Outer join with Oracle syntax

Similar Questions

• Left Outer Join with sum

  Dear elders,
  
  
  
  Firstly, sorry if my question is considered to be 'very novice' but I have this problem of confusion.
  
  Can I use the left outer join with summary?
  For example:
  
  Work table

  JobCode      JobGroupCode      
  111A         1100              
  112B         1100               
  113C         1100
  121A         1200
  333F         3300               

  Table No.

  JobGroupCode    ParentCode
  1100            1000
  1200            1000
  1300            1000
  3300            3000

  Activity table

  Jobcode      Mandays
  111A         5
  112B         7
  113C         3

  I want to choose with this result

  Job.JobCode    Job.JobGroupCode     JobGroup.Parentcode       Mandays
  111A           1100                 1000                      5
  112B           1100                 1000                      7               
  113C           1100                 1000                      3
  121A           1200                 1000                      0
  333F           3300                 3000                      0

  All I did was:

  select j.jobcode, j.jobgroupcode, jg.parentcode, sum(a.mandays)
    from job j
   inner join jobgroup jg on j.jobgroupcode = jg.jobgroupcode
    left join activity a on j.jobcode = a.jobcode
   group by j.jobcode, j.jobgroupcode, jg.parentcode

  and I got was only jobcode activity table, not exactly what I want to get all the work table jobcode. Could you please tell me, where I was wrong?
  
  result

  Job.JobCode    Job.JobGroupCode     JobGroup.Parentcode       Mandays
  111A           1100                 1000                      5
  112B           1100                 1000                      7               
  113C           1100                 1000                      3

  Thank you very much.
  
  Published by: user11197113 on May 25, 2009 03:31
  
  Published by: user11197113 on May 25, 2009 03:32

  Hello (and welcome!).

  Edit

  OK, try this:

  select j.jobcode, j.jobgroupcode, jg.parentcode, NVL(sum(a.mandays),0)
    from job j
   inner join jobgroup jg on j.jobgroupcode = jg.jobgroupcode
    left join activity a on j.jobcode = a.jobcode
   group by j.jobcode, j.jobgroupcode, jg.parentcode
  

  That will give you this:

  JOBC JOBG PARE SUM(A.MANDAYS)
  ---- ---- ---- --------------
  111A 1100 1000              5
  112B 1100 1000              7
  113C 1100 1000              3
  121A 1200 1000              0
  333F 3300 3000              0
  

  These are real results of your test data.

• SSRS for lack of outer join with the Oracle data source

  It seems to be a problem with the Oracle driver used in the Reporting SERVICES query designer.

  When you use an Oracle data source, if I create an outer join in the graphic designer, it automatically inserts '{OJ' before the join and '} ' after her.  This is an incorrect syntax for Oracle and refuses to start.  The curly braces and the JO editable in designer text, but if I go back to the graphic designer and immediately to reintegrate them.

  Only, this has started to happen a year or two ago - before that it worked, but with the old (+) syntax.

  Can it not be healed?  It makes things very difficult.

  -Geoff

  Hi Geoff,

  Thanks for posting in the Microsoft Community.

  However, the question you posted would be better suited in the Forums of the Oracle Support; We recommend that you post your query in Oracle Support Forums to get help:

  https://forums.Oracle.com/forums/main.jspa;JSESSIONID=8d92100c30d8fb401bcbd10b46c38c9ddf1a3242549a.e34SbxmSbNyKai0Lc3mPbhmSc3aNe0? CategoryID = 84

  If you have any other questions or you need Windows guru, do not hesitate to post your questions and we will be happy to help you.

• outer join with syntax plus sign

  To do this in Sql outer joins can be used as grammar:
  ( + )
  What are the standards (ISO, ANSI and what version) created this grammar?
  Or is it only the invention of Oracle?
  
  Published by: CharlesRoos on July 1st, 2010 07:23

  Apparently, he was "there are chandeliers" ;)

  http://asktom.Oracle.com/pls/asktom/f?p=100:11:0:P11_QUESTION_ID:2419891400346921578 #2439016200346544570

• Why left outer join with a table gives me more lines?

  Hi gurus,

  I can see "view_a" and a table 'table_a '.

  view_a a county of 100 lines. Now, when I left outer join that discovers with a 'table_a', I expect all 100 lines.

  However, I'm more than 100 lines. Is it still possible?

  Also even to analyze these situations, how can I move forward?

  Because it is very high volumn of sight and takes longer to run.

