Add new ID/Primary Key column with custom ID generation

Similar Questions

• How to refer to the primary key column of newly inserted rows of tabular form

  Hello

  I use APEX 4.2.0.00.27 with Oracle DB 11.2.0.3.0.

  I work with a tabular presentation wizard-created for insert and update a table using the integrated SRM process (sequence 10).  I'm trying to use a process of anonymous block of PL/SQL (sequence 30) to make another manipulation of table after the records were inserted or updated.  The manual process is associated with my tabular form and I use the variables of name of column binding in my program block.

  My (rsn_test) table has 3 columns: test_id (number), test_nbr (number), test_id2 (number).  Test_id column is identified as the primary key and the type of the source already exists a sequence rsn_test_seq.  Column test_id2 gets its default value 0 to a hidden page element.

  I would use my manual process for updating the value of the test_id2 column.  If it's 0 then I want to put the value of the column test_id.  If it is any other value, then it must remain at this value.  My logic works very well for an existing line, but I'm running into a problem with the newly added lines.  The new lines get inserted, but the test_id2 column remains the default value 0.  I can tell the debugger that the SRM process is triggered first and inserts the line, then my manual dealing with fires.  The problem seems to be that the connection variable: TEST_ID for the primary key column remains NULL after insertion.  I don't know how to get the value of the column test_id of my newly created line to use in my PL/SQL block to my update.

  Process of PL/SQL:

  DECLARE
  BEGIN
     :P7_SHOW := NULL;
     :P7_SHOW := NVL(:TEST_ID2,555) || ' and ' || NVL(:TEST_ID,787) || ' and ' || NVL(:TEST_NBR,9999);
     IF :TEST_ID2 = 0 AND :TEST_ID IS NOT NULL THEN
        UPDATE rsn_test
           SET test_id2 = :TEST_ID
         WHERE test_id = :TEST_ID;
     ELSE
        :TEST_ID2 := :TEST_ID2;
     END IF;
  END;
  
  

  Excerpt from the debugger:

  0.01625 0.00010 Processes - point: ON_SUBMIT_BEFORE_COMPUTATION
  0.01635 0.00008 Branch point: Before Computation
  0.01643 0.00003 Process point: AFTER_SUBMIT
  0.01646 0.00022 Tabs: Perform Branching for Tab Requests
  0.01668 0.00008 Branch point: Before Validation
  0.01676 0.00024 Validations:
  0.01700 0.00135 Perform basic and predefined validations:
  0.01835 0.00020 Perform custom validations:
  0.01855 0.00049 ...Validation "TEST_NBR must be numeric" - Type: ITEM_IS_NUMERIC
  0.01904 0.00007 ......Skip for row 1 because row hasn't changed
  0.01911 0.00016 ......Skip for row 2 because row hasn't changed
  0.01927 0.00012 ...Validation "TEST_ID2 must be numeric" - Type: ITEM_IS_NUMERIC
  0.01939 0.00007 ......Skip for row 1 because row hasn't changed
  0.01945 0.00018 ......Skip for row 2 because row hasn't changed
  0.01964 0.00005 Branch point: Before Processing
  0.01968 0.00004 Processes - point: AFTER_SUBMIT
  0.01972 0.00588 ...Process "ApplyMRU" - Type: MULTI_ROW_UPDATE
  0.02560 0.00154 ...Execute Statement: declare function x return varchar2 is begin begin for c1 in ( select "RSN_TEST_SEQ".nextval pk from sys.dual ) loop return c1.pk; end loop; end; return null; end; begin wwv_flow.g_value := x; end;
  0.02714 0.00140 ......Row 3: insert into "APPPCSRSN"."RSN_TEST" ( "TEST_ID", "TEST_NBR", "TEST_ID2") values ( :b1, :b2, :b3)
  0.02854 0.00011 ...Process "ApplyMRD" - Type: MULTI_ROW_DELETE
  0.02865 0.00004 ......Skip because condition or authorization evaluates to FALSE
  0.02869 0.00015 ...Process "Process Submit" - Type: PLSQL
  0.02884 0.00007 ......Skip for row 1 because row hasn't changed
  0.02891 0.00012 ......Skip for row 2 because row hasn't changed
  0.02903 0.00012 ......Process row 3
  0.02915 0.00429 ...Execute Statement: begin DECLARE BEGIN :P7_SHOW := NULL; :P7_SHOW := NVL(:TEST_ID2,555) || ' and ' || NVL(:TEST_ID,787) || ' and ' || NVL(:TEST_NBR,9999); IF :TEST_ID2 = 0 AND :TEST_ID IS NOT NULL THEN UPDATE rsn_test SET test_id2 = :TEST_NBR WHERE test_id = :TEST_ID; ELSE :TEST_ID2 := :TEST_ID2; END IF; END; end;
  0.03344 0.00013 ...Session State: Saved Item "P7_SHOW" New Value="0 and 787 and 1300"
  0.03356 0.00004 Branch point: After Processing
  0.03360 0.00048 ...Evaluating Branch: "AFTER_PROCESSING" Type: REDIRECT_URL Button: (No Button Pressed) Condition: (Unconditional)
  0.03407 0.00013 Redirecting to f?p=290:7:8717971109610:::::&success_msg=0%20row(s)%20updated%2C%201%20row(s)%20inserted.Success%2FEBD244168556408CBA714E3974918C09%2F
  0.03420 0.00012 Stop APEX Engine detected
  0.03432 0.00007 Stop APEX Engine detected
  0.03439 - Final commit
  
