekend column as 'Y' else 'N' if it's not.

You can use TO_CHAR function with a format mask as "fmDAY" to your needs:

with t_data
as
(
select sysdate + level as col1 from dual connect by level <= 7
)
SELECT
    col1,
    TO_CHAR(col1,'fmDAY') col_day,
    CASE
        WHEN TO_CHAR(col1,'fmDAY') IN ('SATURDAY','SUNDAY')
        THEN 'Y'
        ELSE 'N'
    END AS weekend_ind
FROM
    t_data;

COL1                      COL_DAY                     WEEKEND_IND
------------------------- --------------------------- -----------
08-NOV-11 07:54:12        TUESDAY                     N
09-NOV-11 07:54:12        WEDNESDAY                   N
10-NOV-11 07:54:12        THURSDAY                    N
11-NOV-11 07:54:12        FRIDAY                      N
12-NOV-11 07:54:12        SATURDAY                    Y
13-NOV-11 07:54:12        SUNDAY                      Y
14-NOV-11 07:54:12        MONDAY                      N           

7 rows selected

Classic generic column report - update the list column

Consider a dynamic pivot statement as follows (as described by the AMIS) which
```
select * from table( pivot(  'select deptno,  job, count(*) c from scott.emp group by deptno,job' ) )
```
I used this as a SQL source for the classic report, by selecting "use generic column names.
The report looks like
I can set up the data displayed with a selection list and including a clause where clause
```
where job =v(''P39_JOB'') or v(''P39_JOB'') is null
```
List select not PPR on the region,demonstrated here:
https://Apex.Oracle.com/pls/Apex/f?p=73000:39:
Note: , and all swivel jobs are still visible in the form of columns, data is limited
-other jobs are empty
-departments with no clerks are not shown
Let's say you have another list of selection than the work of groups, say the "Présidents" and "close".
How can you update the region to exclude certain jobs?
Reload the page with the value of P39_JOB already set will not affect the result, while including a literal string made.
It seems maybe ODCI function analyzes the query on Lees null, and then fills with actual results.
Does anyone have any suggestions?
Update: it seems a workaround is to inject the value as a substitution string, for example:
and employment as ' & P39_CEO. »
where P39_CEO is lacking in session state %
I've only had this concept works in my client application (4.2, 11 2 GR), not my demo (5.0, 12 c), to even try to meet the relevant quotes and when null. I kept getting
error report: ORA-1403: no data found
I could not work how to correctly apply sys.dbms_assert.enquote_literal)
Scott

Hi Scott

What approach I described in this post: matrix report?

It eliminates the need for any dynamic SQL and I think filtering options might be implemented using SQL/XML features.

Update the Blob column by replacing a string...

Hello
I have a table T1 with a (large binary BLOB) column, I need (you) to update the Blob string with replacement of a chain (in the Blob) with another string.
The update failed: ORA-00933: not correctly completed SQL command
CREATE table T1
(
   ID                   NOMBRE (12)           PAS NULL,
DELIVERY_CONTENT BLOB ,
)
Update T1 The VALUE DELIVERY_CONTENT = LOB_UTL_PCKG . BLOBREPLACE ()DELIVERY_CONTENT 'old_string' 'new_string'( )
WHERE ID in (...)
ORA-00933: SQL not correctly completed command
I use a function LOB_UTL_PCKG . BLOBREPLACE starting from this package
CREATE OR REPLACE PACKAGE LOB_UTL_PCKG
IS
FUNCTION BLOBREPLACE ()p_blob BLOB p_what VARCHAR2 p_with_what VARCHAR2() RETURN BLOB
FUNCTION BLOB2CLOB (p_blob BLOB) RETURN CLOB ;
FUNCTION CLOB2BLOB (p_clob CLOB) RETURN BLOB ;
END;
/
CREATE OR REPLACE PACKAGE BODY LOB_UTL_PCKG
IS
FUNCTION BLOBREPLACE ()p_blob BLOB p_what VARCHAR2 p_with_what VARCHAR2()RETURN BLOB
IS
START
RETURN CLOB2BLOB () REPLACE (BLOB2CLOB()p_blob), p_what p_with_what( ) );
END BLOBREPLACE ;
------------------------------ BLOB2CLOB -------------------------
FUNCTION BLOB2CLOB (p_blob in BLOB) RETURN CLOB
IS
   v_clob     CLOB ;
dest_offset INTEGER := 1 ;
offset INTEGER := 1 ;
lang_context INTEGER := DBMS_LOB. DEFAULT_LANG_CTX;
warning INTEGER ;
START
DBMS_LOB. CREATETEMPORARY ( ) v_clob TRUE );
DBMS_LOB. CONVERTTOCLOB( )
v_clob ,
p_blob ,
DBMS_LOB. LOBMAXSIZE,
dest_offset ,
offset ,
DBMS_LOB. DEFAULT_CSID,
lang_context ,
warning
);
RETURN v_clob ;
END BLOB2CLOB ;
------------------------------ CLOB2BLOB -------------------------
FUNCTION CLOB2BLOB (p_clob CLOB) RETURN BLOB
ACE
   l_blob     BLOB ;
l_dest_offset INTEGER := 1 ;
l_source_offset INTEGER := 1 ;
l_warning INTEGER ;
lang_context INTEGER := DBMS_LOB. DEFAULT_LANG_CTX;
START
DBMS_LOB. CREATETEMPORARY()l_blob TRUE);
DBMS_LOB. CONVERTTOBLOB( )
l_blob ,
p_clob ,
DBMS_LOB. LOBMAXSIZE,
l_dest_offset ,
l_source_offset ,
DBMS_LOB. DEFAULT_CSID,
lang_context ,
l_warning
);
RETURN l_blob ;
END CLOB2BLOB ;
END;
/
Concerning
Djam