  Select count (*) view_a, view_b

  where view_a.col1 = view_b.col1 (+)

  and view_a.col2 = view_b.col2 (+);

  Thank you

  I can see "view_a" and a table 'table_a '.

  view_a a county of 100 lines. Now, when I left outer join that discovers with a 'table_a', I expect all 100 lines.

  However, I'm more than 100 lines. Is it still possible?

  Also even to analyze these situations, how can I move forward?

  Because it is very high volumn of sight and takes longer to run.

  Select count (*) view_a, view_b

  where view_a.col1 = view_b.col1 (+)

  and view_a.col2 = view_b.col2 (+);

  Which is not necessarily related to the use of an outer join.

  Just join of two tables in general will give you more rows of one table has.

  Scott DEPT table contains ONE row for deptno = 10

  The EMP table has THREE rows of deptno = 10

  The number of rows you plan if you join two tables using an equi-join?

  Three - what is MORE lines the DEPT table has for deptno = 10

  Select * from Department where deptno = 10

  DEPTNO, DNAME, LOC
  10, ACCOUNTING, NEW YORK

  Select * from emp where deptno = 10

  MGR, EMPLOYMENT ENAME, EMPNO, HIREDATE, SAL, COMM, DEPTNO
  7782, CLARK, MANAGER, 7839, 6/9/1981,2450, 10
  7839, KING, PRESIDENT, 17 NOVEMBER 00, 10
  7934, MILLER, CLERK, 7782, 23 JANUARY 00: 10

  Select dept.*, emp.*
  Department, emp
  where dept.deptno = 10
  and dept.deptno = emp.deptno

  DEPTNO, DNAME, LOC, EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO_1
  10, ACCOUNTING, NEW YORK, 7782, CLARK, MANAGER, 7839, 6/9/1981,2450, 10
  10, ACCOUNTING, NEW YORK, 7839, KING, PRESIDENT, 17 NOVEMBER 00, 10
  10, ACCOUNTING, NEW YORK, 7934, MILLER, CLERK, 7782, 23 JANUARY 00: 10

  So if these are the lines ONLY in the table EMP and DEPT the query would give you THREE lines despite the DEPT table only ONE line.

  No do you expect? You get ALL the child rows that belong to the parent company. Otherwise, how could it possibly work?

  The OUTER join includes lines where the parent row exists but there is NO child line as others have shown.

  Outer joins

  Outer join extends the result of a simple join. Outer join returns all rows that satisfy the join condition and also returns some or all rows in a table for which no line of the other meet the join condition.

  Get more lines to exist in one of the paintings is a basic necessity. It usually has NOTHING to with the question of whether you have an outside to join or not.

  See the section on the JOINTS in the Oracle documentation

  http://docs.Oracle.com/CD/B28359_01/server.111/b28286/queries006.htm

• outer join with the additional constraint

  Hello

  With the help of Oracle 11 g R2.

  I would of outer join tables 2 together and put down restrictions on the types of records that are returned in the query result. Here's a mock-up of the tables and data.

  create table aaa (col1 number not null, col2 number not null)

  create table bbb (col1 number not null, col2 number not null)

  insert into values of aaa (1: 80)

  insert into values aaa (2, 90)

  insert into values aaa (3, 80)

  insert into values aaa (4, 90)

  insert into values aaa (5, 80)

  
  insert into bbb values (3, 600)

  insert into values of bbb (4, 700)

  This is the query

  
  select a.col1, a.col2, b.col1, b.col2
  from aaa a, bbb b
  where a.col1 = b.col1 (+)
  and   (a.col2, b.col2) <> ((90, 700))
  

  The result of the query is as follows.

  col1 col1 col2 col2

  1 80

  3 80 3 600

  5 80

  Where col1 = 4 has been deleted, which is an expected result. However, where col1 = 2 has also been removed, which is not a desired outcome. Your response is appreciated.

  Hello

  Here is a way that works for the given sample data:

  SELECT *.

  AAA a

  LEFT OUTER JOIN bbb b ON a.col1 = b.col1

  WHERE the NVL (a.col2, 0) <> 90

  OR NVL (b.col2, 0) <> 700

  ;

  I don't know if that will satisfy your requirements with other data, since you didn't say what your needs are.

  Whenever you have a WHERE clause is applied after the outer join, all columns of the table in option (table bbb in this example) must be used in an NVL, NVL2 or something like a CASE expression that takes into account null values; otherwise, the effect will be the same as an inner join.

