Update of the partition table and update rank with php.

Similar Questions

• 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

• How to upgrade the parent table and child by updating the parent table

  I have a parent EMPLOYEE table that includes columns (sysid, serviceno, employeename...) sysid is the primary key, serviceno is the Unique key and I have DEPENDENT child table includes columns (sysid, employee_sysid, name, date of birth...) there still SYSID is a primary key for the table of dependants, employee_sysid is a foreign key in the EMPLOYEE table.
  
  Now I want to change SYSID (with the help of the sequence) in the EMPLOYEE table that they want an update in the table of people dependent
  
  Note: I have 10000 records in the EMPLOYEE table as I have 5 more children tables that need to update new SYSID.
  
  Please help me

  first disable FOREIGN KEY constraints.
  You can update Parent and child record with the help of the trigger.
  Here I give you an examlpe... It can help u.

  create a parent (id number primary key, name varchar2 (100)) table
  /
  create table child_1 (primary key id, p_id number number, date of birth, date)
  CONSTRAINT FK_id FOREIGN KEY (p_id) REFERENCES parent (ID))
  /
  create table child_2 (key primary id, p_id2, addr varchar2 number number (1000))
  CONSTRAINT FK_id2 FOREIGN KEY (p_id2) REFERENCES parent (ID))
  /

  Insert some test data for the parent tables and children.

  change the constraint to disable child_2 table FK_id2
  /
  change the constraint to disable child_1 table FK_id2
  /

  CREATE OR REPLACE TRIGGER delete_child
  BEFORE parent UPDATE ON
  FOR EACH LINE
  BEGIN
  UPDATE CHILD_1
  P_ID =:NEW.ID SET
  WHERE P_ID =:OLD.ID;
  UPDATE CHILD_2
  SET = P_ID2: NEW.ID
  WHERE P_ID2 =:OLD.ID;
  END;
  /

  then Upadte parent table primary key col and check the children tables.
  do enable constraints...

• 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))
  ;

• Partitioned Tables and indexes

  Hello
  
  
  I have a question on the table and index partitioning. My scenario is:
  
  Charge 2 mio records in table T once a month. Loaded records are added to existing records, and once loaded data is never changed.
  At some point, I want to delete the older recordings, so I intend to this partition table.
  
  T table looks like:
  

  create table t (id       number(10) not null  constraint t_pk primary key,
                  period   number(10) not null,
                  contract number(10) not null,
                  attr     number(10) not null);
  
  create unique index t_ux1 on t(contract,period);
  
  create index t_ix2 on t(period);

  My plan is to partition T over the period, and I'm trying to read through the concepts
  http://download.Oracle.com/docs/CD/B19306_01/server.102/b14220/partconc.htm#g471747
  
  
  My question is now, how to manage the indexes, the t_pk, the t_ux1 and the t_ix2. Concepts of say,
  
  «1. If the table partitioning column is a subset of index keys, use a local index.»
  
  "2. If the index is unique, use a global index. If this is the case, you are finished. »
  
  
  So, that's how I read it
  -t_pk is unique, so this should be global
  -t_ux1 of columns is a subset, unless I have misunderstood (?), which should be local
  -index t_ix2 column is the same as the partitioning column, so it must be local
  
  Is this right, this t_ux1 should be a local partioned index, even if the period is the second column in the index?
  
  If true, what will happen when a partion fell?
  
  
  I am new in this area, so please feel the comment as you wish.
  
  
  Concerning
  Peter
  
  
  BANNER
  ----------------------------------------------------------------
  Oracle Database 10 g Enterprise Edition release 10.2.0.3.0 - 64bi
  PL/SQL version 10.2.0.3.0 - Production
  CORE Production 10.2.0.3.0
  AMT for IBM/AIX RISC System/6000: Version 10.2.0.3.0 - production
  NLSRTL Version 10.2.0.3.0 - Production

  Peter Gjelstrup wrote:

  My question is now, how to manage the indexes, the t_pk, the t_ux1 and the t_ix2. Concepts of say,

  «1. If the table partitioning column is a subset of index keys, use a local index.»