  

  Any suggestions?

  I have run tests on

  https://apex.Oracle.com/pls/apex/f?p=83488:1 demo/demo

  to see your problem.

  I have 2 solution for your problem.
  I add trial NOT tabular just usual block of PL/SQL

  	BEGIN
  	I'm IN (SELECT TEST_ID FROM RSN_TEST WHERE TEST_ID2 = 0)
  	LOOP
  	UPDATE RSN_TEST
  	SET test_id2 = TEST_ID
  	WHERE test_id = i.TEST_ID;
  	END LOOP;
  	END;

  and works very well, you can see in the sample.

  The other solution is to show new generated TEST_ID

  Adding a sequence as a default value for a column in a table field

  And to execute your procedure.

  I get how is with the good luck of time.

  By

• addition of populated table's primary key column

  Hi all
  
  
  I am trying to add a column to a table that is already filled. The column to be added must be the primary key.
  As expected, oracle jumped to the top of the error (ORA-01758) of a column NOT NULL cannot be added, unless the table is empty.
  
  A possible way is to create a new temp table and then transfer all current data... rename the temporary table later.
  
  Is there another way to accomplish this?
  
  Thank you

  Hello

  As error suggested you cannot add a constraint not null on the new column to the table that is already filled. What else you can do next to you, you have a fairly simple and fast to create a temporary table and moving data?
  You can add the new column but null, update the column with values and alter the table of marking column as not null. But your idea of empty temporary table is simple and fast.

  CREATE TABLE my_objects
  AS
     SELECT 'my_new_column' my_new, owner
     FROM all_objects
     WHERE ROWNUM < 1;
  
  _Not Null_
  ALTER TABLE MY_OBJECTS
  MODIFY(MY_NEW  NOT NULL);
  
  _Primary key_
  ALTER TABLE MY_OBJECTS
   ADD CONSTRAINT MY_OBJECTS_PK
   PRIMARY KEY
   (MY_NEW);
  

  Concerning

• How to convert a single column in the primary key column

  Hi all
  
  I have a column that's unique cnstraint, this column contains the value as digital and also null.
  
  So how do you convert to primary key column
  
  Concerning
  Prashant

  Prashant wrote:
  Hi all

  I have a column that's unique cnstraint, this column contains the value as digital and also null.

  So how do you convert to primary key column

  Concerning
  Prashant

  Prashant,

  You must assign new values to null values. You can use a sequence to produce the figures in this column.

  Select max (your_unique_column) from your_table; -find the maximum value of this column, so you can start the sequence of this number.

  create sequence seq_for_your_table with XXXX - value max + 1 you have found

  Update your_table set your_unique_column = seq_for_your_table.nextval () where your_unique_column is null

  You can create a primary key on this column.

  Best regards

  Grosbois

  -------------------------------------------------------
  If you answer this question, please mark appropriate as correct/useful messages and the thread as closed. Thank you

• A composite primary key column cannot be null

  Hi all
  It will be a silly question, but still, I would like to ask if a composite primary key column can be null?
  
  
  Thank you
  Rafi.

  Rafi,

  Why you think he would be allowed?

  SQL> drop table test purge;
  drop table test purge
             *
  ERROR at line 1:
  ORA-00942: table or view does not exist
  
  SQL> create table test as select * from dba_objects;
  
  Table created.
  
  SQL> alter table test add primary key(object_id, owner);
  
  Table altered.
  