• HELP SQL (auto / full outer join with date corresponding)

  I'm having a hard time get this query nailed... hoping someone can help me sorted.

  create table tab1 (identification number,

  date of eff_date,

  Code1 varchar2 (2),

  Code2 varchar2 (2)

  )

  /

  insert into tab1 values (2, to_date('2015-01-14','YYYY-MM-DD'), 'DAT', 'AS');

  insert into tab1 values (2, to_date('2015-03-19','YYYY-MM-DD'), 'DAT', 'AS');

  insert into tab1 values (2, to_date('2015-08-28','YYYY-MM-DD'), 'DAT', 'AS');

  insert into tab1 values (2, to_date('2015-11-12','YYYY-MM-DD'), 'DAT', 'AS');

  insert into tab1 values (2, to_date('2015-01-03','YYYY-MM-DD'), "DAT", "AE");

  insert into tab1 values (2, to_date('2015-03-14','YYYY-MM-DD'), "DAT", "AE");

  insert into tab1 values (2, to_date('2015-04-18','YYYY-MM-DD'), "DAT", "AE");

  insert into tab1 values (2, to_date('2015-09-14','YYYY-MM-DD'), "DAT", "AE");

  insert into tab1 values (2, to_date('2015-01-14','YYYY-MM-DD'), "DAT", "BS");

  insert into tab1 values (2, to_date('2015-02-14','YYYY-MM-DD'), "DAT", "BS");

  insert into tab1 values (2, to_date('2015-03-14','YYYY-MM-DD'), "DAT", "BS");

  insert into tab1 values (2, to_date('2015-05-14','YYYY-MM-DD'), 'DAT', 'BE');

  insert into tab1 values (3, to_date('2015-09-16','YYYY-MM-DD'), 'DAT', 'AS');

  insert into tab1 values (3, to_date('2015-04-16','YYYY-MM-DD'), "DAT", "AE");

  tab1

  ID, date, code 1, code2

  2. DID DAT 2015-01-14

  2. DID DAT 2015-03-19

  2. DID DAT 2015-08-28

  2. DID DAT 2015-11-12

  2 AE DAT 2015-01-03

  2 AE DAT 2015-03-14

  2 AE DAT 2015-04-18

  2 AE DAT 2015-09-14

  2 BS DAT 2015-01-14

  2 BS DAT 2015-02-14

  2 BS DAT 2015-03-14

  BE DAT 2 2015-05-14

  3. DID DAT 2015-09-16

  3 AE DAT 2015-04-16

  What I need to do...

  1 auto join to match EI for each partition ID

  2. THAT the date must be less than or equal to the date of the AE and when there is more then a line corresponding to this criterion has chosen the date of closest EI of the date of the ACE.

  3. it must be a full outer join because I want to show all lines, even if it is not a match.  There is a beginning, but not record end end gold but no record of departure

  4. If there is an AE line for many AS lines (the SA date is less then equals the date of EI) then join this AE line to all 3 rows of ACE

  5. the same rules for BS and BE.

  result should look like this.

  ID, date, code 1, id_1 code2, date_1, code1_1, code2_1

  2 2015-01-14 DAT AS 2 AE DAT 2015-03-14

  2 2015-03-19 DAT AS 2 AE DAT 2015-04-18

  2 2015-08-28 DAT AS 2 AE DAT 2015-09-14

  2 2015-11-12 DAT DID ZERO ZERO ZERO ZERO

  NO NO NO NO 2 AE DAT 2015-01-03

  2015-01-2 14 DAT BS 2 BE DAT 2015-05-14

  2015-02-2 14 DAT BS 2 BE DAT 2015-05-14

  2015 03-2 14 DAT BS 2 BE DAT 2015-05-14

  3 2015-09-16 DAT DID ZERO ZERO ZERO ZERO

  NO NO NO NO 3 AE DAT 2015-04-16

  My attempt was somewhat along these lines (dealing only with SA / combos AE) but it does not manage the many scenarios one (req 4).

  Select a.*, b.* from

  (select row_number () on the rn (partition by a.id order a.eff_date), a.*)

  of tab1 where a.code2 = 'AS') a

  full outer join

  (select row_number () on the rn (b.eff_date order by b.id partition), b.*)

  tab1 b where b.code2 = 'Æ') b

  on a.id = b.id

  and a.rn = b.rn

  and a.eff_date < = b.eff_date

  Hello

  owbdev99 wrote:

  I'm having a hard time get this query nailed... hoping someone can help me sorted.

  create table tab1 (identification number,

  date of eff_date,

  Code1 varchar2 (2),

  Code2 varchar2 (2)

  )

  /

  insert into tab1 values (2, to_date('2015-01-14','YYYY-MM-DD'), 'DAT', 'AS');

  ...

