Become table partitioning

Similar Questions

• Slow running queries on a table partitioned by date

  I have a table partitioned by COLDATE, when I write a query to get the data is slow, even if I try to only get 1 day data. How can I correct this query runs faster?

  Oracle Version: 10g

  Number of lines:

  See the below query

  select COL1, COL2 RANK() OVER (PARTITION BY   COL1, COLDATE ORDER BY OTHER_DATE DESC) SHOW_RANK
  from 
  SAMPLETABLE
  where COLDATE BETWEEN '01-JAN-2015' AND '01-JAN-2015'
  
  
  create table SAMPLETABLE
  (
    COL1   NUMBER(10) not null,
    COL2        NUMBER(10) not null,
    COLDATE       DATE not null,
    OTHER_DATE DATE
  )
  partition by range (COLDATE)
  (
    partition PARTITION_01_2015 values less than (TO_DATE(' 2015-02-01 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN'))
      tablespace MYTABLESPACE_2015
      pctfree 20
      initrans 1
      maxtrans 255
      storage
      (
        initial 64K
        minextents 1
        maxextents unlimited
      ),
    partition PARTITION_02_2015 values less than (TO_DATE(' 2015-03-01 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN'))
      tablespace MYTABLESPACE_2015
      pctfree 20
      initrans 1
      maxtrans 255
      storage
      (
        initial 64K
        minextents 1
        maxextents unlimited
      ),
    .
    .
    .
    .
    .
    .
    partition PARTITION_12_2016 values less than (TO_DATE(' 2017-01-01 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN'))
      tablespace MYTABLESPACE_2016
      pctfree 20
      initrans 1
      maxtrans 255
      storage
      (
        initial 64K
        minextents 1
        maxextents unlimited
      )
  

  Force the session in parallel mode and check with the parallel hint option.

  SQL > alter session force parallel dml.

  SQL > select / * + parallel (table_name, none of the servers) * / colA, colB from table_name;

• DBA_TABLES. NULL for tables partitioned, yet Nom_tablespace DBMS_METADATA. GET_DDL displays a name...

  Hello

  Hence the name of tablespace comes when you make one

  ~ Select DBMS_METATDATA. GET_DDL ('TABLE', 'tname', 'owner')...

  on a table partitioned, where you see things like:

  ~ CREATE TABLE...

  ~ ...

  ~ STORAGE(INITIAL 134217728 NEXT 134217728 MINEXTENTS 1 MAXEXTENTS 2147483645)

  (~ DEFAULT USER_TABLES)

  ~ TABLESPACE "ABCD_LOGD2".

  before all the blocks related to the different partitions of this table, where a

  ~ Select nom_tablespace from dba_tables where table_name = 'TNOM '...

  Returns a null value?

  Whence dbms_metatdata this ABCD_LOGD2 value? (in what dictionary?,)

  DBA_SEGMENTS/extents display only the partitions information...)

  Thank you...

  SEB

  Hello

  What is the value of 'deffered_segment_creation '.

  Take a look of it

  Select def_tablespace_name in the dba_part_tables where table_name = 'tname'--> your table

  Concerning

• Table partitioning

  I have a table in my DB with stored more then 24 million and is growing every day. Growth rate increases with the passage of time. I want this partition table, in such a way that three first partitions such as each partition contains 10 years of data, and the remaining partition such as each partition contains data for 5 years.
  This is the first time that I am any table partitioning, I understand following sql is not correct, I need your help in resolving this issue.
  

  ALTER TABLE cb_pen_pmt_slip_dtl
  ADD
  PARTITION BY range (payment_date)
  INTERVAL (NUMTOYMINTERVAL(5,'YEAR'))
  (PARTITION p1 VALUES LESS THAN (TO_DATE('2000-01-01','YYYY-MM-DD'))
  partition p2 values less than (maxvalue));

  Hello

  This link can help you...
  >

  Kind regards

• Table Partitioned parallelized vs

  Hello
  (11.2 server, OS = OL5)
  
  I have a table that contains more than 50 million lines. I created this table with option 'parallel degree (x).
  
  It seems that the time required to execute a query on this table does not change when I share that, for example partition range dividing the table to 3 partitions.
  
  (access the request a unique index in the table and the query criteria match exactly one of the partitions)
  
  the question:
  1. What is the difference of strategy parallelism between «Parallel degree...» "and the"Division "?
  2. which of them are better in what scenario?
  3. is it better to use PARALLEL_AUTOMATIC_TUNING?
  4 and especially why the performance doesn't change even with a local index on the partitions?
  