  "2. If the index is unique, use a global index. If this is the case, you are finished. »

  So, that's how I read it
  -t_pk is unique, so this should be global
  -t_ux1 of columns is a subset, unless I have misunderstood (?), which should be local
  -index t_ix2 column is the same as the partitioning column, so it must be local

  Is this right, this t_ux1 should be a local partioned index, even if the period is the second column in the index?

  A partitioned index locally can only be defined as unique if the partition key is part of the columns in the index. Imagine what the database would have to do if this is not the case: in order to verify if a newly added or updated value violates the uniqueness, it will have to travel all the partitions in a serialized operation - means that no one else could do the same thing at the same time. Since he is a killer of serious scalability in terms of locking and contention, this is not allowed.

  So: Your T_UX1 index can be defined as a unique index that is local because it contains the partition key. Although the index is not prefixed ("Prefix" means that it is divided by the left side of the columns in the index) which means that there may be access patterns where all partitions should be scanned or the optimizer cannot use a method of size of effective partition according to the way the index is reached.

  Your T_PK index cannot be set as local because it must be unique (you can not use a local non-unique index in this case), but does not contain your partition key. It must be a global index. An overall index can be partitioned as well (different from the underlying table) but it doesn't have to be.

  Depends on how you access your data you have not T_IX2 index when partitioning by this key because it corresponds to the partition key and therefore could not actually be used by the mechanism of partition pruning that limit your query to the scores of individuals.

  If you have more than one MAS environment where running queries are used longer, you should be fine with the index the in general (because they could be analyzed in parallel in parallel operations), but if you have an OLTP environment, then you should avoid local no prefix indexes due to the potential problem that you need to analyze all partitions.

  Be borne in mind that with partitioning adds an important layer of complexity to other areas: in particular the options available to the optimizer and analyze cost optimizer statistics. Depends on how you access your statistical data must be maintained on several levels now (level of score and at the global level, in the case of subpartitioning may be still at this level). If your data is important and you rely on "global" level statistics (these are always the case when the optimizer at the time analysis cannot limit access to a single partition) then in the pre - 11 g databases analyze these "global" level statistics can take a lot of time and resources, since actually , you need data several times (once for the partition and even global level).

  Presenting this partitioning may mean other potential problems in terms of execution that change (not for the better sometimes) plans and how to effectively collect statistics. Note that g 11 addresses the issue of 'statistics' by introducing the so-called "extra" global statistics. Greg Rahn wrote a [blog note | http://structureddata.org/2008/07/16/oracle-11g-incremental-global-statistics-on-partitioned-tables/] on this nice feature.

  >

  If true, what will happen when a partion fell?

  Since you're already on 10g, you can specify the database to update the scores of the local index using the UPDATE of the INDEX clause, while 9i could maintain only an overall index and it is up to you to rebuild the local index partitions after the partition DDL on the table (according to the DDL operation).

  Kind regards
  Randolf

  Oracle related blog stuff:
  http://Oracle-Randolf.blogspot.com/

  SQLTools ++ for Oracle (Open source Oracle GUI for Windows):
  http://www.sqltools-plusplus.org:7676 /.
  http://sourceforge.NET/projects/SQLT-pp/

  Published by: Randolf Geist on Sep 30, 2008 16:39

  Added statistics / optimizer warning when you use the partitioning

• 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.

• 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;

• Left join of the two tables and multiple values into a single value separated by commas

  Hello

  I have following tables with their structures and their data as below.

  CREATE TABLE 'BETODI '. "" BETINFO ".

  (

  VARCHAR2 (8 BYTE) "CURRENTPRESS."