  Thanks for posting the CREATE TABLE and INSERT.  I know it can be a lot of trouble.  You want to get answers that work, not you?  Make sure that the statements you post too much work.  Test (and, if necessary, attach) your statements before committing.  You said code1 be VARCHAR2, but all the instructions insertion have values of 3 characters for code1.

  You are on the right track, with an analytical function, but ROW_NUMBER solves this problem.  1 "THAT line" could correspond to the 1st, 2nd, 3rd or any other line 'AE' and vice versa.  Try to use the analytical MIN function instead or ROW_NUMBER, like this:

  WITH got_next_e_date AS

  (

  SELECT id, eff_date, code1, code2

  MIN (CASE

  WHEN SUBSTR (code2, 2) = 'E '.

  THEN eff_date

  END

  ) OVER (PARTITION BY ID.

  , SUBSTR (code2, 1, 1)

  ORDER BY eff_date DESC

  ) AS next_e_date

  OF tab1

  )

  s AS

  (

  SELECT *.

  OF got_next_e_date

  "WHERE SUBSTR (code2, 2) s ="

  )

  e

  (

  SELECT *.

  OF got_next_e_date

  WHERE SUBSTR (code2, 2) = 'E '.

  )

  SELECT s.id

  s.eff_date

  s.code1

  s.code2

  e.id AS id_1

  e.eff_date AS eff_date_1

  e.code1 AS code1_1

  e.code2 AS code2_1

  S

  FULL OUTER JOIN e ON s.id = e.id

  AND s.next_e_date = e.eff_date

  AND SUBSTR (s.code2, 1, 1) = SUBSTR (e.code2, 1, 1)

  ORDER OF NVL (s.id, e.id)

  , NVL (SUBSTR (s.code2, 1, 1)

  , SUBSTR (e.code2, 1, 1)

  )

  s.eff_date

  ;

  Out (as you asked):

  ID EFF_DATE CODE1, CODE2 ID_1 EFF_DATE_1 CODE1_1 CODE2_1

  --- ---------- ----- ----- ----- ---------- ------- -------

  2 2015-01-14 DAT AS 2 AE DAT 2015-03-14

  2 2015-03-19 DAT AS 2 AE DAT 2015-04-18

  2 2015-08-28 DAT AS 2 AE DAT 2015-09-14

  2. DID DAT 2015-11-12

  2 AE DAT 2015-01-03

  2015-01-2 14 DAT BS 2 BE DAT 2015-05-14

  2015-02-2 14 DAT BS 2 BE DAT 2015-05-14

  2015 03-2 14 DAT BS 2 BE DAT 2015-05-14

  3. DID DAT 2015-09-16

  3 AE DAT 2015-04-16

  I guess code2 is always 2 characters, and the 2nd character is always ' or 'E '.

  I assume that the combination [id, eff_date, code2] is unique.

  If these assumptions are wrong, you need a few minor changes, but nothing big.

• Using Left Outer Join with reference

  I have three tables.
  Table 1: BOOK_DETAILS
  Fields: BOOK_ID, BOOK_NAME
  
  Table 2: BOOK_ISSUE_RECORD
  Fields: BOOK_ID, USER_NAME
  
  Table 3: BOOK_AUTHOR
  Fields: BOOK_ID, AUTHOR_NAME
  
  I must link table 1 and table 2 with a left outer join, because even if the book is not the questions to anyone, his name should come.
  I have once again display the name of the author of books for each book.
  I am able to create a query with the left outer join between table 1 and table 2. However, I am not able to give a reference to Table 3.
  Can someone help me with this please.
  
  Concerning
  Hawker

  select  d.book_name,
          a.book_author,
          i.user_name
    from       book_details d
          join
               book_author a
            on (d.book_id = a.book_id)
          left join
               book_issue_recors i
            on (d.book_id = i.book_id)
  /
  

  SY.

• Outer join with a constant

  Outer join when made between the columns of the two tables is easy to understand and I'm familiar with it. But when it comes to the same join with a constant, my head starts spinning... you guys can help me with this? I would appreciate that show you with examples.
  Thanks a ton.

