overall index using partitioning of the interval.

Can we have an index that is partitioned across the world using range partitioning on the number field in the database table.

>
Can we have an index that is partitioned across the world using range partitioning on the number field in the database table.
>
No - a global index cannot be paritioned of the interval.
You can create a range partitioned or Hash-Partitioned Global index, but it cannot use partitioning interval. The index may be on a table is partitioned apart, but the index itself cannot be partitioned apart.

See Creation of Partitions in the VLDB and partitioning Guide database
http://docs.Oracle.com/CD/E18283_01/server.112/e16541/part_admin001.htm#i1006455

and the CREATE INDEX SQL language doc section
http://docs.Oracle.com/CD/E11882_01/server.112/e26088/statements_5012.htm#i2062403

Tags: Database

Similar Questions

Partitions of the interval

Hi guys,.
I use DBMS_REDEFINITION to partition tables not partitioned in order to facilitate better performance. My question is; These tables already have clues about them. I read scores interval may not work on nested index tables. I need advice here. Can I used partitions of the interval on tables with indexes on the?

by default, oracle creates a table as an array of heap and stores the rows in its blocks without a special order. On a heap table (which can be partitioned), you can create indexes. An index organized table (IOT) is basically an index without an additional table of the segment (but you can also define a segment of overflow, but it is another story). There are some cases of use of IOT, but in most situations, a heap table is quite sufficient.

By the way: here's an example for the redefinition of a table with partitions of the interval: http://uhesse.com/2010/02/15/partitioning-a-table-online-with-dbms_redefinition/

Range of partitioning of the interval
Hello

Using Oracle 11.20.3
We want to have a table partitioned composite range-hash, a partition for each month.

However, want to esnure are still a tehrfore partition would use range interval partitioning.

Have a few questions, if I say 201301 to 201312 but get intellectualization rosw partition with a value date in June 2014 it create
a new partition for each month IE 201401, 201402 etc. or just a 201406?

Wanyt esnure system can automatically dial with a scenario

Daty1 have partitions 201301 to 201312

Day 2 get record for 201406 (June 2014)

Day 3 get record for 201403 (March 2014)

We have no control over the date range, we were able to get but don't want to tpo have to manually create partitions BOF tehse all dates in advance - want system she managed automatically.

Also you can just rename it the names of system partition generated to something more useful later.
Yes, the creation of partitions of the interval is 'rare' - a given partition will be created only when necessary. (This should be easy to try, by the way).

Query for the partitions of the interval, system and reference

Hi, I want to ask some of range partitions, system and reference information.

In the partition of the interval, the script looks like this

CREATE TABLE interval_sales

(prod_id NUMBER (6))

cust_id NUMBER

time_id DATE

)

PARTITION OF RANGE (time_id)

INTERVAL (NUMTOYMINTERVAL (1, 'MONTH'))

(PARTITION p0 VALUES LESS THAN (TO_DATE (' 01/01/2005 ',' DD-MM-YYYY "")));

I would like to write a query to find the expression of the interval "MONTHS" but could not find what he makes.

In the reference partition, the script looks like this

CREATE TABLE child_tab)

ID NUMBER NOT NULL,

parent_tab_id NUMBER NOT NULL,

code VARCHAR2 (10),

Description VARCHAR2 (50).

CREATED_DATE DATE,

CONSTRAINT child_tab_pk PRIMARY KEY (id),

CONSTRAINT child_parent_tab_fk FOREIGN KEY (parent_tab_id)

REFERENCES parent_tab (id)

)

PARTITION BY REFERENCE (child_parent_tab_fk);

I want to write a query to find the foreign key name 'child_parent_tab_fk' after the PARTITION BY REFERENCE to the last line, but have no luck so far.

In the end, I want to achieve this goal:

If (interval_partitioning_type)

Assign the interval expression to a string variable

Another yew (reference_partitioning_type)

Set the foreign key to a variable of type of foreign key

Is there a way to get the type of partitioning? (interval and reference here)

Please advise,

Jack

You can query ALL_PART_TABLES

`REF_PTN_CONSTRAINT_NAME`	`VARCHAR2(30)`		Name of the referential partitioning constraint for partitioned tables reference
`INTERVAL`	`VARCHAR2(1000)`		The value of the interval chain

Partition wise joined possible with partitions of the interval?