  SQL> insert into test(object_id, owner) values(null, 'aman');
  insert into test(object_id, owner) values(null, 'aman')
                                            *
  ERROR at line 1:
  ORA-01400: cannot insert NULL into ("SYS"."TEST"."OBJECT_ID")
  
  SQL> insert into test(object_id, owner) values(1,null);
  insert into test(object_id, owner) values(1,null)
                                              *
  ERROR at line 1:
  ORA-01400: cannot insert NULL into ("SYS"."TEST"."OWNER")
  
  SQL>
  

  HTH
  Aman...

• How to find the primary key columns in the tables in MS Access using SQL queries

  How to find the primary key columns in the tables in MS Access using SQL queries

  Hello

  This is the forum for Windows Vista programs related issues.

  For better assistance, please try instead the Forums in SQL Server .

  Thank you! Vincenzo Di Russo - Microsoft MVP Windows Internet Explorer, Windows Desktop Experience & security - since 2003. ~ ~ ~ My MVP profile: https://mvp.support.microsoft.com/profile/Vincenzo

• Diagram of all the tables list of primary key column names

  Hi people,
  
  I have a Scott user for example, I want to retrieve all tables primary key column names in the user Scott.
  
  can someone help me please.
  
  Thanks in advance,
  karmaya

  You can log in to SCOTT and try this

  select c.constraint_name
       , cc.table_name
       , cc.column_name
    from user_constraints c
    join user_cons_columns cc
      on c.constraint_name = cc.constraint_name
   where c.constraint_type = 'P';
  

• How can I update the primary key column

  Hello
  Can you suggest me best workaround/algorithm for task below:
  
  (Oracle 10g, Solaris OS.)
  Location:
  Table a primary key column P 'Code', tables children F1, F2,..., reference to F15 "Code_P' foreign key column"P.Code"column, and we do not know which of the child tables data available for a particular value"P.Code.
  Task:
  Change the value "P.Code" from 100 to 200. So that the result would be that record P [Code = 100] should be updated:

  update P set
  Code = 200
  where Code = 100;

  And child tables 'Code_P' column should be updated as:

  update F1, F2, .., F15 set
  P_code = 200
  where P_code = 100;

  The best solution would be that you can easily repeat this task.
  
  Edited by: CharlesRoos the 28.12.2010 12:10

  Will there be FK constraints in place?
  If so, you may need to modify them so that their application may be temporarily deferred.

  The gameplan would be something like this:

  -Making of CF can be deferred, immediate departure.
  -Alter FK set delay.
  -Update of Parent table.
  -Update of the children tables.
  -Immediate constraints set
  s ' engage;

• I cannot pair the new Apple TV Remote app with my first generation Apple TV.

  I cannot pair the new Apple TV Remote app with my first generation Apple TV. Someone knows what to do?

  You can use the old, now called iTunes Remote? A new one seems really targeted to ATV 4.

• update of a primary key column

  Hello

  I have a problem I want to expose you for your opinion.

  In a table of the source, I have a primary key consisting of four columns (let's call them A, B, C and D).

  Columns D must be updated with a different value (a Mappin with another table)

  My ODI interface, I set the update keys a, B, C, but of course, it does not work because there is violation of the primary key

  I thought, the first value of D in a new column (D_OLD) and use as a 4th update key.

  What do you think?

  Please suggest me the right way to do it

  Thank you for your return

  Hello

  Maybe I did not understand your problem correctly, but you say that you want to update the value of col D what happens to the part b of your primary key. You cannot update the row using columns A, B and C, because they are not unique. Am I wrong? If this is the case, then how will you identify which line you want to update the value of col D for. Tell if you have 2 lines with the same value for columns A, B and C, and a different value of D so how do you know which line you want to update? If you have the answer to this question, then you should be able to update as well.

  Thank you

  Ajay

• By specifying the Composite primary key (columns 4 and 5)

  Hi all

  I am new to Oracle Apex,

  I have a few paintings with a composite primary key (4 or 5 columns)... When I try to create a primary key constraint, it gives me only 3 columns!

  Any help please in how can I solve this problem?

  Thanks in advance

  You are probably using the workshop for this sql Object Explorer

  which seems to be limited to 4 columns on APEX 4.2.3

  best thing to do is to create manually using the sql commands in sql workshop, here's an example:

  {code}

  CREATE UNIQUE INDEX myui ON JC_ACCOUNT

  (TITLE_ID, FAMILY_NAME GIVEN_NAME, GENDER_ID, TELEPHONE_NUMBER)

  /

  {code}

  I hope this helps you

  KR

  Martin

• How to use a Select list on a second primary key column in a tabular presentation?