It works well for me.

Update T1 SET DELIVERY_CONTENT is LOB_UTL_PCKG. BLOBREPLACE (DELIVERY_CONTENT, 'old_string', 'new_string')

Where IDSQL > 2 = 1;

0 lines to date.

Y at - there any API in EBS for updating the attributes columns on lines IN... ?
Hi all

I'm looking for some API which will help me to update the columns attribute on the lines of PO. I searched in metalink also but I have not found any idea for that.
Help, please.

Thank you
Ravi Raj.
API in general do not update or manipulate ATTRIBUTE columns from these customer data defined in the store. You will need to update these columns directly.

year https://forums.Oracle.com/forums/Search.jspa?threadID=&q=attribute+and+API+and+Update&objid=C3&DateRange=Last & userID = & numResults = 15 & rankBy = 10001

HTH
Srini

Cannot update the END_DATE_ACTIVE column in the HZ_CODE_ASSIGNMENTS Table through the API
Hello

I'm not able to update the column END_DATE_ACTIVE HZ_CODE_ASSIGNMENTS Table through the API: hz_classification_v2pub.update_code_assignment.
After running with all the facings that he managed, but it is not properly updated the end_date_active to the date given as parameters when executing the API defines it rather end_date_active to null.

Please help.

Thanks and greetings
Salvation;

What is your version of the EBS? If you are on r12.x you can hit bug. Please see:
Unable to update HZ_CODE_ASSIGNMENTS end_date_active with January 1, 2012 [1166367.1 ID]

Respect of
HELIOS

time to update the index
I created a new table and inserted the number of lines (for example near 1000), the index database is updated to take into account (or view) the newly inserted rows. How to find the time to update the taken by the index of prior State to a new State, referring to the newly inserted rows of 1000?
You can try to use the runstats tool to compare 2 different scenarios SQL and deduct the time required for maintenance of indexes.

Scenario1:
Insert 1000 rows in a table without indexes on the table

scenario2:
Insert 1000 rows in the table with the index on the table

time for scenario2 - time for scenario1 is time for index maintenance, provided that you run these tests on hosts/databases without concurrent activity.

See http://asktom.oracle.com/pls/asktom/ASKTOM.download_file?p_file=6551378329289980701

Cannot insert or update the password column
For the site that I am building, I have problems with the Member to add, change, and change password pages. I determined that the problem lies in the insert or update the password for the database column. If I remove the field of password in both add and edit pages members, they work very well (in doing so, to change the password page would go on to defeat its purpose, so have not tried it). I have Dream Weaver 8 allows to design and was wondering if there is something in the log in user and restrict access (user, password and level) it is tool that creates some how would lock the column preventing the modification of the server behaviors. Can someone help me with this you problem?
It's probably a reserved word, try to escape the domain name with brackets
(for example [password]), or rename it.
HTH
--
Tim Carley
www.recfusion.com
[email protected]

update the column of type char to stem after insert or update the same column
I have the table "people":