Hello
I want to know the score wise join (NTC) is possible with interval partitioning - I can't find an explicit statement that he isn't, but I can't make it work - I did a simple test case to illustrate the issue.
below, I have 2 create table scripts - 1 for the case of interval and 1 for the case of hash - I then a simple query on these 2 objects which should produce a NTC.
In the case of hash, it works very well (see screenshot 2nd with a set of slaves), the first screenshot shows the case of the interval where I find myself with 2 sets of slaves and no NTC.
No idea if this is possible and I just missed something?
(for the test case choose the names of schema/storage appropriate for your system)
Oh and version (I almost forgot... :-))-East 11.2.0.4.1 SLES 11)
See you soon,.
Rich
-case interval
CREATE TABLE 'SB_DWH_IN '. "' TEST1 '.
TABLESPACE "SB_DWH_INTEGRATION".
PARTITION BY RANGE ("OBJECT_ID") INTERVAL (10000)
(PARTITION 'LESS_THAN_ZERO' VALUES LESS THAN (0) TABLESPACE "SB_DWH_INTEGRATION")
in select * from DBA_OBJECTS where object_id is not null;
CREATE TABLE 'SB_DWH_IN '. "" TEST2 ".
TABLESPACE "SB_DWH_INTEGRATION".
PARTITION BY RANGE ("OBJECT_ID") INTERVAL (10000)
(PARTITION 'LESS_THAN_ZERO' VALUES LESS THAN (0) TABLESPACE "SB_DWH_INTEGRATION")
in select * from DBA_OBJECTS where object_id is not null;
-case of hash
CREATE TABLE 'SB_DWH_IN '. "' TEST1 '.
TABLESPACE "SB_DWH_INTEGRATION".
8 partitions PARTITION OF HASH ("OBJECT_ID")
store in ("SB_DWH_INTEGRATION")
in select * from DBA_OBJECTS where object_id is not null;
CREATE TABLE 'SB_DWH_IN '. "" TEST2 ".
TABLESPACE "SB_DWH_INTEGRATION".
8 partitions PARTITION OF HASH ("OBJECT_ID")
store in ("SB_DWH_INTEGRATION")
in select * from DBA_OBJECTS where object_id is not null;
-query to run
Select / * + PARALLEL(TEST2,8) PARALLEL(TEST1,8) * / *.
of 'SB_DWH_IN '. "" TEST2 ","SB_DWH_IN ". "' TEST1 '.
where TEST1.object_id = test2.object_id

It is planned and a consequence of the estimate of the number of parallel slaves.

To the parallel 41 each slave made 3 passes (i.e. sleeves 3 partitions).

Add a partition (by table), and a set of slaves will have to manage a 4th pass: the cost of the query using NTC would increase from 33 percent even if the modification of the data is less than 0.8%.

I guess that in the production Oracle distributes your lines of 1 M for a hash join.

Because the decision is encrypted, it is possible that a very extreme tilt in partition in the table sizes billion line might overthrow the optimizer in a non - NTC join - but I have not tested that.

If you want to force the plan John Watson suggestion for a hint of pq_distribute is relevant. To cover all the bases and call your tables SMALL and LARGE

/*+

leading (FAT kid)

USE_HASH (large)

no_swap_join_inputs (large)

PQ_DISTRIBUTE (wide none none)

*/

If it's legal, that should do it.

Concerning

Jonathan Lewis

How to search the dictionary of data for the partitions of the interval?

Oracle 11.2.0.3
When I have a DBA_PART_TABLES request. PARTITIONING_TYPE he says "beach". I want to be able to differential an interval of other partitions of range based in the data dictionary?

used this example. Watch "range".
http://www.rittmanmead.com/2008/09/investigating-Oracle-11g-interval-partitioning/
Oracle Docs list 'SYSTEM' as a partitioning_type that I thought would be intervals.
ALL_PART_TABLES
[code]
1 CREATE TABLE interval_sales
2 (prod_id NUMBER (6))
3, cust_id NUMBER
4, time_id DATE
5, channel_id char (1)
6, promo_id NUMBER (6)
7, quantity_sold NUMBER (3)
8, amount_sold NUMBER (10.2)
9 )
10 PARTITION BY RANGE (time_id)
INTERVAL OF 11 (NUMTOYMINTERVAL (1, 'MONTH'))
12 (PARTITION p0 VALUES LESS THAN (TO_DATE (' 01/01/2005 ',' DD-MM-YYYY "")),)
13 PARTITION p1 VALUES LESS THAN (TO_DATE (' 01/01/2006 ',' DD-MM-YYYY ""));
14 p2 PARTITION VALUES LESS THAN (TO_DATE (' 07/01/2006 ',' DD-MM-YYYY ""));
15 * PARTITION p3 VALUES LESS THAN (TO_DATE (' 01/01/2007 ',' DD-MM-YYYY "")))
SQL > /.
Table created.
Elapsed time: 00:00:00.01
SQL > select partitioning_type from the user_part_tables where table_name = 'INTERVAL_SALES ';
PARTITION
---------
RANGE
[/ code]
```
CREATE TABLE junk1
   ( id         NUMBER )
   PARTITION BY RANGE (id)
   INTERVAL(5)
     ( PARTITION p0 VALUES LESS THAN (5),
       PARTITION p1 VALUES LESS THAN (10),
       PARTITION p2 VALUES LESS THAN (15),
       PARTITION p3 VALUES LESS THAN (20) )
/

CREATE TABLE junk2
   ( id         NUMBER )
   PARTITION BY RANGE (id)
     ( PARTITION p0 VALUES LESS THAN (5),
       PARTITION p1 VALUES LESS THAN (10),
       PARTITION p2 VALUES LESS THAN (15),
       PARTITION p3 VALUES LESS THAN (20) )
/

select owner, table_name, partitioning_type, interval
  from dba_part_tables
 where table_name in ('JUNK1', 'JUNK2')
/
```
TABLE_NAME OWNER PARTITIONING_TYPE INTERVAL