  ENABLE 'TYPEIDCONTAINER' VARCHAR2 (30 BYTE) NOT NULL

  )

  INSERT INTO Betinfo (Currentpress, typeidcontainer) VALUES ('A24G', 'PMC');

  INSERT INTO Betinfo (Currentpress, typeidcontainer) VALUES ('A24D', 'Pensky-MARTENS');

  INSERT INTO Betinfo (Currentpress, typeidcontainer) VALUES ("A25D", "CMP");

  INSERT INTO Betinfo (Currentpress, typeidcontainer) VALUES ('A25G', 'PMC');

  INSERT INTO Betinfo (Currentpress, typeidcontainer) VALUES ('A26D', 'PMC');

  INSERT INTO Betinfo (Currentpress, typeidcontainer) VALUES ('A26G', 'PMC');

  INSERT INTO Betinfo (Currentpress, typeidcontainer) VALUES ("A32G", "V-BFC3");

  INSERT INTO Betinfo (Currentpress, typeidcontainer) VALUES ('A32D', "V-BFC2");

  CREATE TABLE 'BETODI '. "" BETMASTER ".

  (

  ACTIVATE THE "CUREPRESS" TANK (5 BYTES) NOT NULL,

  ACTIVATE THE "TYPE" VARCHAR2 (5 BYTE) NOT NULL,

  NUMBER (5.0) "LASTPCIRIM".

  )

  INSERT INTO BetMaster (Curepress, type, lastpcirim) VALUES ('A24', '45 M 8', 15);

  INSERT INTO BetMaster (Curepress, type, lastpcirim) VALUES ('A25', 42 16', 15);

  INSERT INTO BetMaster (Curepress, type, lastpcirim) VALUES ("A26", 16' 45, 15);

  INSERT INTO BetMaster (Curepress, type, lastpcirim) VALUES ("A27", '45 M 34', 16);

  INSERT INTO BetMaster (Curepress, type, lastpcirim) VALUES ('A28', '45 M 34', 16);

  INSERT INTO BetMaster (Curepress, type, lastpcirim) VALUES ('A29', '45 M 34', 16);

  INSERT INTO BetMaster (Curepress, type, lastpcirim) VALUES ('A30', '45MCH', 15);

  INSERT INTO BetMaster (Curepress, type, lastpcirim) VALUES ("A31", "45MCH", 16);

  INSERT INTO BetMaster (Curepress, type, lastpcirim) VALUES ('A32', '45MCH', 16);

  INSERT INTO BetMaster (Curepress, type, lastpcirim) VALUES ('A33', '45MCH', 16);

  INSERT INTO BetMaster (Curepress, type, lastpcirim) VALUES ("A34", "45MCH", 16);

  These two tables have left join as

  BETMASTER. CUREPRESS = substr (BETINFO. CURRENTPRESS, 1, 3)

  now I want to have the data in the two tables with fields Curepress, Lastpcirim, typeidcontainer.

  Also something like

  Make a group of typeidcontainer if this value is greater than 1 by press separated the values of semicolon (;)

  So, for example above, we should be given as

  A24 PMC 15; PENSKY-MARTENS

  A25 15 PMC

  A26 15 PMC

  A27 16 (NULL)

  A28 16 (NULL)

  A30 15 (NULL)

  A31 16 (NULL)

  A32 16 BFC2-V; V BFC3

  A33 16 (NULL)

  A34 16 (NULL)

  How could do?

  My current request is as

  Select distinct Curepress, lastpcirim, typeidcontainer

  BETMASTER STD left join INF BETINFO

  on the trim (STD. CUREPRESS) = substr (trim (INF. CURRENTPRESS), 1, 3)

  but I am unable to get the values separated by commas.

  Any help would be appreciated.

  Thank you

  Mahesh.

  Hi, Mahesh,

  If you want to only 1 row of output for each distinct combination of currentpress and lastpcirim?  This sounds like a job for GROUP BY.

  And you want the row to contain a list of all different typidcontainers-delimited?  This sounds like a job for the aggregate LISTAGG function.

  WITH joined_data AS

  (

  SELECT DISTINCT

  MST.curepress, mst.lastpcirim, inf.typeidcontainer

  OF betmaster STD

  LEFT JOIN betinfo ON TRIM (mst.curepress) inf = SUBSTR (TRIM (inf.currentpress)

  1

  3

  )

  )

  SELECT curepress, lastpcirim

  LISTAGG (typeidcontainer, ',')

  THE Group (ORDER BY typeidcontainer) AS container_list

  OF joined_data

  Curepress GROUP, lastpcirim

  ;

  Unfortunately, you can't say LISTAGG (DISTINCT ...), so you should always get the separate containers how you already are.  (Note that the subquery is just what you posted).

