Statisctis for the partition table

Similar Questions

• How to configure the partition table for SSD?

  Drive: 32 GB SSD

  I accidentally delete without backup partition table information and would like to know how to put the partition table, I can access the tools of partition table, but do not know how to define.

  Does anyone have any suggestions?

  Thanks in advance for your suggestions

  Hello

  Thanks for posting your query in Microsoft Community.

  The SSD could be seen in the disk management window, and you could name and set up as another hard drive internal. To create a partition or volume on a hard disk, you must be logged in as an administrator, and there must be unallocated disk space or free space in an extended hard disk partition. To repartition your hard drive, please consult the following link and check if it helps.

  I can I repartition my hard disk?

  Additional information:

  Create a new Partition on a hard disk in Windows 7

  Hope this information is useful. Let us know if you need more help, we will be happy to help you.

• Assistance button works do not (= does ' t start VAIO Care Rescue) after having lost and recover the partition table / MBR

  Hello

  For some reason that I lost my patition table but I was able to get it back and I'm rewriting MBR. I'll be back all my partition and files contain the VAIO recovery partition except files on the operating system partition ("C:\ »). It's an unformatted partition. VAIO Care rescue didn't work so I had to format "C:\". "and reinstall windows 7 and now all is good.

  I have VAIO Care Rescue. I think that it is enough to change the partition table / MBR. What exactly is the problem and is it possible to fix the problem of button Assist?

  
  And if one removes all xxxx (include VAIO Recovery) and creates new ones, is it possible, software or image of the recovery restore partition and Assist button start VAIO Care Rescue?

  My laptop model: VPCEA25FG

  Thank you

  You can restore your computer back to factory with all the pre program and software installed by using recovery discs. For further assistance, please recommend contacting Sony's Support Center in your area at http://www.sony-asia.com/support/product/VPCEA25FGfor your model of Asia-Pacific computer. Thank you.

  If my post answered your question, please mark it as "accept as a Solution.

• What is advised to collect statistics for the huge tables?

  We have a staging database, some tables are huge, hundreds GB in size.  Auto stats tasks are performed, but sometimes it will miss deadlines.

  We would like to know the best practices or tips.

  Thank you.

  Improvement of the efficiency of the collection of statistics can be achieved with:

  1. Parallelism using
  2. Additional statistics

  Parallelism using

  Parallelism can be used in many ways for the collection of statistics

  1. Parallelism object intra
  2. Internal parallelism of the object
  3. Inner and Intra object jointly parallelism

  Parallelism object intra

  The DBMS_STATS package contains the DEGREE parameter. This setting controls the intra parallelism, it controls the number of parallel processes to gather statistics. By default, this parameter has the value is equal to 1. You can increase it by using the DBMS_STATS.SET_PARAM procedure. If you do not set this number, you can allow oracle to determine the optimal number of parallel processes that will be used to collect the statistics. It can be achieved if you set the DEGREE with the DBMS_STATS. Value AUTO_DEGREE.

  Internal parallelism of the object

  If you have the 11.2.0.2 version of Oracle database you can set SIMULTANEOUS preferences that are responsible for the collection of statistics, preferably. When there is TRUE value at the same TIME, Oracle uses the Scheduler and Advanced Queuing to simultaneously manage several jobs statistics. The number of parallel jobs is controlled by the JOB_QUEUE_PROCESSES parameter. This parameter must be equal to two times a number of your processor cores (if you have two CPU with 8 cores of each, then the JOB_QUEUE_PROCESSES parameter must be equal to 2 (CPU) x 8 (cores) x 2 = 32). You must set this parameter at the level of the system (ALTER SYSTEM SET...).

  Additional statistics

  This best option corresponds to a partitioned table. If the INCREMENTAL for a partitioned table parameter is set to TRUE and the DBMS_STATS. GATHER_TABLE_STATS GRANULARITY setting is set to GLOBAL and the parameter of DBMS_STATS ESTIMATE_PERCENT. GATHER_TABLE_STATS is set to AUTO_SAMPLE_SIZE, Oracle will scan only the partitions that have changes.

  For more information, read this document and DBMS_STATS

• Create index partition in the partition table tablespace

  Hello
  
  I am running a work custom that
  
  * Creates a tablespace by day
  * Creates the daily table partition in the created tablespace
  * Removes the days tablepartition X
  * Removes the storage space for this partition of X + 1 day.
  
  The work above works perfectly, but it has problems with the management of the index for these partitioned tables. In the old database (10g - single node), all indexes and partitions exist in a BIG tablespace and when I imported the table creation script in the new database, I changed all the partitions table & index to go in their respective space.
  
  For example:
  
  Table_name... Nom_partition... Index_Part_name... Tablespace_name
  ============...================............====================...........=================
  TABL1... TABL1_2012_07_16... TABL1_IDX_2012_07_16... TBS_2012_07_16
  TABL1... TABL1_2012_07_15... TABL1_IDX_2012_07_15... TBS_2012_07_15
  
  
  But now, when the job is run, it creates the index in the tablespace TBS_DATA default.
  