------------------------------ ------------------------------ ----------------- --------------------

GREGS JUNK1 RANGE 5

GREGS JUNK2 RANGE

2 selected lines.

can I use partition with the following

Hi, I wanted to know if I can use or partition the row to the GET the necessary results below. If Yes can someone help me with it.
If there is another way to get results as you please let me know.

I use Oracle 10.2.0.4.0 version

Create a table script:

CREATE TABLE DMM.QUES
(
  CASENAME        VARCHAR2(100 BYTE),
  CASENUMBER      VARCHAR2(20 BYTE),
  COVERAGE        VARCHAR2(10 BYTE),
  DIVISIONNUMBER  VARCHAR2(6 BYTE),
  CLASSNUMBER     number,
  TIER            VARCHAR2(2 BYTE),
  MONTHLYRATE     NUMBER
  )

Insert manuscripts of the declaration:

insert into QUES (CASENAME,CASENUMBER,  COVERAGE,  DIVISIONNUMBER,  CLASSNUMBER,  TIER,MONTHLYRATE)
values ('ABC','123J','health', '456J','987',null,.25)
insert into QUES (CASENAME,CASENUMBER,  COVERAGE,  DIVISIONNUMBER,  CLASSNUMBER,  TIER,MONTHLYRATE)
values ('ABC','123J','health', '456J','123',null,.25)
insert into QUES (CASENAME,CASENUMBER,  COVERAGE,  DIVISIONNUMBER,  CLASSNUMBER,  TIER,MONTHLYRATE)
values ('ABC','123J','health', '456J','453',null,.25)
insert into QUES (CASENAME,CASENUMBER,  COVERAGE,  DIVISIONNUMBER,  CLASSNUMBER,  TIER,MONTHLYRATE)
values ('ABC','123J','dental', '456J','453','ee',.29)
insert into QUES (CASENAME,CASENUMBER,  COVERAGE,  DIVISIONNUMBER,  CLASSNUMBER,  TIER,MONTHLYRATE)
values ('ABC','123J','dental', '456J','453','es',.23)
insert into QUES (CASENAME,CASENUMBER,  COVERAGE,  DIVISIONNUMBER,  CLASSNUMBER,  TIER,MONTHLYRATE)
values ('ABC','123J','dental', '456J','453','ec',.44)
insert into QUES (CASENAME,CASENUMBER,  COVERAGE,  DIVISIONNUMBER,  CLASSNUMBER,  TIER,MONTHLYRATE)
values ('ABC','123J','dental', '456J','453','fa',.33)
insert into QUES (CASENAME,CASENUMBER,  COVERAGE,  DIVISIONNUMBER,  CLASSNUMBER,  TIER,MONTHLYRATE)
values ('ABC','123J','dental', '456J','123','ee',.45)
insert into QUES (CASENAME,CASENUMBER,  COVERAGE,  DIVISIONNUMBER,  CLASSNUMBER,  TIER,MONTHLYRATE)
values ('ABC','123J','dental', '456J','123','es',.45)
insert into QUES (CASENAME,CASENUMBER,  COVERAGE,  DIVISIONNUMBER,  CLASSNUMBER,  TIER,MONTHLYRATE)
values ('ABC','123J','dental', '456J','123','ec',.46)
insert into QUES (CASENAME,CASENUMBER,  COVERAGE,  DIVISIONNUMBER,  CLASSNUMBER,  TIER,MONTHLYRATE)
values ('ABC','123J','dental', '456J','123','fa',.49)

Select * of question:

ABC     123J     health     456J     987          0.25
ABC     123J     health     456J     453          0.25
ABC     123J     health     456J     123          0.25
ABC     123J     dental     456J     453     ee     0.29
ABC     123J     dental     456J     453     es     0.23
ABC     123J     dental     456J     453     ec     0.44
ABC     123J     dental     456J     453     fa     0.33
ABC     123J     dental     456J     123     ee     0.45
ABC     123J     dental     456J     123     es     0.45
ABC     123J     dental     456J     123     ec     0.46
ABC     123J     dental     456J     123     fa     0.49

I would get the following result

ABC     123J     health     456J     123          0.25
ABC     123J     dental     456J     123     ee     0.45
ABC     123J     dental     456J     123     es     0.45
ABC     123J     dental     456J     123     ec     0.46
ABC     123J     dental     456J     123     fa     0.49

I would like the code to take the minimum of classnumber for each combination of divisonnumber/cover and give me all the records in this classnumber.