  Thanks for posting the CREATE TABLE and INSERT statements; It is very useful.  Don't forget to tell what version of Oracle you are using.  LISTAGG was new in Oracle 11.2.

  Why not add CHECK constraints (and perhaps triggers) to your tables, so that curepress and currentpress are not stored with the head or trailing spaces?  Then you wouldn't need to use the PAD in queries like this, and your code would be simpler and more effective.

• Dimension to build using the SQL table and process to populate the SQL table

  I have a dimension in a cube that is manually* built by one of our power users. Now, I have to get all the information of members of this dimension in a SQL table (example: with columns... Level0, level0property, level1, level1property etc...) to use this table in the STUDIO for the responsibility of the Member.
  
  Is there any easy process to do this? Currently I am building each line manually in the SQL table and there are 1100 + members in that manually built dimension. Please advice.

  Look at the outline of applied olap Extractor is a free utility

• How to assign values to the nested table and passes as a parameter for the procedure?

  How to assign values to the nested table and passes as a parameter for the procedure?
  
  Here are the object and its type
  
  create or replace type test_object1 as an object
  (
  val1 varchar2 (50).
  val2 varchar2 (50).
  VARCHAR2 (50) val3
  );
  
  
  create or replace type test_type1 is table of the test_object1;
  
  
  create or replace type test_object2 as an object
  (
  val1 varchar2 (50).
  val2 varchar2 (50).
  VARCHAR2 (50) val3
  );
  
  
  create or replace type test_type2 is table of the test_object2;
  
  
  GRANT ALL ON test_object1 to PUBLIC;
  
  
  GRANT ALL ON test_type1 to PUBLIC;
  
  
  GRANT ALL ON test_object2 to PUBLIC;
  
  
  GRANT ALL ON test_type2 to PUBLIC;
  
  Here is the table object type:
  
  create the table test_object_tpe
  (
  sl_num NUMBER,
  Description VARCHAR2 (100),
  main_val1 test_type1,
  main_val2 test_type2
  )
  
  
  NESTED TABLE main_val1 STORE AS tot1
  NESTED TABLE main_val2 STORE AS earlier2;
  
  
  -----------------------------------------------------------------------------------------------------------
  
  Here is the procedure that inserts values into the nested table:
  
  PROCEDURE INSERT_TEST_DATA (sl_num in NUMBER,
  Description in VARCHAR2,
  p_main_val1 IN test_type1,
  p_main_val2 IN test_type2
  )
  IS
  BEGIN
  
  FOR rec in p_main_val1.first... p_main_val1. Last
  LOOP
  
  INSERT INTO xxdl.test_object_tpe
  (
  sl_num,
  Description,
  main_val1,
  main_val2
  )
  VALUES
  (
  sl_num
  description
  test_type1 (test_object1)
  p_main_val1 .val1 (CRE),
  p_main_val1 .val2 (CRE),
  p_main_val1 .val3 (rec)
  )
  )
  test_type2 (test_object2 (p_main_val2 .val1 (CRE),
  p_main_val2 .val2 (CRE),
  p_main_val2 .val3 (rec)
  )
  )
  
  );
  
  END LOOP;
  
  commit;
  
  END INSERT_TEST_DATA;
  
  -------------------------------------------------------------------------------------------
  
  Here are the block anonymoys what values attributed to the object type and pass values in the procedure:
  
  Set serveroutput on;
  
  declare
  
  p_sl_num NUMBER: = 1001;
  p_description VARCHAR2 (50): = 'Test Val1;
  
  inval1 test_type1: = test_type1();
  inval2 test_type2: = test_type2();
  
  Start
  
  
  inval1 (1) .val1: = "testx1";
  inval1 (1) .val2: = "testx2";
  inval1 (1) .val3: = "testx3";
  
  inval2 (1) .val1: = "testy1";
  inval2 (1) .val2: = "testy2";
  inval2 (1) .val3: = "testy3";
  
  CSI_PKG. INSERT_TEST_DATA (sl_num = > p_sl_num,)
  Description = > p_description,
  p_main_val1 = > inval1,
  p_main_val2 = > inval2
  );
  
  end;
  /
  Someone can correct me.
  