  Hello

  >
  Ok... I want to understand the output of the last sql statement given here.
  ...
  Select * from a, b
  where col_a (+) = col_b
  and col_a (+) = 1

  It shows put it is as follows
  col_a, col_b
  1 1
  2 NULL
  NULL NULL

  Joining means: 'display all rows in table b and the corresponding rows in the table one '.
  Since it is an outer join, it also means "view b lines even if there is no corresponding row in a.

  Thus, the output contains 3 lines, which correspond to the 3 lines in b.
  Only one of them (the one with col_b = 1) had a game in a.col_a, if the column is NULL in the other lines.

  With the sample data you posted, you will get the same results without condition

  and     col_a (+) = 1
  

  If you want to understand what makes this condition, use some examples of data where this condition would fail.
  For example, add this line to the table a:

  insert into a values (2);
  

  and run your original query (including the)

  and     col_a (+) = 1
  

  ). what output you get? You get the same exact results you did without the new line in a.

  In other words, the b line where col_b = 2 has still no matches in the table has.
  True, there is now a line of table where col_a = 2, but which does not constitute a game. A match is defined by two conditions:

  where   col_a (+) = col_b
  and     col_a (+) = 1
  

  in order to have a pair of lines which meets only the first condition

  where   col_a (+) = col_b
  

  is not enough.

• Outer join with lpad

  Hello
  
  We have this basic question that works very well:
  Select pp.upc, a.barcode from the ALBUM, PHYSICALPRODUCT RP where lpad (pp.upc, 13, '0') = lpad (a.barcode, 13, '0')
  
  But we wanted to do an outer join. If we remove the lpad, it works. But if we as below, we have ORA-00936, lack of expression:
  
  Select pp.upc, a.barcode from the ALBUM, PHYSICALPRODUCT RP where lpad (pp.upc, 13, '0') (+) = lpad (a.barcode, 13, '0')
  
  If anyone can help us in this regard.
  
  Thks.

  Outer join can be used outside the functions of text
  It will be like

  select pp.upc, a.barcode from ALBUM a, PHYSICALPRODUCT pp where lpad(pp.upc (+), 13, '0')  = lpad(a.barcode, 13, '0')
  

• outer join on query with the GOLD clause

  Hi all, I have a problem outerjoining a clause or a request with
  Here's my data
  

  WITH table1 AS
  (
   SELECT  'test' txt1, 'pak' txt2, 'ced' txt3, 'su' txt4 FROM dual UNION ALL
   SELECT  null txt1, 'pak' txt2, 'ced2' txt3, 'su2' txt4 FROM dual UNION ALL
   SELECT  null txt1, NULL txt2, 'ced3' txt3, 'su3' txt4 FROM dual UNION ALL
   SELECT  null txt1, NULL txt2, null txt3, 'su3' txt4 FROM dual UNION ALL
   SELECT  'text5' txt1, NULL txt2, null txt3, 'su3' txt4 FROM dual UNION ALL
   SELECT  null txt1, NULL txt2, null txt3, null txt4 FROM dual 
  
  
  )
  ,table2 AS
  (
   SELECT 111 pid, 'test' txt1, 'pak4' txt2, 'ced' txt3, 'su' txt4 FROM dual UNION ALL
   SELECT 222 pid, 'test1' txt1, 'pak' txt2, 'ced2' txt3, 'su2' txt4 FROM dual UNION ALL
   SELECT 333 pid, 'test2' txt1, 'pak3' txt2, 'ced3' txt3, 'su4' txt4 FROM dual UNION ALL
    SELECT 444 pid, 'test2' txt1, 'pak3' txt2, 'ced4' txt3, 'su3' txt4 FROM dual 
  
  )
  SELECT b.pid, a.*
  from table1 a, table2 b
  WHERE (a.txt1 = b.txt1 OR
         a.txt1 IS NULL AND a.txt2=b.txt2 OR
         Nvl(a.txt2, a.txt1) IS NULL AND a.txt3 = b.txt3 OR
         Nvl(a.txt2, a.txt1) IS NULL  AND a.txt3 IS NULL AND a.txt4 = b.txt4
         ) 

  as you can see I am attaching the table 1 and table 2. I join with txt1, if txt1 is null, then join by txt2, if null, then reach by txt3 and so on.
  the above produces this output code
  

  PID     TXT1     TXT2     TXT3     TXT4
  ===     ====     ===   ==== ====
  111     test     pak      ced     su
  222             pak      ced2     su2
  333                     ced3     su3
  444                          su3

  This output is partially correct. only 4 lines were in display and two has been left out

   SELECT  'text5' txt1, NULL txt2, null txt3, 'su3' txt4 FROM dual UNION ALL
   SELECT  null txt1, NULL txt2, null txt3, null txt4 FROM dual 

  I tried to use the outer join, but oracle will complain that I can't use outerjoin with the GOLD clause.
  
  can someone modify my query to display the output below?
  