Thanks in advance.

Hello

This is an example of a Query of Top - N , and Yes, the analytic functions are a good way to do it:

WITH     got_r_num     AS
(
     SELECT     ques.*
     ,     RANK () OVER ( PARTITION BY  coverage
                           ,          divisionnumber
                      ORDER BY          classnumber
                    ) AS r_num
     FROM    ques
)
SELECT  casename, casenumber, coverage, divisionnumber, classnumber, tier, monthlyrate
FROM     got_r_num
WHERE     r_num     = 1
;

Thanks for posting the CREATE TABLE and INSERT, it is very useful!

"PARTITION OF ' is a term often used with analytic functions. "PARTTIION" means also at least two other things in Oracle, so people you better understand if you say "analytic functions", rather than "partition by.

For more info about the RANK function, see this thread:
Re: A "working group"double"SQL query

Impact of setting the interval partitioning off and then on again

Hi all
Have discovered that to split the partitions of a partitioned table Beach-interval, in that we need to define the partitioning interval all issue the command split then on again.
e, g.
ALTER TABLE retailer_transaction SET INTERVAL ();
ALTER TABLE POR2_retailer_transaction
SPLIT PARTITION PART_400012 AT (TO_DATE (' 2020-09-01 00:00:00 ',' SYYYY-MM-DD HH24:MI:SS ',' NLS_CALENDAR = GREGORIAN '))
INTO (PARTITION PART_202008,
PART_400012 PARTITION)
ALTER TABLE retailer_transaction SET INTERVAL (NUMTOYMINTERVAL(1,'MONTH'))
Everything we need to pay attention to that?
Using oracle 11.2.0.3
Thank you

You can NOT split the LAST (most HIGH_VALUE) partition of a partitioned table interval interval. You can only separate partitions BEFORE the last. Which converts also all partitions from the bottom (split and below) for the partitions of the RANGE.

If interval partitioning will only apply to data added above the transition point (the score higher.

To return to your original question. To split the LAST interval follow the procedure described in your original post

1. range partitioning

2 split the partitioning

3 set the partitioning interval on

Conclusion with this however when adding data back to a nine month doesn't automatically create the partition again.

This "new month" MUST BE above the point of passage for a new partition to create. In your code example, this point of transition is for the year 4000.

This works very well for me:

ALTER TABLE sales PARTITION RENAME SYS_P1122 to SALE_400012

ALTER TABLE sale SET INTERVAL ();

ALTER TABLE SPLIT PARTITION for sale

SALE_400012 TO (TO_DATE('01/02/2020','DD/MM/YYYY')) IN

(

SALE_202001 PARTITION,

SALE_400012 PARTITION

);

ALTER TABLE sale SET INTERVAL (NUMTOYMINTERVAL(1,'MONTH'))

INSERT INTO sales

(SALES_DATE, STORE_NUM, QTY_SOLD)

values (to_date('31/12/4001','DD/,MM/YYYY'), 1, 10);

This last INSERT statement above creates a NEW partition because the date (31/12/4001) is above the transition point.

Looks like you want to move the transition point downwards. That is - you were using this year there date of 4000 as a 'catch all '.

1. set the partitioning interval

2 split that last partition into two NEW partitions (new names) by using a VALUE that is high enough to contain ALL the data you have in this last partition.

3. remove the last partition - it should now be empty

4. turn on range partitioning

Now, any data that you insert above this NEW point of transition (the limit that you used for the new partition) it will create a new partition of the interval. Make sure the border is really an end of month, if that's what you want.

If your data sample had this

INSERT INTO sales

(SALES_DATE, STORE_NUM, QTY_SOLD)

values (to_date('05/10/2016','DD/,MM/YYYY'), 1, 10);

INSERT INTO sales

(SALES_DATE, STORE_NUM, QTY_SOLD)

values (to_date('31/12/4000','DD/,MM/YYYY'), 1, 10);

So for the step #2 above use split date 06/2016 so that the line above for 10/05/2106 goes to the new partition and 12/31/4000 stays in the old.

Then, drag this old and lines in it (31/12/4000).

You can NOT split the last partition if interval is always on because it is the last partition interval.

And as long as you have given for 12/31/4000, you will get no new partitions unless you insert was more TOP than this.

So if you want partitioning interval to activate again 06/2016 that higher data would be deleted and the score he falls.

You have discovered only among "traps" with the help of partitioning of the interval. Oracle will happily create new partition needed to keep the data. If ANYONE, even accidentally, between a date as 12/31/4000 that nobody would ever know until the new data seem to DISAPPEAR for some use cases.