  Table_name... Nom_partition... Index_Part_name... Tablespace_name
  ============...================.............====================...........=================
  TABL1... TABL1_2012_08_16... TABL1_IDX_2012_08_16... TBS_DATA
  TABL1... TABL1_2012_08_15... TABL1_IDX_2012_08_15... TBS_DATA
  
  
  I can issue alter index rebuild to move the index to its tablespace default, but how can I make sure that the index is created in the designated tablespace?
  
  NOTE: the partition/tablespace management work that I run only creates the partition of the table and not the index.
  
  
  The new env is a cluster of CARS of 2 nodes 11 GR 2 on Linux x86_64.
  
  
  Thanks in advance,
  aBBy.

  try something like this

  ALTER table tab_owner.tab_name add the partition v_new_part_nm
  values less (to_date('''|| v_new_part_dt_formatted ||'') ((', "DD-MON-YYYY)) tablespace ' | part_tbs
  update the index (ind1_name (partition ind_partition_name tablespace ind_part_tbs)
  ind2_name (partition tablespace ind_part_tbs ind_partition_name))
  ;

• Desperately need help - "the partition table on device sda was unreadable." Erase all data?

  I'm forced to install CentOs on my Macbook using a machine virtual (VMWare, which is what I downloaded my teacher recommended). It's for my first class of operating systems, so bare with me because I don't know anything about operating systems (except about how to use Windows and my Macbook). I am also a daughter, what makes my same technical knowledge that much more rare. However, I am eager to learn...

  In any case, I downloaded them both. When I launch Fusion, I choose the file> New> install Windows or another OS on a virtual machine> continue without disk> click on the button that says 'Operating system installation disc using image file'> then I select the appropriate iso CentOS (CentOS - 5.4 - i386-bin-1of6) file> continue> drop down lists, I choose LINUX as operating system and Red Hat Enterprise Linux 5 as my Version> continue-> finish (no customization settings). It appears the screen of CentOS and I am invited by a text asking me if I want to install in text or graphics mode. (Don't know what to do here, but I select the graph). A bunch of white text scroll a screen of black background, then a blue screen shows up saying: 'Welcome to CentOS' at the top and a gray screen that says "to start testing CD media before installation OK press. Choose Ignore to ignore the media test and launch the installation. "I click on the red button (even once, if it is incorrect, let me know). A few rows of white text on black background (from anaconda... something on a video card...) then a nice graphic display of CentOS which I click NEXT. I then select English as my tongue (twice) and I gives me the message:

  "The partition table on device sda (VMware, VMware Virtual S 20473 MB) was unreadable. To create new partitions, it must be initialized, cauing the loss of all DATA on this drive. This operation will replace all choice on what players to ignore previous installation. You would like to initialize this drive, erasing all THE DATA? »

  I have no idea what to do here. I googled the message and others have been too, but they can talk about some hard file (and I have no idea what it is, and I can't seem to find mine).

  I can't lose what's on my hard drive (my school, music, photos, applications, files etc.). Clicking YES will erase my HD? I'm terribly afraid. I need to get this thing installed so I can do my operating systems class work, but I don't know if I did the installation correctly so far and especially, what to click at this point. Can someone help me out please (I spent 4 hours on this one today. "I have no one else to ask)? Finally, the installation is almost done after this step?

  Any help at all would be so greatly appreciated.

  I do not use Fusion, but I have VMware Workstation on Windows. I went through the same process that you went through. The answer to your question is that when you choose the operating system you install and you choose the option for the installation disc, it creates a virtual hard drive for you, size 20 GB. When they ask you to erase the hard drive, it's clear that virtual you create, not your actual computer hard drive. You can check by looking at the properties of the hard drive on your computer. It's probably 250 GB or 500 GB or something big like this. The CentOS WARNING speaks of a 20 GB hard drive, which is the size of the disk that VMware Fusion (and Workstation) create for CentOS.

• Not enough space on disk for the Partition: Windows Bootcamp

  iMac (retina 4K, 21.5 inches, end of 2015)

  Processor: Intel Core i5 to 3.1 GHz

  Memory: 6 GB 1867 MHz DDR3

  Graphics card: Intel Pro Iris Graphics 6200 1536 MB

  The OSX Version: 10.11.4

  I tried to download windows for my iMac. I bought the full version via USB for it. I followed these instructions: create an ISO image for Windows - Apple Support installation media Boot Camp to create my own ISO image. I did, and I ejected from disk image Windows "10" as he told me too.

  When I go to download the support software and partition it, he will get after downloading the Windows Support software, then to halfway in partition the drive, it will say "disk does not of enough space to be partitioned. (PS) The "BOOTCAMP" image will appear on the desktop until it says the message and then will disappear and the message will occur.

  I fixed the disc and it defragmented. It's a lot I put 300 concerts for the Partition. I don't know what is happening.

  I've also entered the recovery disc and repaired this way.

  I'm on the forums before I decided to call Apple.

  Thank you.

  You have a Fusion drive?

• query the partition tables

  Hi all
  
  I want to query the partition tables that are over 1 GB in size?
  
  How can I ask questions?