  PID     TXT1     TXT2     TXT3     TXT4
  ===    ====      ===   ====  ====
  111     test     pak      ced     su
  222             pak      ced2     su2
  333                     ced3     su3
  444                          su3
  NULL  NULL   NULL    NULL   NULL
       test5 

  Not sure you can do it with the style of the Oracle of outer joins, but open ANSI style joins is simple, in fact exactly as you had it.

  SQL> set null null;
  SQL> WITH table1 AS (
    2   SELECT  'test' txt1, 'pak' txt2, 'ced' txt3, 'su' txt4 FROM dual UNION ALL
    3   SELECT  null txt1, 'pak' txt2, 'ced2' txt3, 'su2' txt4 FROM dual UNION ALL
    4   SELECT  null txt1, NULL txt2, 'ced3' txt3, 'su3' txt4 FROM dual UNION ALL
    5   SELECT  null txt1, NULL txt2, null txt3, 'su3' txt4 FROM dual UNION ALL
    6   SELECT  'text5' txt1, NULL txt2, null txt3, 'su3' txt4 FROM dual UNION ALL
    7   SELECT  null txt1, NULL txt2, null txt3, null txt4 FROM dual),
    8  table2 AS (
    9   SELECT 111 pid, 'test' txt1, 'pak4' txt2, 'ced' txt3, 'su' txt4 FROM dual UNION ALL
   10   SELECT 222 pid, 'test1' txt1, 'pak' txt2, 'ced2' txt3, 'su2' txt4 FROM dual UNION ALL
   11   SELECT 333 pid, 'test2' txt1, 'pak3' txt2, 'ced3' txt3, 'su4' txt4 FROM dual UNION ALL
   12   SELECT 444 pid, 'test2' txt1, 'pak3' txt2, 'ced4' txt3, 'su3' txt4 FROM dual)
   13  SELECT b.pid, a.*
   14  from table1 a
   15     LEFT JOIN table2 b
   16        ON (a.txt1 = b.txt1 OR
   17            a.txt1 IS NULL AND a.txt2=b.txt2 OR
   18            Nvl(a.txt2, a.txt1) IS NULL AND a.txt3 = b.txt3 OR
   19            Nvl(a.txt2, a.txt1) IS NULL  AND a.txt3 IS NULL AND a.txt4 = b.txt4);
  
         PID TXT1   TXT2   TXT3   TXT4
  ---------- ------ ------ ------ ------
         111 test   pak    ced    su
         222 null   pak    ced2   su2
         333 null   null   ced3   su3
         444 null   null   null   su3
  null       text5  null   null   su3
  null       null   null   null   null
  

  John

• BAD RESULTS WITH OUTER JOINS AND TABLES WITH A CHECK CONSTRAINT

  HII All,
  Could any such a me when we encounter this bug? Please help me with a simple example so that I can search for them in my PB.
  
  
  Bug:-8447623
  
  Bug / / Desc: BAD RESULTS WITH OUTER JOINS AND TABLES WITH a CHECK CONSTRAINT
  
  
  I ran the outer joins with check queries constraint 11G 11.1.0.7.0 and 10 g 2, but the result is the same. Need to know the scenario where I will face this bug of your experts and people who have already experienced this bug.
  
  
  Version: -.
  

  SQL> select * from v$version;
  
  BANNER
  --------------------------------------------------------------------------------
  Oracle Database 11g Enterprise Edition Release 11.1.0.7.0 - 64bit Production
  PL/SQL Release 11.1.0.7.0 - Production
  CORE    11.1.0.7.0      Production
  TNS for Solaris: Version 11.1.0.7.0 - Production
  NLSRTL Version 11.1.0.7.0 - Production

  Why do you not use the description of the bug test case in Metalink (we obviously can't post it here because it would violate the copyright of Metalink)? Your test case is not a candidate for the elimination of the join, so he did not have the bug.

  Have you really read the description of the bug in Metalink rather than just looking at the title of the bug? The bug itself is quite clear that a query plan that involves the elimination of the join is a necessary condition. The title of bug nothing will never tell the whole story.