This is because all data will start to enter this new partition ACCIDENTALLY created instead of a partition for the month, it's part.

And, and you have discovered, you cannot split this partition directly.

Delayed segment creation can allow you to pre-create the partitions ahead without actually affecting the segment for them until the first row is inserted.

Interval partitioning will create the appropriate partitions but not if someone accidentally or intentionally, data will be a valuable capitalized in the table.

is an overall index provides fast response in OLTP times?

Hi all
I read a doc
http://docs.Oracle.com/CD/E18283_01/server.112/e16541/partition.htm#insertedID5
and it is said that If the application is an OLTP we and users needs quick response time, use an overall index
I don't know how an overall index provides quick access.
Assume there is a table and that is partitioned by hash.
This table has a local index on column owner - no partition key column.
To query the information of owner of SCOTT, Oracle will scan all index partitions that have given Scott.
On the other hand, there is an overall index on column 'owner '.
To query the information of SCOTT, Oracle probes also all index partitions that have given Scott.
So, here's the question.
Why everyone says that an overall index has good performance (response time) in OLTP?
Supplementary question
Similarly with above, there is a table copied from object. And the table is partitioned by hash (object_id).
To the query with where clause (where owner = xxx), If there is a local index, Oracle scans RANGE SCAN on the t_idx_local index in the first partition through the last partition (take a look at Pstart, Pstop)
But when Oracle uses the overall index, Pstart. Pstop column values are ' KEY | KEY "."
What does that mean??
Thank you.

The KEY to the formulation of the plan means that Oracle will use pruning partitiong dynamic rather than static.

Only it Oracle knows he can partition plum but it determines real partitions during execution, rather than during the analysis.

It's EXACTLY what you'd expect with an OVERALL index. Oracle search the index keys in the overall index and at this time there may find that some lines are in partition 13, others in the bulkhead 27, etc..

With a local index Oracle uses more generally STATIC pruning. You have a local index with a MONTHLY distribution key and your query wants all data of MAY. Oracle knows at the time you want the data for MAY and can use the data dictionary to determine what partitions have CAN data since the index INCLUDES partitiong keys.

How to create the INTERVAL partitioning using EM?

Hi guys
I'm still not able to use the INTERVAL partitioning using EM. Is it possible to do?
Oracle 11 g 2 on Linux
Thank you
John

Hi John,.

I don't think it's possible, that the option of partition of the range seems to have things that make you think of it would generate the definition as the partitioning interval but it doesn't.

I'm on 12.1.0.3 of the SGC and 12.1.0.5 DB plugin so I'm sure it's 'missing' features for the moment.

See you soon,.

Rich

the question of the overall index was updated
I have a partitioned table and the local partitioned indexes in 9i. In the past, when I add a new partition, partitioned indexes have become "unusable" so I have to rebuild the index to the "exploitable": State

ALTER INDEX REBUILD PARTITION myindex TYPE my_partition nologging;

I just found out that when you use the clause to UPDATE GLOBAL INDEXES, there is no need to rebuild the local index in the underlying table. It seems that updating the GLOBAL INDEX also update local index to the status of "exploitable".

create tablespace you see

ALTER TABLE my_table SPLIT PARTITION of VALUES 'TYPE_DEFAULT' ('new') INTO (PARTITION "TYPE_NEW" TABLESPACE "you see", "TYPE_DEFAULT" PARTITION) UPDATE GLOBAL INDEXES;

I wonder if UPDATE GLOBAL INDEXES also update local indexes as it appears in this case. I thought that the UPDATE GLOBAL INDEXES only updated global index.

Published by: user7435395 on March 8, 2013 14:33

Published by: user7435395 on March 8, 2013 14:36

Published by: user7435395 on March 8, 2013 15:15
>
I just found out that when you use the clause to UPDATE GLOBAL INDEXES, there is no need to rebuild the local index in the underlying table. It seems that updating the GLOBAL INDEX also update local index to the status of "exploitable".
>
Interesting! Can you post the DDL that you use for the table and index, so we can try to reproduce it?

This seems to contradict what said documentation, but that would not be something new.

See "Split of Partitions" in the Guide DBA 9i
http://docs.Oracle.com/CD/B10501_01/server.920/a96521/partiti.htm#6736
>
Behavior of the index

Regular (TAS)
Oracle UNUSABLE brand new partitions (there are two) in each local index.

Unless you specify GLOBAL INDEXES to UPDATE, all index global, or all partitions of global partitioned indexes are marked UNUSABLE and must be rebuilt
>
Also you not "add" a partition you split one and creates two new partitions, which both can have data. These two partitions of the local index must be rebuilt.

When you "add" a partition it lacks all the data so there is not need to rebuild the local index for this 'new' partition.

Try to convert the partitioned Table of interval in the range... Swap partition...

