Encryption of the columns and their corresponding indices

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  Hi everyone, I am using oracle 10g consider the following data:

  WITH the data as {}

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  can someone help me write a query that displays the output in oracle 10g above?

  Hello

  elmasduro wrote:

  It's great Frank. Thank you very much.

  I have a request.  What happens if I want to add total main grad, wages, the pc? How would I do that.

  output

  Total general AFFID CTID DESCRIPTION MAIN PAY PC

  XTR AAA hi tech 234.46 1554,46 1788.92

  XTR2 BBB Hi tech2 555,46 555.46

  CCC XTR3 Hi Tech 3 777,23 777.23

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  SELECT Mntrid

  ctid

  Description

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  , SUM (CASE code WHEN = "SALARY" THEN rise END) AS pay

  , SUM (CASE WHEN code = 'PC', THEN rise END) AS pc

  , The SUM of (amount) AS grand_total-* NEW *.

  FROM the data

  GROUP BY mntrid

  ctid

  Description

  ;

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  PDF does not contain columns, lines, formats, styles or other aspects of word processing or spreadsheet file formats.

  This is because the PDF is decidedly not a file format of word processor or a spreadsheet or something 'like' one of these.

  (see ISO 32000 for what PDF "is")

  Which allows to optimize the export of the content of a page PDF is to start with that tag well formed PDF (ISO 14289-1, PDF/UA-1).

  Without that export is what it is and we performs the cleanup of any necessary content using the native application for the export of files (MS Word or Excel).

  Be well...

• a list of all the tables, columns and their consntraints in the database

  Hello

  I need to write sql code which considers a list of all tables in the database, their columns and the primary or foreign key for each column (if any) and the refrence for each foreign key column.

  Any ideas how ro?

  DROP TABLE table_b;
  DROP TABLE table_a;
  --
  CREATE TABLE table_a (
   CONSTRAINT pk_table_a PRIMARY KEY (column_b)
  ,column_a INTEGER
  ,column_b INTEGER
  );
  --
  CREATE TABLE table_b (
   column_c INTEGER
  ,column_d INTEGER CONSTRAINT fk_table_a REFERENCES table_a(column_b)
  );
  --
  
  SELECT  utc.table_name
         ,utc.column_name
         ,CASE WHEN uc.constraint_type = 'P' AND position IS NOT NULL THEN uc.constraint_name END primary_key
         ,CASE WHEN uc.constraint_type = 'R' AND position IS NOT NULL THEN uc.constraint_name END foreign_key
         ,CASE WHEN uc.constraint_type = 'R' AND position IS NOT NULL THEN uc.r_constraint_name END referenced_key
  FROM    user_tab_cols utc
          LEFT JOIN user_constraints uc
            ON  (utc.table_name = uc.table_name)
          LEFT JOIN user_cons_columns ucc
            ON(ucc.table_name = utc.table_name
                 AND utc.column_name = ucc.column_name
                 AND uc.constraint_name = ucc.constraint_name)
  WHERE   utc.table_name IN ('TABLE_A','TABLE_B');
  

  TABLE_NAME COLUMN_NAME PRIMARY_KEY FOREIGN_KEY REFERENCED_KEY

  TABLE_A COLUMN_B PK_TABLE_A

  TABLE_A COLUMN_A

  TABLE_B COLUMN_D FK_TABLE_A PK_TABLE_A

  TABLE_B COLUMN_C

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• Find the Agent and their Manager - connect by

  Hello
  version database Oracle 11 g Enterprise Edition Release 11.2.0.3.0 - 64 bit Production
  Please help me to know Agent and their Manager. also published the desired output
  

  CREATE TABLE AGENT
  (
    AGENT_NAME  VARCHAR2(100 BYTE),
    SID         NUMBER,
    MGR1        VARCHAR2(100 BYTE),
    MGR2        VARCHAR2(100 BYTE),
    DEPT_ID     NUMBER,
    START_DT    DATE,
    END_DT      DATE
  );
  
  CREATE TABLE PERSON
  (
    PERSON_NAME  VARCHAR2(100 BYTE),
    SID          NUMBER,
    DEPT_ID      NUMBER,
    START_DT     DATE,
    END_DT       DATE
  );
  