  Thank you
  SMSK.

  >
  It seems that the time required to execute a query on this table does not change when I share that, for example partition range dividing the table to 3 partitions
  . . .
  (access the request a unique index in the table and the query criteria match exactly one of the partitions
  . . .
  @Adam Martin: thanks, but I was explained to question 4 that the performance does not change when I toggle locality index.
  >
  I do not see why it would be necessarily change just of partitioning. What is your reason to think so?

  If the index access is used it is basically to get the IDENTIFIER of a line of interest. The line can then be retrieved using the ROWID. The recovery time will be essentially the same way regardless of the line in a partitioned table or non-partitioned table, and regardless of whether the line of partition. This is because the ROWID is (simple explanation here) a FILE ID (which file?), a BLOCK number (relative block in the file) and a LINE NUMBER (which line in the block. It doesn't matter if the segment that is the block is a segment of the table, partitioned table segment, segment index, etc.

  It is the only factor that COULD change the time of the query when the index is used if the access of the index itself is more effective when the table is partitioned when the table is not partitioned.

  Clearly if it is an overall index, there is a change since the index access itself will be the same.

  The only factor now that MIGHT change the time of the request is so if 1) the index is changed locally and 2) the local index has a different structure than the old index that allows to find the best performing required index entries.

  You have not published information on the index of structure, so let me give you an analogy which shows a scenario where it is clear that the access to the index would not change.

  Assume that the overall index is analogous to a catalogue of cards with 26 entries - one for each letter of the alphabet. My query accesses the entries beginning with the letter 'F' As discussed above in table access to aid a ROWID obtained from the index will be the same, if the table is partitioned or not.

  So now I have the partition table and have 26 partitions - one for each letter of the alphabet.

  With the help of the global index of the access to the partition of 'F' will be the same.

  Now, if I create a local index (instead of global) it's almost as if I now 26 index, one for each letter of the alphabet. The first entry for the 'F' in the global index can be found by reading the root node and perhaps one or two other nodes - then that an index scan can be performed.

  The first entry for the 'F' in the local index is at the beginning of the index - no need to jump or to do a binary search to ignore the 'A', 'B', etc. entered.

  This time difference is microscopic.

  Once I found the first entry for the 'F' either in the overall index, or local 'F' remaining entries in the index are consecutive, so access time will be the same.

  I don't know if there is a structural difference between the indices and the that you tried, but the above should show if the structure (order) has not changed it would not be a noticeable difference in access.

• Why cannot create the table partitioned successfully?

  Why cannot create the table partitioned successfully?

  SQL> create table hr.gps_log_his
  (id number,
  name varchar2(10),
  time date)
  tablespace ts_log_his
  PARTITION BY RANGE (TIME)
  (PARTITION udp_part09110707 VALUES LESS THAN (TO_DATE('09110708','yymmddhh24')),
  PARTITION udp_part09110708 VALUES LESS THAN (TO_DATE('09110709','yymmddhh24')),
  PARTITION udp_part09110709 VALUES LESS THAN (TO_DATE('09110710','yymmddhh24')),
  PARTITION udp_part09110710 VALUES LESS THAN (maxvalue)
  );
  
  (PARTITION udp_part09110707 VALUES LESS THAN (TO_DATE('09110708','yymmddhh24')),
  Error on line 7: 
  ORA-14120: DATE columns did not specify the complete partitioning limits.

  The detailed error message is as follows:

  ORA-14120: incompletely specified partition bound for a DATE column
  
  Cause: An attempt was made to use a date expression whose format does not fully
  (i.e. day, month, and year (including century)) specify a date as a partition bound
  for a DATE column. The format may have been specified explicitly (using TO_DATE()
  function) or implicitly (NLS_DATE_FORMAT).
  
  Action: Ensure that date format used in a partition bound for a DATE column supports
  complete specification of a date (i.e. day, month, and year (including century)). If
  NLS_DATE_FORMAT does not support complete (i.e. including the century) specification
  of the year, use TO_DATE() (e.g. TO_DATE('01-01-1999', 'MM-DD-YYYY') to fully
  express the desired date. 
  

  Action: Change to_date('09110708','yymmddhh24') to
  to_date('2009110708','yyyymmddhh24')

• How to encrypt a USB, when I get the message "(GPT) GUID Partition Table partitioning scheme is required? '."