  Hello
  
  I created a tabular presentation located on the details page of the user (form), that allows to manage roles for this user.
  
  The shape of table contains 3 elements:
  -> a checkbox
  -> user (PK) column is hidden, and its default value is generated by a function pl/sql (to submit the time), which refers to the element to username.
  -> the role of the column is displayed as a named select list LOV: add you a line, choose a role, submit, etc..
  
  But my ROLES table (not created by me...) contains 2 columns: USER and GRANTED_ROLE.
  The USER is the primary key, but a user can have multiple roles (= one line per role).
  GRANTED_ROLE is the second primary key, so that we can differentiate between one line of another.
  
  At the moment my tabular form has only one primary key (USER), so the GRANTED_ROLE can display as a list select.
  But, for this reason, in my opinion, remove the button do not work: all the lines of the user have the same key, so when I select a few lines to remove these, all lines in the form of tables are deleted.
  
  I don't know how my button Delete could recognize the subkey to remove only selected rows...
  Or maybe I should use a different type of region?
  
  Thanks for your help!
  Fanny

  In the process of ApplyMRD, try to specify the 'secondary key' column in the section "Source: Multi line Update and Delete"to the second column key in tabular form.»

  I have not tested this with your situation, but worth it.

• Add new storage "opération TimedOut" with RDM LUN on host. Pls Help

  Hello

  Installation program:

  4 ESX 3.5 hosts each with 2 HBA adapter connected to the HP MSA1000.

  Question:

  I set up MSCS SQL 2000 with RDM through host, MSCS failover without problem, performance wise is not a problem at all too. But the problem is that if the MSCS SQL resource is managed by the Active virtual machine that is running on host1, I can navigate to storage / adding LUNs, set in shape of new LUN with no problems.

  If I want to add new storage on the other host "host2" I received the error "Tor time to ask"is it because of the resource ROW being busy serving the machine virtual active it is why the host not allowing do not change with the addition of new storage?

  Guests see these LUNs. I can add the data store or to do a new analysis also long the LUNS are not presented to the host. Also I can browse the data store, able to see the addition of Storge Wizard when adding new

  Data store and make a new analysis only on the host where the operation of the active node. Suppose the preseneted RDM LUN on host1 and host2 and VM SQL running on host1, I can only browse the data warehouses, a new analysis and add new storge to this host. I can't do the same thing with

  hosts2.

  But if I take these "RDM LUN" LUNS in host2, host3 and host4. I can do a rescan, adding other LUNS as the RDM presented to the virtual machine, able to see the add storage wizard.

  I googled the error and found that the Ontario server is question, or DNS

  Best regards

  Hussain Al Sayed

  Post edited by: habibalby

  I had similar questions to those two problems here where I couldn't add a data store as he would expire and I got the long startup time. I also use RDM which I use for MSCS in the whole of boxes 2. Here's a quote from my previous post in which I found a response that helped me. Changingthe SCSI retry time increased my boot time and allowed me to add one more time without the question of the time-out, data warehouses.

  Response to previous Post:

  I did some research and I think I've made some progress. I found that I get lots of SCSI errors in the VMkernel newspaper. I did some more research and found out that I can change the time retrying SCSI 80 to 10 and it has done wonders for my time to reboot. Now, instead of taking 20 minutes to start, it takes less than 5 minutes now. Much better. I made the change in the host-> advance-> SCSI--> SCSI configuration try again. 80 a the default and 10A was suggested as being a good value. It helped and I will keep an eye on what can make the effect, but so far it has helped with startup times.

  http://communities.VMware.com//thread/203122?TSTART=0

• Problem with the primary key

  Hi all
  I have a table with the same data repeat 2 times. I mean that the lines are depating twice. so now, I create a primary key for the table. He throws me an error saying voilation key primary. I'll give you the clear script.
  