  Salvation;

  You can also try with dba_tab_partitions view

  Respect of
  HELIOS

• How to find the size of the partitioned tables?

  How to find the size of the partitioned tables?

  Select nom_segment, sum (bytes) /(1024*1024) 'Size in MB' from dba_segments
  where owner = 'owner name' and segment_type like '% PARTITION % '.
  Group by nom_segment;

• Why the partitioned table cannot create?

  Why the partitioned table cannot successfully create?

  SQL> create table hr.gps_log_his
    2  (id number,
    3  name varchar2(10),
    4  time date)
    5  tablespace ts_log_his
    6  PARTITION BY RANGE (TIME)
    7  (PARTITION udp_part2009110707 VALUES LESS THAN (TO_DATE('2009110708','yyyymmddhh24')),
    8  PARTITION udp_part2009110708 VALUES LESS THAN (TO_DATE('2009110709','yyyymmddhh24')),
    9  PARTITION udp_part2009110709 VALUES LESS THAN (TO_DATE('2009110710','yyyymmddhh24')),
   10  );
  )
  *
  Error in line 10:
  ORA-14004: Missing PARTITION key word.

  Hello

  Your problem is the comma ending line 9. Delete this.

  Concerning
  Peter

• Look for the partition for the fact table

  Oracle version: Oracle 10.2
  
  I have a table of facts with daily partitions.
  I'm insertion of test data in this table for the old date 20100101.
  
  I am able to insert that record into this table as below
  insert into fact_table values (20100101,123,456);
  
  However I noticed that the partition for this date does not exist in the table (all_tab_partitions) more I'm not able to select the data using
  
  Select * from facT_table partition (d_20100101)
  
  but I am able to extract the data using
  
  Select * from facT_table where date_id = 20100101
  
  could you get it some please let me know how to find the partition in which these data could be inserted
  and if the partition date 20100101 is not present so why put so that the date doesn't work?

  user507531 wrote:
  However I noticed that the partition for this date does not exist in the table (all_tab_partitions) more I'm not able to select the data using

  Select * from facT_table partition (d_20100101)

  Wrong approach.

  but I am able to extract the data using

  Select * from facT_table where date_id = 20100101

  Correct approach.

  could you get it some please let me know how to find the partition in which these data could be inserted
  and if the partition date 20100101 is not present so why put so that the date doesn't work?

  Who says that the date is invalid. ? It is a partition of the range - which means that each partition covers a range. And if you bothered to read in the SQL Reference Guide on how the definition of a partition of the range, you will notice that each partition is defined with the value end of the range it covers. There is no starting value - as the end of the previous partition is the "+ border +" between this and the previous partition.

  I suggest that before you use a database feature first become familiar you with it. Another evil to use and making assumptions wrong about it, more than likely outcomes.

• How to recover the partition table (corrupt) for the partition in good health?

  Partition (drive H) was working fine, then I rebooted my system normally and the drive is now inaccessible.

  "An error occurred reading the file: the file or directory is corrupted and unreadable." (1392).

  Window CMD CHKDSK says NTFS file system and "unable to determine the State and version volume.»

  [This is a disk external drive with 2 partitions on it - H and Mr. the second partition (drive M) discs works fine.]
  [Vista Ultimate OS]

  Is there a way I can get it to work or get the data off of it?

  Try the procedure described in the article below

  http://www.ehow.com/how_7500292_recover-bad-partition-table.html

• Create the partition table

  Hello Experts,

  I only have 4 distinct values in my column of company. I would just create a table with partition depend on these values (201402, 201403,201404, 201405). Should I use LIST or RANGE?

  I just want the table for each distinct value of the part = 201402, 201403,201404, 201405

  So, should what partition key word I use? LIST or RANGE? or OTHERS?

  Create table MY_TABLE NOLOGGING

  as

  SELECT company_id, name...

  Of...

  WHERE THE...

  LIST BY SCORE (company_id);

  However when I use the above, that it does not work, should we specify the values? I want he's leaving for each distinct value? Even if so, should we indicate?

  Furthermore, I want to use partitioning because once I created this table, I could delete e.g. company_id = 201404, rather than delete, I can drop all of the part belonging to the 201404 does make sense? However, at the same time, I want to insert as new campanyid insert into select... where company_id = 201406;  When I insert a new campany_id the Oracle creates a new partition for it?

  Thanks in advance

  Concerning

  Charlie

  Of course you can. It would take less time to test that in order to post and wait for the reply:

  Scott@ORCL > CREATE TABLE company)
  2                       id,
  3                       name
  4                      )
  PARTITION of 5 PER LIST (id)
  6      (
  7 VALUES of company 1 PARTITION (201402),
  8 PARTITION company2 VALUES (201403).
  9 PARTITION company3 VALUES (201404)
  10)
  11 NOLOGGING
  12 AS
  13. SELECT deptno case
  14 when 10 then 201402
  15. when 20 then 201403
  16 when 30 then 201404
  end 17
  ename 18
  19 FROM emp
  20.

  Table created.

  Scott@ORCL >

  SY.

• 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

• 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