  I just tried to encrypt a USB drive with my Mac book (latest version) and got the message "A GUID Partition Table (GPT) partitioning schΘma required." What does this mean and suggestions on how to proceed?

  The drive must be formatted for Mac. Start in the Recvery mode. Erase the drive using disk utility (select the top icon which represents the entire disk and click clear. Partition scheme must be Mac OS (Extended) first partition GUID, which is the default value.

  If you want to install a system on it follow these steps:

  How to reinstall OS X on your Mac - Apple Support

• Is it possible to copy a table partitioned to another different partitioned table statistics?

  Hello

  Using oracle 11.2.0.3 and study the possibility to copy his stats of a hash range partitioned table to another partitioned table range-hash.

  New table is based on the old table with the same structure, about right. of lines.

  estimate_percent auto not practical sample - size took 10 hours in tests and careful to just down for example 1% percent estimate.

  Is it possible to simply copy the stats on our old table

  Thank you

  Mismatch / typo in statid?

  statid-online "CURRENT_STATS."

  statid => ' CURRENT_STATUS

• Create no partition Table partitioning

  Dear all,

  I have a table that is not partition, and it has about 20 G data... If I want that table with say DATE_M partition column

  Please can anyone suggest the best way to do it.

  Thank you

  To create a partitioned table

  (1) you must make a backup of the existing one you can take expdp or DEC

  (2) remove the table and create the same table now with partitions (a minimum score is required)

  you want a range of ex-based partition if you use a date column

  (3) loading data from the previous table into the new partitioned table, either by DEC or impdp, the inserts will be automatically in their respective partitions and you don't mention explicitly.

  Their is no other possible way to partition a non partitioned table, you can do than workarounds to minimize the downtime of the table trying different ways to load data into the partitioned table.

  You can use the enable row movement clause of create table and also use local indexes for easy maintenance partition.

• Number of rows in each partition is displayed with a NULL value for a table partitioned in user_tab_partitions. Why?

  I created a table and partitioned on the date of the entry and added a local partitioned index.

  Now, I use a query to extract "num_rows" of user_tab_partitions to know the number of rows in each partition.

  Getting this value as null num_rows, wonder why?

  After looking to explain the Plan after interrogation ("select * from my_table1 where entry_date = 1 January 2015" ;))

  to find out if she actually partitioned table and its data in different partitions, I interpreted in effect because the query plan had a line like Partition_range (Single).

  My Question is:

  (a) is actually partitioned data (have I misinterpreted the Explain plan)

  (b) why is the num_rows null column in the query (Pasted below)

  (c) also in addition what difference it would have been if I had created a Global Index instead of the Local Index in my case?

  The following code Snippet:

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

  create the table my_table1
  (
  roll_no number constraint my_table1_pk primary key,
  date of entry_date
  )
  partition of range (entry_date)
  (
  PARTITION data_p1 VALUES LESS THAN (TO_DATE (December 31, 2014 ',' DD-MM-YYYY ""));
  PARTITION data_p2 VALUES LESS THAN (MAXVALUE)
  );

  
  create an index only my_table1_indx on my_table1 (entry_date) local;

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

  I now insert two lines:
  insert into my_table1 values (1, to_date ('01-01-2015', ' dd-mm-yyyy'));
  insert into my_table1 values (2, to_date('01-02-2015','dd-mm-yyyy'));

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

  These have been inserted successfully, now using the query below shows num_rows column as null. I don't know why?

  SELECT table_name, num_rows, high_value, nom_partition
  Of user_tab_partitions
  where table_name = 'MY_TABLE1 '.
  ORDER BY table_name, nom_partition;

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

  
  

  (a) is actually partitioned data (have I misinterpreted the Explain plan)

  Yes, it is partitioned. You can query this particular partition SELECT * FROM my_table1 (data_p1) PARTITION to check that.

  (b) why is the num_rows null column in the query (Pasted below)

  As already mentioned that you have not collected statistics.

  (c) also in addition what difference it would have been if I had created a Global Index instead of the Local Index in my case?

  In fact, you have created two types of indexes without knowing (can be)! One is not partitioned (although this column is not partition key) and another is partitioned (LOCAL). They are MY_TABLE1_PK and MY_TABLE1_INDX. You can check that USER_INDEXES.

  You can read this article to get an early jump on the partitioning of decision. Partition: Partition decisions

• table partitioned and unique keys

  I really don't understand why Oracle does not apply in some cases, unique keys.