  create table employee)
  ID VARCHAR2 (4 BYTE) NOT NULL,
  First name VARCHAR2 (10 BYTE),
  VARCHAR2 (10 BYTE) last_name,
  START_DATE DATE,
  End_date DATE,
  Number (8.2) of salary.
  City VARCHAR2 (10 BYTE),
  Description VARCHAR2 (15 BYTE)
  )
  
  
  insert into employee values ('01', 'Jason', 'Martin', to_date('19960725','YYYYMMDD'), to_date('20060725','YYYYMMDD'), 1234.56, 'Toronto', 'Program');
  insert into employee values ('02', "Alison", "Mathews', to_date('19760321','YYYYMMDD'), to_date('19860221','YYYYMMDD'), 6661.78, 'Vancouver', 'Tester');
  insert into employee values ('03', 'James', 'Smith', to_date('19781212','YYYYMMDD'), to_date('19900315','YYYYMMDD'), 6544.78, 'Vancouver', 'Tester');
  insert into employee values ('04', 'Celia', 'Rice', to_date('19821024','YYYYMMDD'), to_date('19990421','YYYYMMDD'), 2344.78, "Vancouver", "Manager");
  insert into employee values ('05', 'Robert', 'Black', to_date('19840115','YYYYMMDD'), to_date('19980808','YYYYMMDD'), 2334.78, 'Vancouver', 'Tester');
  insert into employee values ('07', 'David', 'Larry', to_date('19901231','YYYYMMDD'), to_date('19980212','YYYYMMDD'), 7897.78, 'New York', 'Manager');
  insert into employee values ('06', 'Linda', 'Green', to_date('19870730','YYYYMMDD'), to_date('19960104','YYYYMMDD'), 4322.78, 'New York', 'Tester');
  insert into employee values ('08', 'James', 'Cat', to_date('19960917','YYYYMMDD'), to_date('20020415','YYYYMMDD'), 1232.78, 'Vancouver', 'Tester')
  insert into employee values ('07', 'David', 'Larry', to_date('19901231','YYYYMMDD'), to_date('19980212','YYYYMMDD'), 7897.78, 'New York', 'Manager');
  insert into employee values ('06', 'Linda', 'Green', to_date('19870730','YYYYMMDD'), to_date('19960104','YYYYMMDD'), 4322.78, 'New York', 'Tester');
  
  
  The table is now with 10 columns
  
  ID FIRST_NAME LAST_NAME START_DAT END_DATE SALARY CITY DESCRIPTION
  ---- ---------- ---------- --------- --------- ---------- ---------- ---------------
  Jason Martin 25 July 96 01 25 July 06 1234.56 Toronto programmer
  02 Alison Mathews 21 March 76 February 21, 86 6661.78 Vancouver Tester
  03 James Smith December 12, 78 March 15, 90 6544.78 Vancouver Tester
  04 Celia rice 24 October 82 21 April 99 2344.78 Vancouver Manager
  Robert Black 05 January 15, 84 8 August 98 2334.78 Vancouver Tester
  Linda Green 06 July 30, 87 January 4, 96 4322.78 tester of New York
  Manager of New York 07 David Larry 31 December 90 February 12, 98 7897.78
  08 James Cat 17-AUG-96 April 15 02 1232.78 ester of Vancouver
  
  
  Create a primary key for the table
  
  "Employee to ALTER TABLE PRIMARY KEY (ID, First_Name, Last_Name, start_date, end_date,
  Salary, city, Description); »
  
  I get error voilation primary key
  
  IS IT POSSIBLE TO CREATE THE KEY PRIMARY ON THE TABLE WITH ON THE DELETION OF THE DATA?
  Is IT POSSIBLE CREATING PRIMARY KEY USING 'GROUP BY' FUNCTION or OTHER FUNCTIONS?
  
  Published by: user11872870 on September 23, 2010 17:41

  Primary key columns cannot contain NULL values. If you tried to create a primary key constraint, you will get an error like this:

  ERROR at line 1:
  ORA-01449: column contains NULL values: cannot alter to NOT NULL
  

  However, you can add a unique constraint by doing something like the following:

  ALTER TABLE test ADD CONSTRAINT test_unq UNIQUE (column1, column2);
  

  I still wonder why you take the approach you take. I have a vague feeling that're missing us the key pieces of information to recommend the correct solution.

• Update primary key with waterfall

  Lets say I have 10 tables that have a foreign key defined, that everything points to a single table primary key.
  Now, I want to update the value of the primary key to another value, and I want all 10 tables to get this new value in their columns FK.
  Also, note also that some of these 10 tables can have this FK-value that is updated to PK-table.
  I can do this task with a good script?
  
  I would like to somehow in this way:

  Update PkTable set
  Pkcol=NewAvailableValue
  CASCADE;

  Right now I use the following approach:
  1. from table "user_constraints" find FK constraints binded to the primary change being key to find the list of tables.
  2 query each table if it has the PK value in their columns-FK, if has, then update the value allowed another PK-value.
  3. Update PK-value of primary key column of the table to the new value.
  4 validation.

  Maybe this will be useful: Tom Kyte: UPDATE CASCADE PACKAGE