  Thank you
  Lavan

  Thanks for posting the DOF and the sample code but whenever you post provide your Oracle version 4-digit (result of SELECT * FROM V$ VERSION).
  >
  How to assign values to the nested table and passes as a parameter for the procedure?
  >
  Well you do almost everything bad that could be hurt.

  Here is the code that works to insert data into your table (the procedure is not even necessary).

  declare
  p_sl_num NUMBER := 1001;
  p_description VARCHAR2(50) := 'Testing Val1';
  inval1 test_type1 := test_type1();
  inval2 test_type2 := test_type2();
  begin
  inval1.extend();
  inval1(1) := test_object1('testx1', 'testx2', 'testx3');
  inval2.extend();
  inval2(1) := test_object2('testy1', 'testy2', 'testy3');
  
  INSERT INTO test_object_tpe
  (
  sl_num,
  description,
  main_val1,
  main_val2
  )
  VALUES
  (p_sl_num, p_description, inval1, inval2);
  commit;
  end;
  /
  

  See example 5-15 making reference to an element of nested Table Chapter 5 using PL/SQL collections and records in the PL/SQL doc
  http://docs.Oracle.com/CD/B28359_01/AppDev.111/b28370/Collections.htm#CJABEBEA

  1. you don't even have the procedure because it is a simple INSERTION in the table you can do directly (see my above code)

  inval1(1).val1 := 'testx1';
  

  Since you have not yet created all the elements, there is no element 1 "inval1". You need EXTEND the collection to add an element

  inval1.extend();
  

  And then, there is an empty element, but "inval1" is a container for objects of type 'test_object1' not for scalars as "val1", "val2", and "val3".
  If you can not do

  inval1(1).val1 := 'testx1';
  

  You must create an instance of 'test_object1 '.

  inval1(1) := test_object1('testx1', 'testx2', 'testx3');
  

  And so on for the other collection

  You don't need the procedure (as my code example shows), but once you fill in the variables correctly it will work.

• 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

• 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

• Select for a partitioned table, it will work with > sign?

  Hello
  I partitioned table and select for that:

  ...
  PARTITION BY RANGE  ( "INDATE"  )
    (   PARTITION "PRT_20091201" VALUES LESS THAN (TO_DATE(' 2009-12-02 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN')...
  ....
     SELECT COLUMN_LIST FROM TABLE1   WHERE
             ( indate >= to_date(t_indate || t_timefrom, 'mm/dd/yyyy HH24:MI:SS'))                       AND
             ( indate <= to_date(t_indate || t_timeto,   'mm/dd/yyyy HH24:MI:SS'))                        AND... 

  so by selecting all of the Info for the given date specified inside time, and I use > = condition on the column partitioned, do you think that it will work correctly learn directly to the single partition or Oralce requires me to put strictly = sign for this column and then additioanly I can probably choose time, something like:
  

  SELECT COLUMN_LIST FROM TABLE1   WHERE
            ( indate = to_date(t_indate , 'mm/dd/yyyy'))                        AND
                 ( trim_to_time(indate) >= to_char(t_timefrom, ' HH24:MI:SS'))              AND
                 ( trim_to_time(indate) <= to_char( t_timeto,   ' HH24:MI:SS'))              AND... 

  All TX
  TR

  Trento wrote:
  Thanks to all the guys,
  Always learn to follow the plan. I can't use SELECT FROM TABLE of PARTITION (P - NAME) because of dynamic political anti on our site.

  Not a bad policy as dynamic sql is a blade against double edge that can cut you, and the problem with its employees.

  But do not use the option of PARTITION in a dynamic query - use manually in a one-time ad hoc when query data. Decide what partition to get the data is done automatically and is what is the process for