  Let's say I have a table

  create table t (t_type not null, t_key not null, id not null, not null value)

  list partition (t_type)

  (

  type1 partition values ('ppp1"),

  rest of partition (default) values

  ) as

  Select "ppp" | rownum, rownum, rownum, rownum of double connect by rownum < = 100;

  And following changes in the indices for unique keys

  -1

  create index idx_1 on t (t_key, id);

  ALTER table t add constraint t_uk unique (t_key) using index idx_1;

  ALTER table drop constraint t_uk drop index;

  -2

  create index idx_1 on t (t_key, id, t_type);

  ALTER table t add constraint t_uk unique (t_key) using index idx_1;

  ALTER table drop constraint t_uk drop index;

  -3

  create index idx_1 on t (t_key, id, t_type) local;

  ALTER table t add constraint t_uk unique (t_key) using index idx_1;

  Drop index idx_1;

  -4

  create an index only idx_1 on t (t_key, id);

  ALTER table t add constraint t_uk unique (t_key) using index idx_1;

  Drop index idx_1;

  -5

  create an index only idx_1 on t (t_key, id, t_type);

  ALTER table t add constraint t_uk unique (t_key) using index idx_1;

  Drop index idx_1;

  -6

  create an index only idx_1 on t (t_key, id, t_type) local;

  ALTER table t add constraint t_uk unique (t_key) using index idx_1;

  Drop index idx_1;

  -7

  create index idx_1 on t (t_key, t_type, id) local;

  ALTER table t add t_uk unique constraint (t_key, t_type) using index idx_1;

  ALTER table drop constraint t_uk drop index;

  -8

  create an index only idx_1 on t (t_key, t_type, id) local;

  ALTER table t add t_uk unique constraint (t_key, t_type) using index idx_1;

  Drop index idx_1;

  I got the set of questions:

  1. first and second cases were performed with success while 4th and 5th failed.

  The only difference is that the overall index in 4th and 5th cases are unique.

  It is unclear why ' ORA-14196: specified index cannot be used to apply the constraint. ' has occurred.

  2. in the third case index is similar to that in second place with the only difference that it is local.

  Still, it is not clear why it failed with ORA-14196.

  3 and the last question why ORA-14196 arose in 8 cases, while the 7th, it didn't.

  The only difference is the local index in 8 cases is unique.

  Thank you

  Hi Alex

  (1) index 4 and 5 cannot be used to police the unique constraint as a unique index must have the same list of column under duress. Index 4, to establish in a non-unique index or to remove the column id of the index. To index 5, you will need to make a non-unique index or remove the columns id and index t_type.

  (2) 3 index cannot be used to the unique constraint of police that it is a local index and is not partitioned column (t_type) in both the definition of index/constraint. If allowed, this would require Oracle having to visit each partition to ensure that any new value of t_key is truly unique, that would not scale. Include the t_type column in the constraint and the index to both t_type and t_key as of the columns (so move the id column 3rd position) and it would be successful.

  (Index 3) 8 fails for the same reason as in 1). It's a unique index with a different column from the constraint list. Remove the id column of the index or to make it a non-unique index and it will be successful.

  In summary, a unique index MUST have the same list of column than the constraint and a local index MUST have the partitioning columns in both the definition of index/constraint.

  See you soon

  Richard Foote

  http://richardfoote.WordPress.com/

• Move table Partition Composite

  Hello

  I use 11.2.0.3.0 version of oracle. I intend to pass an array to a different tablespace composite partition(range-hash). And I was under the assumption by moving the subpartition / the index of the table, it will move the wholw table. Here's the scenario.

  Now I have moved all the subpartition of a table / index in the new tablespace, using below command.

  ALTER table t1 move subpartition t1_sub1 tablespace new_tablespace;

  ALTER index rebuild subpartition id1_sub1 tablespace new_ind1_tablespace id1;

  but when I ask dba_tab_subpartition I see the tablespace again against the subpartition name but when I the dba_tab_partitions data dictionary query I see the name of the old tablespace against the name of the partition. And I am running under instruction to move the partition

  ALTER table t1 move the partition t1_p1 tablespace new_tablespace;

  It gives error saying: ORA-14257: cannot move the partition other than a partition of the range, list, system, or hash.

