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

Tags: Database

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.

Query on partitioned Table
I have two partition
upto_mar_2011, upto_jun_2011
table_name: order
1 > if I run the query as below
Select * command where order_date = to_date (' 23/03/2011 ',' dd/mm/yyyy');

Oracle will automatically search the data in the Partition upto_mar_2011
or he fetches the data in the entire table.

2 > or I give the query in the form below
Select * enforcement of partition (upto_mar_2011) where order_date = to_date (' 23/03/2011 ',' dd/mm/yyyy');

Is there a difference between this two query
OraFighter wrote:

Is this enough to write as below

Select * command where order_date = to_date (' 23/03/2011 ',' dd/mm/yyyy')
or order_date = to_date (' 23/05/2011 ',' dd/mm/yyyy');

and CBO fetch data, only the partition where it is available.
Wheather it is spread in a two or three partition
It will ignore all other partitions automatically

Fix. 'S called it the size of partition. It is the approach you use to application code. Requests should not be concerned with partitioning - should not need to know the names and other physical attributes of the table physical partition it uses.

You can also see the size of the partition when looking at such a SQL statement execution plan.

The only code that will focus on names of partition and the physical aspects of the table's code maintenance. For example, you maintain that a sliding window of 32 days of the date was the partitioned table. This code must determine which partitions are older than 32 days and drop them. And then to add new partitions for future treatment. Copy the following code deals with the physical layer and not the logical layer.

It is important to keep the two apart and not to mix the physical things in the logical layer which is using the app. He's mixing means that changes to optimize performance and to take advantage of the new features which may not be at the physical layer, as it will break the application layer code.

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.

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.

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

How to query the partition of plsql table
Hello
I have a table tab1 and tab2 partition created on it.

I want to write a plsql program to check if a value is present in my default partition. Here I want to hard-code the value of the partition in my sql because this value is coming as INPUT in my procedure when I try it doesn't work. can someone help me on this pls...

For example.
procedure p1 (l_part_string varchar2)
is
number of l_cnt_part;
Start
Select count (1)
in l_cnt_part
tab1 partition (l_part_string).

dbms_output.put_line ('l_cnt_part -' | l_cnt_part);

Select count (1)
in l_cnt_part
tab2 partition (l_part_string).
dbms_output.put_line ('l_cnt_part -' | l_cnt_part);

end;
rocedure p1 (l_part_string varchar2)
is
number of l_cnt_part;
Start
immediate execution
' Select count (1).
tab1 partition ('| l_part_string |') '
in l_cnt_part;

dbms_output.put_line ('l_cnt_part -' | l_cnt_part);

immediate execution
' Select count (1).
tab2 partition ('| l_part_string |') '
in l_cnt_part;

dbms_output.put_line ('l_cnt_part -' | l_cnt_part);
end;
/

-----------
Sybrand Bakker
Senior Oracle DBA

end;

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

Querying the oracle table that has a column with the name of "FILE".

Hi all
I need to have an oracle table that has the column with the name "FILE".
I can query all columns with the query "select * from table".
But I'm not able to use the query "select the table file.
This table is converted from ms access to oracle and I need to have this column with the name "FILE".
Any suggestions?
Thank you
Abdellaoui

Hello

FILE is a keyword from Oracle, so it's not a good name,

Use FILEDATE, or DATE_FILED, or something else that is not in V$ RESERVED_WORDS. KEYWORD as the name column.

If you need to use the FILE, then you must use quotation marks.

Unique key on the partitioned table range

Hello

We use a range of composite range-hash partitioned table

Use index - try to do this have same tablespace to the local partitions but not liking it

ALTER table RETAILER_TRANSACTION_COMP_POR
Add primary key constraint RETAILER_TRANSACTION_COMP_PK (DWH_NUM)
using index
LOCAL

ORA-14039: partitioning columns must be a subset of the columns in a unique index key

Without local then fine but does not have same tablespace to walls and don't want to make this part of the partition key.

Range Tbal partitioned - it's just a UK to avoid duplicates

[oracle@localhost ~]$ oerr ora 14039
14039, 00000, "partitioning columns must form a subset of key columns of a UNIQUE index"
// *Cause:  User attempted to create a UNIQUE partitioned index whose
//          partitioning columns do not form a subset of its key columns
//          which is illegal
// *Action: If the user, indeed, desired to create an index whose
//          partitioning columns do not form a subset of its key columns,
//          it must be created as non-UNIQUE; otherwise, correct the
//          list of key and/or partitioning columns to ensure that the index'
//          partitioning columns form a subset of its key columns

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

How to find the partitioned tables and the number of sheets in each table
Hello friends,

I have a scheme called ICS_OWNER where I partitioned and tables not partitoined.

I want to list all the tables that are partitioned.

and also

I want to know what partitions exists in every partitioned table.

with respective schema specified

Thank you/Kumar
When you have questions like this, you don't ask here.
but you take your (or should I say 'your') keyboard
and type
Select *.
dict
where table_name like ' % PART %.
/

All the dictionary views are listed in the DICT.

----------
Sybrand Bakker
Senior Oracle DBA

Update of the partition table and update rank with php.

Hello
I have a partition table on my site which is
Scores: id - userid - ranking - points-
Get points for doing things while using the site users. Points are updated, but I don't know how
to update this user's position by comparing everyone elses points and rank numbers update
based on the users with most points in php.
Hope you can help. Thank you

Weird, the link I provided is not what I wanted; It seems very different and is difficult to read. Anyway, you don't mention what goes wrong with the code you provided. I also don't think that you want to prepend rank alias with the table name 'notes' because it is not a table column. I do not use MySQL, but I think you want to use ' @' for your SQL local variables, so something like:

$sqlgo = mysql_query ("SET @rank = 0; SELECT @rank: = @ row + 1 rank Users.id, Users.username, Users.profilepic, Scores.scoreid, Scores.userid, Scores.level, Scores.points, rank, Scores.badges OF Scores INNER JOIN users ON Scores.userid = Users.id ORDER BY invoking DESC LIMIT 30 ");

query the partition tables

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