Requirement:

Interval of replacement partitioned Table by range partitioned Table

DROP TABLE A;

CREATE TABLE A
(
   a              NUMBER,
   CreationDate   DATE
)
PARTITION BY RANGE (CreationDate)
   INTERVAL ( NUMTODSINTERVAL (30, 'DAY') )
   (PARTITION P_FIRST
       VALUES LESS THAN (TIMESTAMP ' 2001-01-01 00:00:00'));


INSERT INTO A
     VALUES (1, SYSDATE);

INSERT INTO A
     VALUES (1, SYSDATE - 30);

INSERT INTO A
     VALUES (1, SYSDATE - 60);

I need to change this partitioned Table apart to a partitioned range Table. I can do using the EXCHANGE PARTITION. Like if I use the classic method to create another table range partitioned, then:

DROP TABLE A_Range
CREATE TABLE A_Range
(
a NUMBER,
CreationDate DATE
)
PARTITION BY RANGE (CreationDate)
   (partition MAX values less than (MAXVALUE));

Insert  /*+ append */  into A_Range Select * from A; --This Step takes very very long..Trying to cut it short using Exchange Partition.

Problems:

I can't do

 ALTER TABLE A_Range
  EXCHANGE PARTITION MAX
  WITH TABLE A
  WITHOUT VALIDATION;

ORA-14095: ALTER TABLE CHANGE requires a not partitioned table nonclustered
This is because the tables are partitioned. So it does not allow me.

If I instead:

Create a table that is not partitioned for exchanging data by partition.

  Create Table A_Temp as Select * from A;
  
   ALTER TABLE A_Range
  EXCHANGE PARTITION MAX
  WITH TABLE A_TEMP
  WITHOUT VALIDATION;
   
  select count(*) from A_Range partition(MAX);

-The problem is that all the data is in MAX Partition.
Even after the creation of a large number of partitions by walls of separation, the data is still in MAX Partition only.

So:

-What we cannot replace a partitioned Table to the Table partitioned using the EXCHANGE PARTITION range interval. that is, we have to insert in...
-We can do it, but I'm missing something here.
-If all the data is in MAX Partition due to "WITHOUT VALIDATION", can say us be redistributed in the right type of range partitions.

You must pre-create the partitions in a_range and then swap one for one for a tmp, and then to arange. With the help of your sample (thanks to proviing code, incidentally).

SQL> CREATE TABLE A
  2  (
  3     a              NUMBER,
  4     CreationDate   DATE
  5  )
  6  PARTITION BY RANGE (CreationDate)
  7     INTERVAL ( NUMTODSINTERVAL (30, 'DAY') )
  8     (PARTITION P_FIRST
  9         VALUES LESS THAN (TIMESTAMP ' 2001-01-01 00:00:00'));

Table created.

SQL> INSERT INTO A VALUES (1, SYSDATE);

1 row created.

SQL> INSERT INTO A VALUES (1, SYSDATE - 30);

1 row created.

SQL> INSERT INTO A VALUES (1, SYSDATE - 60);

1 row created.

SQL> commit;

Commit complete.

You can find the form of existing partitions assistance:

SQL> select table_name, partition_name, high_value
  2  from user_tab_partitions
  3  where table_name = 'A';

TABLE_NAME PARTITION_NAME HIGH_VALUE
---------- -------------- --------------------------------------------------------------------------------
A          P_FIRST        TO_DATE(' 2001-01-01 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIA
A          SYS_P44        TO_DATE(' 2013-01-28 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIA
A          SYS_P45        TO_DATE(' 2012-12-29 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIA
A          SYS_P46        TO_DATE(' 2012-11-29 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIA

You can then create the table a_range with apporopriate partitions. Note that you may need to create additional in a_range partitions because the partitioning interval does not create the partitions has no data for, even if that leaves 'holes' in the partitioning scheme. So, on that basis:

SQL> CREATE TABLE A_Range (
  2     a NUMBER,
  3     CreationDate DATE)
  4  PARTITION BY RANGE (CreationDate)
  5     (partition Nov_2012 values less than (to_date('30-nov-2012', 'dd-mon-yyyy')),
  6      partition Dec_2012 values less than (to_date('31-dec-2012', 'dd-mon-yyyy')),
  7      partition Jan_2013 values less than (to_date('31-jan-2013', 'dd-mon-yyyy')),
  8      partition MAX values less than (MAXVALUE));

Table created.

Now, create a regular table to use in the constituencies:

SQL> CREATE TABLE A_tmp (
  2     a              NUMBER,
  3     CreationDate   DATE);

Table created.

and all partitions in Exchange:

SQL> ALTER TABLE A
  2    EXCHANGE PARTITION sys_p44
  3    WITH TABLE A_tmp;

Table altered.

SQL> ALTER TABLE A_Range
  2    EXCHANGE PARTITION jan_2013
  3    WITH TABLE A_tmp;

Table altered.