  If you try to work through a few tens of thousands of bugs in 11.1.0.7, of which many are not published, trying to determine whether your application would be affected by the bug? Wouldn't be order of magnitude easier to upgrade the application to 11.1.0.7 in a test environment and test the application to see what, if anything, breaks? Understand that the vast majority of the problems that people experience during an upgrade are not the result of bugs - they are the result of changes in behaviour documented as changes in query plans. And among those who encounter bugs, a relatively large fraction of the new variety. Even if you have completed the Herculean task of verifying each bug on your code base, which would not significantly easier upgrade. In addition, at the time wherever you actually performed this analysis, Oracle reportedly released 3 or 4 new versions.

  And at this stage would be unwise to consider an upgrade to 11.2?

  Justin

• [8i] need help with full outer join combined with a cross join...

  I can't understand how to combine a full outer join with a different type of join... is it possible?
  
  Here are some create table and insert for examples of database:

  CREATE TABLE     my_tab1
  (     record_id     NUMBER     NOT NULL     
  ,     workstation     VARCHAR2(4)
  ,     my_value     NUMBER
       CONSTRAINT my_tab1_pk PRIMARY KEY (record_id)
  );
  
  INSERT INTO     my_tab1
  VALUES(1,'ABCD',10);
  INSERT INTO     my_tab1
  VALUES(2,'ABCD',15);
  INSERT INTO     my_tab1
  VALUES(3,'ABCD',5);
  INSERT INTO     my_tab1
  VALUES(4,'A123',5);
  INSERT INTO     my_tab1
  VALUES(5,'A123',10);
  INSERT INTO     my_tab1
  VALUES(6,'A123',20);
  INSERT INTO     my_tab1
  VALUES(7,'????',5);
  
  
  CREATE TABLE     my_tab2
  (     workstation     VARCHAR2(4)
  ,     wkstn_name     VARCHAR2(20)
       CONSTRAINT my_tab2_pk PRIMARY KEY (workstation)
  );
  
  INSERT INTO     my_tab2
  VALUES('ABCD','WKSTN 1');
  INSERT INTO     my_tab2
  VALUES('A123','WKSTN 2');
  INSERT INTO     my_tab2
  VALUES('B456','WKSTN 3');
  
  CREATE TABLE     my_tab3
  (     my_nbr1     NUMBER
  ,     my_nbr2     NUMBER
  );
  
  INSERT INTO     my_tab3
  VALUES(1,2);
  INSERT INTO     my_tab3
  VALUES(2,3);
  INSERT INTO     my_tab3
  VALUES(3,4);

  And, the results that I want to get:

  workstation     sum(my_value)     wkstn_name     my_nbr1     my_nbr2
  ---------------------------------------------------------------
  ABCD          30          WKSTN 1          1     2
  ABCD          30          WKSTN 1          2     3
  ABCD          30          WKSTN 1          3     4
  A123          35          WKSTN 2          1     2
  A123          35          WKSTN 2          2     3
  A123          35          WKSTN 2          3     4
  B456          0          WKSTN 3          1     2
  B456          0          WKSTN 3          2     3
  B456          0          WKSTN 3          3     4
  ????          5          NULL          1     2
  ????          5          NULL          2     3
  ????          5          NULL          3     4

  I tried a number of different things, google my problem and no luck yet...

  SELECT     t1.workstation
  ,     SUM(t1.my_value)
  ,     t2.wkstn_name
  ,     t3.my_nbr1
  ,     t3.my_nbr2
  FROM     my_tab1 t1
  ,     my_tab2 t2
  ,     my_tab3 t3
  ...

  So, what I want, it's a full outer join of t1 and t2 on workstation and a cross join of one with the t3. I wonder if I can't find examples of it online because it is not possible...
  
  Note: I'm stuck dealing with Oracle 8i
  
  Thank you!!

  Hello

  The query I posted yesterday is a little more complex that it should be.
  My_tab2.workstation is unique, there is no reason to make a separate subquery as mt1. We can join my_tab1 to my_tab2 and get the SUM in a subquery.