  SET DEFINE OFF;
  Insert into PERSON
     (PERSON_NAME, SID, DEPT_ID, START_DT, END_DT)
   Values
     ('DEEPAK', 200, NULL, NULL, NULL);
  Insert into PERSON
     (PERSON_NAME, SID, DEPT_ID, START_DT, END_DT)
   Values
     ('VIJAY', 300, NULL, NULL, NULL);
  Insert into PERSON
     (PERSON_NAME, SID, DEPT_ID, START_DT, END_DT)
   Values
     ('MICHELLE', 400, NULL, NULL, NULL);
  Insert into PERSON
     (PERSON_NAME, SID, DEPT_ID, START_DT, END_DT)
   Values
     ('NATLO', 500, NULL, NULL, NULL);
  COMMIT;
  
  
  SET DEFINE OFF;
  Insert into AGENT
     (AGENT_NAME, SID, MGR1, MGR2, DEPT_ID, 
      START_DT, END_DT)
   Values
     ('AGENT4', 103, 'NATLO', 'MISHKA', NULL, 
      NULL, NULL);
  Insert into AGENT
     (AGENT_NAME, SID, MGR1, MGR2, DEPT_ID, 
      START_DT, END_DT)
   Values
     ('AGENT1', 100, 'DEEPAK', 'MICHELLE', 10, 
      TO_DATE('01/01/2013 00:00:00', 'MM/DD/YYYY HH24:MI:SS'), TO_DATE('12/31/9999 00:00:00', 'MM/DD/YYYY HH24:MI:SS'));
  Insert into AGENT
     (AGENT_NAME, SID, MGR1, MGR2, DEPT_ID, 
      START_DT, END_DT)
   Values
     ('AGENT2', 101, 'MICHELLE', 'VIJAY', 55, 
      TO_DATE('01/01/2013 00:00:00', 'MM/DD/YYYY HH24:MI:SS'), TO_DATE('05/01/2013 00:00:00', 'MM/DD/YYYY HH24:MI:SS'));
  Insert into AGENT
     (AGENT_NAME, SID, MGR1, MGR2, DEPT_ID, 
      START_DT, END_DT)
   Values
     ('AGENT3', 102, 'VIJAY', 'NATLO', 105, 
      TO_DATE('05/12/2013 00:00:00', 'MM/DD/YYYY HH24:MI:SS'), TO_DATE('12/31/9999 00:00:00', 'MM/DD/YYYY HH24:MI:SS'));
  COMMIT;

  output Required:
  Agent1  100 DEEPAK MICHELLE VIJAY NATLO MISHKA
  Agent2  101 MICHELLE VIJAY NATLO MISHKA
  Agent3  102 VIJAY NATLO MISHKA
  Agent4  103 NATLO MISHKA

  Hello

  You can simplify the solution Al's using pseudo-column CONNECT_BY_ISLEAF:

  SELECT    CONNECT_BY_ROOT(agent_name) agent_name
          , CONNECT_BY_ROOT(sid) sid
          , SYS_CONNECT_BY_PATH(mgr1,' ')||' '||mgr2 path
  --      , ROW_NUMBER() is not needed
       FROM  agent
       WHERE CONNECT_BY_ISLEAF     = 1     -- *****  Added  *****
    CONNECT BY PRIOR mgr2 = mgr1
  ;
  

  In addition, mixture of analytical functions and CONNECT BY in the same query does not always. Before Oracle 11.2, it almost never works. I have not seen documentation on this, but I am always wary. In any case, CONNECT_BY_ISLEAF is simpler and more efficient, if you want even use ROW_NUMBER here.

  Remember that Oracle 11.2 has recursive WITH clauses, which can do everything that CONNECT AT the time, and much more. For this particular problem, CONNECT BY is always the best solution, but don't forget that you now have a choice.

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  Hello

  new to oracle/mark. I need to write a query that will pull the customer details and on the columns, it will have the month-year: Jan2015, Feb2015... Dec2015.

  on the lines, it will contain the details of the client. the information it will provide are the categories of risk to each month.