DESC PEOPLE
Name Null? Type
------------------------------- -------- ------------
ID NOT NULL NUMBER
NAME VARCHAR2 (25) NOT NULL

and I need to make sure that the NAME column is always in uppercase letters, so I did as he illustrated in this link:
http://asktom.Oracle.com/tkyte/mutate/

I did this: -.
==================================
CREATE OR REPLACE TRIGGER parent_bi
BEFORE THE INSERT OR UPDATE
ON people
BEGIN
state_pkg.newrows: = state_pkg.empty;
END;
/
-----------------------------------------------------------------
CREATE OR REPLACE TRIGGER parent_aifer
AFTER INSERT OR UPDATE
ON PEOPLE
FOR EACH LINE
BEGIN
state_pkg.newrows (state_pkg.newrows.COUNT + 1): =: NEW. ROWID;
END;
/
------------------------------------------------------------------
CREATE OR REPLACE TRIGGER HR. PARENT_AI
AFTER INSERT OR UPDATE
ON HUMAN RESOURCES. PEOPLE
REFERRING AGAIN AS NINE OLD AND OLD
FOR EACH LINE
BEGIN
BECAUSE me in 1... state_pkg.newrows.count
LOOP
UPDATE of the people
NAME = UPPER (NAME)
People WHERE. ROWID = state_pkg.newrows (i);
END LOOP;
END;
/
======================================

but nothing is coming, as the same that these triggers do not exist.

So, what is wrong?
and how I workaround to make this statement:-
Setting a DAY people SET NAME = UPPER (NAME);
whenever I update or insert in the name column.
It is much simpler that that - you just need a BEFORE INSERT trigger
```
CREATE OR REPLACE TRIGGER your_trigger_name
  BEFORE INSERT OR UPDATE ON persons
  FOR EACH ROW
BEGIN
  :new.name := UPPER( :new.name);
END;
/
```
As you can see from this example, which will force all uppercase
```
SQL> create table persons (
  2    name varchar2(10)
  3  );

Table created.

SQL> CREATE OR REPLACE TRIGGER your_trigger_name
  2    BEFORE INSERT OR UPDATE ON persons
  3    FOR EACH ROW
  4  BEGIN
  5    :new.name := UPPER( :new.name);
  6  END;
  7  /

Trigger created.

SQL> insert into persons values( 'justin' );

1 row created.

SQL> insert into persons select lower(ename) from emp;

14 rows created.

SQL> select * from persons;

NAME
----------
JUSTIN
SMITH
ALLEN
WARD
JONES
MARTIN
BLAKE
CLARK
SCOTT
KING
TURNER
ADAMS
JAMES
FORD
MILLER

15 rows selected.
```
Also, you generally want to add a check constraint to ensure that NAME = UPPER (NAME).

Justin

Compare and update the same column
Hello

SQL & gt; Select * from
version of v$.

BANNER
-----
Oracle Database 10g Express
Edition Release 10.2.0.1.0 - product
PL/SQL Release 10.2.0.1.0-
Production
CORE 10.2.0.1.0 Production
AMT for 32-bit Windows:
Version 10.2.0.1.0 - Production
NLSRTL Version 10.2.0.1.0-
Production

I'm having a problem to update the column in a Table...

What I'm trying to do is... .compare 2 column and then update one column by report...

Table 1 name: TEST1 _

Bq. LNID------NEW_AN8------OLD_AN8------OLD_DESC------2--------123002------100001------C2370------125846------CS24 100001------123024------100028------C2392------125868------CS25 100028------123025------C2393 100036------125869------100036 123026 CS26------100116------C2394------125870------100116------CS
_ Test2

SZAN8 SZAN82 SZAN86 SZAT1
123 002 100 001 100 001 C
125 846 CS 100 001 100 001
123 024 100 028 100 028 C
125 868 100 028 100 028 CS
125 869 100 036 100 036 CS
123 025 100 036 100 036 C
125 870 CS 100 116 100 116
123 026 100 116 100 116 C

I wanted to update the table SZAN82 of TEST2... based on the value of SZAT1...

I shud be able to choose the record type CS first, then the C type registration and the value of SZAN82 and check if it is present in the TEST1 table if she then I shud update the 82 value updated.

UPDATE LPLSCHEMA. LPL_F0101Z2_CUSTOMER C SET (SZAN82) =
(SELECT NEW_AN8 FROM LPL_CROSS_CUSTOMER B, LPL_F0101Z2_CUSTOMER C WHERE 'C.SZAT1 =' CS AND B.OLD_DESC = 'CS' and C.szan82 = B.old_an8);

I tried like tis... I received a request to void error more than a ranks...