SQL> ALTER TABLE A
  2    EXCHANGE PARTITION sys_p45
  3    WITH TABLE A_tmp;

Table altered.

SQL> ALTER TABLE A_Range
  2    EXCHANGE PARTITION dec_2012
  3    WITH TABLE A_tmp;

Table altered.

SQL> ALTER TABLE A
  2    EXCHANGE PARTITION sys_p46
  3    WITH TABLE A_tmp;

Table altered.

SQL> ALTER TABLE A_Range
  2    EXCHANGE PARTITION nov_2012
  3    WITH TABLE A_tmp;

Table altered.

SQL> select * from a;

no rows selected

SQL> select * from a_range;

         A CREATIOND
---------- ---------
         1 23-NOV-12
         1 23-DEC-12
         1 22-JAN-13

John

Using PARTITION FOR / SUBPARTITION FOR the syntax to insert

We try to use PARTITION syntax for (or, better, SUBPARTITION of syntax) to insert into a table subpartitioned.
http://docs.Oracle.com/CD/E11882_01/server.112/e26088/sql_elements009.htm#i165979

08:26:46 GM_CS_CDR@oradev02> select * from v$version;

BANNER
--------------------------------------------------------------------------------
Oracle Database 11g Enterprise Edition Release 11.2.0.3.0 - 64bit Production
PL/SQL Release 11.2.0.3.0 - Production
CORE    11.2.0.3.0      Production
TNS for Solaris: Version 11.2.0.3.0 - Production
NLSRTL Version 11.2.0.3.0 - Production

5 rows selected.

This table is interval partitions DATA_ORIGIN_ID and list subpartitions on TABLE_NAME.

First of all, an insert in base without specifying the partitions:

08:10:05 GM_CS_CDR@oradev02> insert into key_ids2 (
08:11:51   2  DATA_ORIGIN_ID         ,
08:11:51   3  TABLE_NAME             ,
08:11:51   4  DUR_UK                 ,
08:11:51   5  DUR_UK_ID              )
08:11:51   6  select
08:11:51   7  12         ,
08:11:51   8  TABLE_NAME             ,
08:11:51   9  DUR_UK                 ,
08:11:51  10  DUR_UK_ID              
08:11:51  11  from key_ids where table_name = 'PART' 
08:11:51  12  and rownum <= 10;

10 rows created.

Check SUBPARTITION to a SELECT statement syntax:

08:19:43 GM_CS_CDR@oradev02> 
08:26:45 GM_CS_CDR@oradev02> 
08:26:45 GM_CS_CDR@oradev02> 
08:26:45 GM_CS_CDR@oradev02> 
08:26:45 GM_CS_CDR@oradev02> select count(*) from key_ids2 
08:26:45   2  subpartition for(12,'PART');

  COUNT(*)
----------
        10

1 row selected.

But if we add a subpartition specification (to limit the locking to a single subpartition), we get a syntax error:

08:14:57 GM_CS_CDR@oradev02> insert into key_ids2 subpartition for(12,'PART') (
08:14:57   2  DATA_ORIGIN_ID         ,
08:14:57   3  TABLE_NAME             ,
08:14:57   4  DUR_UK                 ,
08:14:57   5  DUR_UK_ID              )
08:14:57   6  select
08:14:57   7  14         ,
08:14:57   8  TABLE_NAME             ,
08:14:57   9  DUR_UK||'!'                 ,
08:14:57  10  DUR_UK_ID              
08:14:57  11  from key_ids 
08:14:57  12  where table_name = 'PART' 
08:14:57  13  and rownum <= 10;
insert into key_ids2 subpartition for(12,'PART') (
                                     *
ERROR at line 1:
ORA-14173: illegal subpartition-extended table name syntax

Specify the partition level did not work either:

08:14:58 GM_CS_CDR@oradev02> insert into key_ids2 partition for(14) (
08:15:23   2  DATA_ORIGIN_ID         ,
08:15:23   3  TABLE_NAME             ,
08:15:23   4  DUR_UK                 ,
08:15:23   5  DUR_UK_ID              )
08:15:23   6  select
08:15:23   7  14         ,
08:15:23   8  TABLE_NAME             ,
08:15:23   9  DUR_UK||'!'                 ,
08:15:23  10  DUR_UK_ID              
08:15:23  11  from key_ids 
08:15:23  12  where table_name = 'PART' 
08:15:23  13  and rownum <= 10;
insert into key_ids2 partition for(14) (
                                  *
ERROR at line 1:
ORA-14108: illegal partition-extended table name syntax

But by specifying explicit partition and subpartition works:

08:17:45 GM_CS_CDR@oradev02> insert into key_ids2 partition (SYS_P15127) (
08:18:23   2  DATA_ORIGIN_ID         ,
08:18:23   3  TABLE_NAME             ,
08:18:23   4  DUR_UK                 ,
08:18:23   5  DUR_UK_ID              )
08:18:23   6  select
08:18:23   7  12         ,
08:18:23   8  TABLE_NAME             ,
08:18:23   9  DUR_UK||'!'                 ,
08:18:23  10  DUR_UK_ID              
08:18:23  11  from key_ids 
08:18:23  12  where table_name = 'PART' 
08:18:23  13  and rownum <= 10;

10 rows created.