  I create a query, but it uses a function to update and is very slow. There are approximately 3000 documents. An update of column required 15 minutes or even more. so an update of column takes 15 minutes * 9 to update until September.

  the best ways to manage this demand?

  my query is:

  UPDATE tmpRiskGrades one

  SET a.feb2015 =)

  Select l.risk_grade

  of rep_loanportfolio l

  where a.agreementno = l.agreeement_no AND l.dtuserdatetime = February 28, 2015 ")

  Certainly, that looks like a pivot base for me.  Assuming that dtuserdatetime is actually a date data type, something like that will not work in your version of non-recognition:

  Select customer_no, agreeement_no, customer_address, loan_type region,

  Max (case when dtuserdatetime = to_date (Jan 31, 2015 "," DD month yyyy ' "))

  end of the era risk_grade) jan2015,.

  Max (case when dtuserdatetime = to_date (February 28, 2015 "," DD month yyyy ' "))

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  ...

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  account_status = 'LIVE' and

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  TO_DATE (31 Dec 2015 "," DD month yyyy ' ")

  Group of customer_no, agreeement_no, customer_address, region loan_type

  order of region loan_type, customer_no, agreeement_no, customer_address

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• How to upgrade are the value of the column and the new column value created

  Hi all

  create the table xxc_temp

  as

  Select donnees_1 'AB103', 'AAA' data_2 data_3 '123', 'RED' data_4, NULL status in all the double union

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  AS

  vstatus VARCHAR2 (8);

  BEGIN

  SELECT status in vstatus FROM xxc_base WHERE donnees_1 = pDATA_1;

  RETURN vstatus;

  END;

  /

  DROP TABLE 'XXC_TEMP ';

  CREATE TABLE 'XXC_TEMP '.

  ('DONNEES_1' CHAR (5),

  "DATA_2" TANK (3).

  "DATA_3" TANK (3).

  "DATA_4' VARCHAR2 (4).

  'STATUS' VARCHAR2 (8) GENERATED ALWAYS AS (SUBSTR (GET_STATUS (DATA_1), 1, 8)) VIRTUAL

  ) ;

  INSERT INTO 'XXC_TEMP' (data_3, donnees_1, data_2, data_4))

  Select "AB103' donnees_1, data_2 'AAA', '123' data_3, 'RED' data_4 of any double union

  Select donnees_1 'AB105', 'BBB' data_2, '222' data_3 data_4 'BLUE' of all the double union

  Select donnees_1 'AB106', 'CCC' data_2 data_3 '333', 'BLUE' double data_4);

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  I found this:
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  Thanks in advance for any advice or help - we appreciate it! :)

  Hello Matt,

  for what you do with files .ini, java provides the java.util.Properties class.
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    JPanel myPanel1= new JPanel(...);
  ...
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• Query to get the data of the column and the metadata in the same set of results.

  Is it possible to build a query to get the values of the columns in a table and also be able to get some metadata (data type, data_length, data_precision, data_scale) for columns in the same set of results.

  If I use a join, have a common value to join on the two tables?

  you use a cross join, not requiring common values.

  create table T (n number, d date, v varchar2(30));
  insert into T values (1,sysdate,'ABC');
  commit;
  
  select C.column_name, c.data_type, c.data_length,
  case c.column_id
   when 1 then to_char(T.N)
   when 2 then to_char(T.D)
   when 3 then T.V
  end VALUE
  from USER_TAB_COLUMNS C, T
  where C.table_name='T'
  order by c.column_id;
  

• How to view the columns (and the sequence and sort order) an index?

  Suppose I have an INDEX for an existing TABLE.
  
  How can I know the columns covered by this INDEX (and the sequence and sort order)?
  
  Which table contains this information?
  
  ALL_INDEXES does not work.
  
  Peter

  user559463 wrote:
  Suppose I have an INDEX for an existing TABLE.

  How can I know the columns covered by this INDEX (and the sequence and sort order)?

  Which table contains this information?

  ALL_INDEXES does not work.

  Peter

  --

  Select the table table_name, column_name, index_name, and position_colonne
  of user_ind_columns
  order by table_name, index_name, position_colonne;