Thanks in advance!

Ananda

Published by: Ananda on January 2, 2009 18:30

Published by: Ananda on January 2, 2009 18:30

Published by: Ananda on January 2, 2009 18:34

Published by: Ananda on January 2, 2009 18:37
```
UPDATE  LPLSCHEMA.LPL_F0101Z2_CUSTOMER C
  SET   SZAN82 = (
                  SELECT  NEW_AN8
                    FROM  LPL_CROSS_CUSTOMER B
                    WHERE C.SZAT1='CS'
                    AND B.OLD_DESC='CS'
                    AND C.szan82=B.old_an8
                 )
  WHERE EXISTS (
                SELECT  1
                  FROM  LPL_CROSS_CUSTOMER B
                  WHERE C.SZAT1='CS'
                    AND B.OLD_DESC='CS'
                    AND C.szan82=B.old_an8
               );
```
SY.

Display warning message if the user tries to update the existing column of trans_dt

Form of Oracle 10g
Hi gurus
I appreciate if someone help out me. I need to display the warning message if the user try to update the existing trans_dt, and I don't know what trigger I use. Thank you
Concerning
Matt

Hello

You can select the value entered by the user and then compare it with the data from the db table

in WHEN-VALIDATE-ITEM trigger

SELECT trans_dt

V_ INTOtrans_dt

table

WHERE THE... -PKs;

Then

IF: trans_dt <> v_trans_dt

THEN

-warning-

-u he could leave it as what or

raise form_trigger_failure;

END IF;

Amatu Allah.

Try to update the new column in the TABLE one by reading another table

I had to add another column to a table and for updating this column, I need to read two other paintings, according to the REF_TYPE_CD.

Update CUST_RQST_DROP c
Set crd_mktr_no = (select mka_mktr_no from marketer_account
where mka_acct_no = c.crd_ref_no
and c.entry_dt between mka_eff_dt and mka_exp_dt
and c.crd_ref_type_cd = 'A');

Update CUST_RQST_DROP c
Set crd_mktr_no = (select aeq_mktr_no from acct_enroll_que
where aeq_work_q_no = c.crd_ref_no
and c.crd_ref_type_cd = 'P');

I don't know how to combine in a single query.

Any suggestions would be greatly appreciated

Well, I still had to build all of the DDL and DML statements me:(dans l'avenir, je recommanderais ajouter ceci dans votre post, vous aurez BEAUCOUP plus de gens prêts à aider si vous le faites.) Just a suggestion for your benefit.

It should work assuming you have a relationship with each other between tablea and b and c... AND you never have more than one row in the child table (b or c to a given value in one).

create table tablea
(
account_no      number,
entry_dt     date,
ref_type     varchar2(1),
refno          number,
mkt_no          number
);

create table tableb
(
id               number,
account_no      number,
mkt_no          number
);

create table tablec
(
id               number,
account_no      number,
mkt_no          number,
eff_dt          date,
exp_dt          date
);

insert into tablea values (123, to_date('04/01/2010','mm/dd/yyyy'), 'P', 9983, null);
insert into tablea values (124, to_date('04/02/2010','mm/dd/yyyy'), 'A', 3452, null);
insert into tablea values (125, to_date('04/02/2010','mm/dd/yyyy'), 'P', 9984, null);

insert into tableb values (9983, 123, 10001);
insert into tableb values (9984, 125, 11232);

insert into tablec values (3452, 124, 10333, to_date('01/01/2010','mm/dd/yyyy'), to_date('05/19/2010','mm/dd/yyyy') );

ME_XE?Update TableA a
set mkt_no =
(
     select mkt_no from tableb b where b.account_no = a.account_no and a.ref_type = 'P'
     union all
     select mkt_no from tablec c where c.account_no = a.account_no and a.ref_type = 'A' and a.entry_dt between c.eff_dt and c.exp_dt
)
where mkt_no is null;

3 rows updated.

Elapsed: 00:00:00.12
ME_XE?select * from tablea;

        ACCOUNT_NO ENTRY_DT             R              REFNO             MKT_NO
------------------ -------------------- - ------------------ ------------------
               123 01-APR-2010 12 00:00 P               9983              10001
               124 02-APR-2010 12 00:00 A               3452              10333
               125 02-APR-2010 12 00:00 P               9984              11232

3 rows selected.