  933257 wrote:

  Hello

  I use 11.2.0.3.0 version of oracle. I intend to pass an array to a different tablespace composite partition(range-hash). And I was under the assumption by moving the subpartition / the index of the table, it will move the wholw table. Here's the scenario.

  Now I have moved all the subpartition of a table / index in the new tablespace, using below command.

  ALTER table t1 move subpartition t1_sub1 tablespace new_tablespace;

  ALTER index rebuild subpartition id1_sub1 tablespace new_ind1_tablespace id1;

  but when I ask dba_tab_subpartition I see the tablespace again against the subpartition name but when I the dba_tab_partitions data dictionary query I see the name of the old tablespace against the name of the partition. And I am running under instruction to move the partition

  ALTER table t1 move the partition t1_p1 tablespace new_tablespace;

  It gives error saying: ORA-14257: cannot move the partition other than a partition of the range, list, system, or hash.

  I think that it is essentially a bug: a partition with subpartitions is not a segment, so it can ' t have a tablespace. The dba_tab_partitions.tablespace_name column should not be filled at all in this circumstance.

  The ora-14257 is also silly: you really need a message saying that you can not move a partition with subparts, because she has no physical existence.

• modeling of the table partitions

  Hello

  How to define partitions in oracle data maker? If it is defined in the relational or physical model (11g)?

  The table will store millions of cities linked to about 20,000 documents. I want to create a partition on the name of the city?

  Please advice...

  
  

  Hello

  Partitioning is defined in the physical model of the Oracle.

  Open the Table Properties dialog box in the physical model and set its property to partitioned on YES.

  You can then go to the partition tab in the dialog box.  You must set the Partition Type (of the LIST) and add the column containing the name of the city to the selected columns list.

  After you close this dialog box, you can, if you wish, define individual partitions.

  To do this, you should expand the node of the Table in the physical model tree and then right-click on its node of Partitions and select the new option.

  This will display the properties of Partition dialog box, where you can provide a list of one or more city names in the list of values property.

• On an oracle table partitioning. Please answer as soon as POSSIBLE.

  Hello

  I partitioned a table, but now I want to check whether or not the partition are created successfully. Can someone help me to know the way to find a partition table in the database?

  Regards,

  Rachid

  query these tables to find relevant information

  DBA_PART_TABLES

  ALL_PART_TABLES

  USER_PART_TABLES

  mark, if this can help

  Make sure you post in the right forum of next time

• Table partitioning/indexing strategy

  Hello
  
  I have a data warehouse containing medical data. Normally, we develop queries against a small number of patients (e.g. where patient_id < 100) and adjust upward to run on a greater number (usually the entire base of 26 000 patients least some due to exclusion criteria).
  
  The largest table contains about 200 million lines, and I'm working on the question of whether I can change the partitioning and indexing in order to improve the performance of queries.
  
  The table contains 'patient_id', 'visit_id', 'item_id', 'chart_time', 'value' and 'valuenum' column and a variety of other columns that are rarely used. The id id columns are integers, pointrefers to the type of value is saved and value (varchar2) and valuenum (number) contain the data. chart_time is a "timestamp with time zone '.
  
  Currently, there are clues on patient_id, visit_id and item_id and a composite index on (visit_id, itemid). There is no partitioning for the moment.
  
  1. is the current index redundent? that is visit_id, item_id and composite (visit_id, item_id) do I need?
  2. we started to run queries of time based for specific patients. that is, we want data for specific items, for a specific set of visits, in the first day. An index on (visit_id, chart_time, item_id) would help with that?
  3. the table of patient_id partitioning help? We "never" queries between patient and therefore the data of each patient are independent. The database is running on 2 raid controllers, one for data, one for the index, perhaps it would be better to put half of patients on a single controller, and the other half on the other?
  
  Thank you
  
  Dan Scott

  Hello

  1. the index on visit_id sounds redundant as it is the main column in the composite index (visit_id, itemid).
  2. added chart_time to the composite index could be useful that the pair of columns (visit_id, item_id) is not quite selective. You must put in place a realistic model of your data distribution, choose some queries that you intend to run often watch their performance. It is not possible to predict performance in function purely thereotical reasoning.
  3 bear partitioning functions: maintenance of data (for example truncating, deleting or moving old partitions) and eliminating the irrelevant data in queries (partition size). You imagine you having to truncate, delete, or move a partition that is defined by a range of patient_id? Probably not. Can you imagine the need to restrict your query to a certain range of patient_id? Unlikely. That answers your question.

  Best regards
  Nikolai