08:18:24 GM_CS_CDR@oradev02> insert into key_ids2 subpartition (SYS_SUBP15126) (
08:19:42   2  DATA_ORIGIN_ID         ,
08:19:42   3  TABLE_NAME             ,
08:19:42   4  DUR_UK                 ,
08:19:42   5  DUR_UK_ID              )
08:19:42   6  select
08:19:42   7  12         ,
08:19:42   8  TABLE_NAME             ,
08:19:42   9  DUR_UK||'!#'                 ,
08:19:42  10  DUR_UK_ID              
08:19:42  11  from key_ids 
08:19:42  12  where table_name = 'PART' 
08:19:42  13  and rownum <= 10;

10 rows created.

We have succeeded by using the syntax of PARTITION for tables partitioned, but not sous-partitionnee.

Any ideas?

Thank you
Mike

Support of Oracle reproduced this issue. The resolution was to ensure cursor_sharing is set to "EXACT", we happened to use "SIMILAR" in this case due to a migration of 10 g and 11 g.

Please visit 1481564.1 for more information.

Partitioning of the network using the API of LOD
Hi all,

as the continuity of this thread.
Difference between partition and partition of BLOBs in NDM
I'm going to make my network partitioning; My network has links 7 817 372 and 6 662 079 nodes, so I think that that is classified as a large network. For this reason, I will exercise the procedures sdo_net.spatial_partition and sdo_net.generate_partition_blobs. I initially put max * 10000 * nodes for the partition.

Relatively to the sdo_net.spatial_partition procedure, I see the tutorial on hillsborough network, where:

-of partition HILLSBOROUGH_NETWORK on link 1 (default link_level) in partitions with at most
knots - 2000
EXEC sdo_net.spatial_partition ('HILLSBOROUGH_NETWORK', ' HILLSBOROUGH_NETWORK_PART$ ', 2000, "WORK_DIR', 'hillsborough.log', 'a'");

-of partition HILLSBOROUGH_NETWORK on link 2
EXEC sdo_net.spatial_partition ('HILLSBOROUGH_NETWORK', ' HILLSBOROUGH_NETWORK_PART$ ', 2000, "WORK_DIR', 'hillsborough.log', 'a', 2 ');

All link level setting, as shown on the guide, topology and designer of the model data network link has been defined as the reflects the level of priority for the link and is used for network analysis, so that links with the higher priority levels can be considered first in the computer path.

My question is:
Link level setting refers to insert value in the table LINK_LEVEL to $ LINK?
If Yes, if I did not insert into the table LINK_LEVEL field of $ LINK, do I put this link level parameter must be 1 (default link_level)?

Thank you in advance.
```
UserData minutes = link.getUserData(2); //2 is the category_id of UserData
```
A UserData object contains a list of data values from the user to the link or node. For example, a link can have a name of user data "MINUTES" and another named "km/h" in the same category of the user data. So to get the value of a specific user data, you must call UserData.get (int userDataIndex). See the following example:
UserData linkUserData = link.getUserData (2); 2 is the stuff of UserData
Assuming that you only have a single user data recorded for the links in the category of user 2 data, then you get the MINUTES value using:
Double minutes = (Double) linkUserData.get (0); 0 is the index of the data of the user "MINUTES".

If you have several user data stored for the link in the same category, then you can get the index for each user's data by querying USER_SDO_NETWORK_USER_DATA:
Select rownum-1 data_index, user_sdo_network_user_data data_name
where table_type = 'LINK' and stuff = 2
by data_name orderY;

Also, after registering new user of your network data, remember to regenerate the blobs of partition if you want to load blob partitions during the analysis of network.

How expand you a healthy (C:\) partition in the unallocated disk space - used to be partitioned as D:\?

At some point in the past, someone partitioned my hard drive in C:\ = 15 GB and D:\ = 95 GB. It is the source of the problems. I can't afford to pay someone to fix it as I became unemployed. SO after some research, I gutted and then deleted the partition D:\. Then I went to "diskpart" in command prompt and typed "extend", thinking that would extend the C drive in the now unallocated space that used to be the partition D:\. Got mssg "part of disk cannot extend the volume. Please make sure the volume is valid for the extension. "How do I do that? Thank you.

As you discovered, you can't extend the system (C) partition using tools provided with Windows. You need a third-party disk partitioning application.

This one is free: http://www.partition-tool.com/personal.htm and this one is $9.95: http://www.paragon-software.com/home/pm-personal/

overall index using partitioning of the interval.

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