Elapsed: 00:00:00.15

Update the same column in the same table
Hello

How do update you a records of column in the same table?

I have purchase_order of the table. Consists of column ID, color, Purchase_No, Sub purchase.

< pre >

Create table Purchase_Order)
Identification number,
color varchar2 (10),
purchase_No varchar2 (5).
purchase_sub varchar2 (2));
< / pre >

< pre >

ID color Purchase_No purchase Sub
6416 S1406 PURPLE 3
6415 S1406 GREEN 2
6414 S1406 GREEN 1
6419 S1406 3
6417 S1406 1
6418 S1406 2

< pre >

6 unique ID records is in the same site. But 2 ID will be confined to a purchase_sub.

For example

6416,6419 purchase_no S1406 but has the same purchase_sub who need 3.I 6419 color "Purple".

Expected results:
< pre >

ID color Purchase_No purchase Sub
6416 S1406 PURPLE 3
6415 S1406 GREEN 2
6414 S1406 GREEN 1
6419 S1406 PURPLE 3
6417 S1406 GREEN 1
6418 S1406 GREEN 2

< / pre >

Thank you!

Published by: CrackerJack on May 4, 2009 15:04
Hello

Try this,
```
UPDATE Purchase_Order a
   SET a.Colour = (SELECT colour
                     FROM Purchase_Order b
                    WHERE a.Purchase_Sub = b.Purchase_Sub
                      AND b.colour Is Not Null)
 WHERE a.Colour Is Null
```
Kind regards
Christian Balz

An update on an index column with the same value generates an index to the top
An update on an index column with the same value generates an update of the index?

Thank you
In addition to my previous answer, see also

http://orainternals.WordPress.com/2010/11/04/does-an-update-statement-modify-the-row-if-the-update-modifies-the-column-to-same-value/

Riyaj Shamsudeen has this to say:
"+ We have an index on this column v1 and we update this column indexed too." Oracle was updating the indexed column? N ° if the values match the level of the indexed column, then the code of RDBMS isn't up-to-date index, a feature for optimization again. Only the row of table is updated, and the index is not updated. + "

Hemant K Collette

update of NULL in the column with the values of the adjacent column

Examples of data

with test_data as
(select 1 as one, null as two, 2 as three,5 as four, 6 as five, null as six from dual
union all
select 1 as one, null as two, 2 as three,5 as four, 6 as five, null as six from dual)
select * from test_data;

This is one of those cases, the case may be where any value of a column can be null

or two similar columns can be null. If the column is null then I want to update the adjacent column

average value of the column, if the first column is null so I want to take the next non-null column and update, if the last column is null

so I want to take prev not zero column and to update.

In this case would be my expected output

Un	Two	Three	Four	Five	Six
1	1.5	2	5	6	6
1	2	3	5	6	6

Prospects for the future the suggesion or advice.

Or, using Analytics:

SQL> with test_data (id, one, two, three, four, five, six) as (
  2    select 1, 1   , null, 2   , 5, 6, null  from dual union all
  3    select 2, 1   , null, 3   , 5, 6, null  from dual union all
  4    select 3, 1   , null, null, 5, 6, null  from dual union all
  5    select 4, null, null, null, 5, 6, null  from dual
  6  )
  7  select *
  8  from (
  9    select id
 10         , cell
 11         , case when next_nn is not null and prev_nn is not null
 12             then (next_nn + prev_nn)/2
 13             else nvl(next_nn, prev_nn)
 14           end val
 15    from (
 16      select id
 17           , cell
 18           , val
 19           , last_value(val) ignore nulls over(partition by id order by cell) as prev_nn
 20           , first_value(val) ignore nulls over(partition by id order by cell range between current row and unbounded following) as next_nn
 21      from test_data
 22      unpivot include nulls (val for cell in (one as 1, two as 2, three as 3, four as 4, five as 5, six as 6) )
 23    )
 24  )
 25  pivot ( min(val) for cell in (1 as "ONE", 2 as "TWO", 3 as "THREE", 4 as "FOUR", 5 as "FIVE", 6 as "SIX") )
 26  ;

        ID        ONE        TWO      THREE       FOUR       FIVE        SIX
---------- ---------- ---------- ---------- ---------- ---------- ----------
         1          1        1,5          2          5          6          6
         2          1          2          3          5          6          6
         3          1          3          3          5          6          6
         4          5          5          5          5          6          6

update the indexed column

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