  SELECT       foj.workstation
  ,       foj.sum_my_value
  ,       foj.wkstn_name
  ,       mt3.my_nbr1
  ,       mt3.my_nbr2
  FROM       (     -- Begin in-line view foj for full outer join
            SELECT        mt1.workstation
            ,        SUM (mt1.my_value)     AS sum_my_value
            ,        mt2.wkstn_name
            FROM        my_tab1   mt1
            ,        my_tab2   mt2
            WHERE        mt1.workstation     = mt2.workstation (+)
            GROUP BY   mt1.workstation
            ,        mt2.wkstn_name
                          --
                  UNION ALL
                          --
            SELECT      workstation
            ,      0      AS sum_my_value
            ,      wkstn_name
            FROM      my_tab2
            WHERE      workstation     NOT IN (     -- Begin NOT IN sub-query
                                             SELECT      workstation
                                     FROM      my_tab1
                                     WHERE      workstation     IS NOT NULL
                                   )     -- End NOT IN sub-query
         ) foj     -- End in-line view foj for full outer join
  ,       my_tab3  mt3
  ORDER BY  foj.wkstn_name
  ,       foj.workstation
  ,       mt3.my_nbr1
  ,       mt3.my_nbr2
  ;
  

  Thanks for posting the CREATE TABLE and INSERT statements, and very clear expected results!

  user11033437 wrote:
  ... So, what I want, it's a full outer join of t1 and t2 on workstation and a cross join of one with the t3.

  She, exactly!
  The trickiest part is when and how get SUM (my_value). You could address the question of exactly what my_tab3 must be attached to a cross that's exactly what should look like the result set of the full outer join between my_tab1 and my_tab2 to. To do this, take your desired results, remove columns that do not come from the outer join complete and delete duplicate rows. You will get:

  workstation     sum(my_value)     wkstn_name
  -----------     -------------   ----------
  ABCD          30          WKSTN 1
  A123          35          WKSTN 2
  B456          0          WKSTN 3
  ????          5          NULL          
  

  So the heart of the problem is how to get these results of my_tab1 and my_tab2, which is done in the subquery FOJ above.

  I tried to use auto-documenté in my code names. I hope you can understand.
  I could spend hours explaining the different parts of this query more in detail, but I don't know that I would lose some of that time, explain things that you already understand. If you want an explanation of the specific element (s), let me know.

• He had to know the right outer join using...

  He had to know the right outer join using...

  For example: first query left outer join for the emp table: SELECT EMPNO, ENAME, D.DEPTNO FROM EMP E, Department D WHERE the E.DEPTNO = D.DEPTNO)

  Second query left outer join for the Dept table: SELECT EMPNO, ENAME, D.DEPTNO FROM EMP E, Department D WHERE the D.DEPTNO = E.DEPTNO)

  In the example above I just Exchange where condition condition to get an outer join of two table with a left outer join itself. Wat is use right outer join, instead, I can swap the status of table name for the result. Please suggest...

  Hello

  chan001 wrote:

  He had to know the right outer join using...

  For example: first query left outer join for the emp table: SELECT EMPNO, ENAME, D.DEPTNO FROM EMP E, Department D WHERE the E.DEPTNO = D.DEPTNO)

  Second query left outer join for the Dept table: SELECT EMPNO, ENAME, D.DEPTNO FROM EMP E, Department D WHERE the D.DEPTNO = E.DEPTNO)

  In the example above I just Exchange where condition condition to get an outer join of two table with a left outer join itself. Wat is use right outer join, instead, I can swap the status of table name for the result. Please suggest...

  The two examples above use the old syntax outer join of Oracle. (I guess there should be a sign inside the parentheses, e.g.. +)

  ...  WHERE E.DEPTNO = D.DEPTNO (+)

  )

  The LEFT OUTER JOIN and RIGHT OUTER JOIN terms apply only to the ANSI join syntax, e. g.

  .

  .. FROM EMP E

  DEPT LEFT OUTER JOIN D ON E.DEPTNO = D.DEPTNO

  As Blushadow said above, there's no real point in having LEFT OUTER JOIN and RIGHT OUTER JOIN;  What you can do with one (or a combination of both) can be done with the other.  Most people use LEFT OUTER JOIN systematically and never use RIGHT OUTER JOIN.

  There are situations where using a combination of the two would mean a little less striking, but only a little less and this kind of situation is very common, and one may wonder if the somewhat shorter code is more specific.  I suggest that forget you RIGHT OUTER JOIN.

• Problem with Oracle performance

  Hi all
  
  I use Oracle database 11 g 2.
  
  I am facing a problem with one of my queries.
  
  
  Scenario is:
  I associate myself with 3 tables.
  The engine is saa_hist.saa_rep_product_activity (70 million lines).
  The second table is scmsa_hist.scmsa_sub_feature_trueup (900 million lines).
  The third table is saa_hist.saa_rep_subscriber_activity (35 million lines).
  
  The output of the query will be only about 120K.
  
  I'm majoriy of the columns in the first table. On the other hand, we get 2 columns and in table 3, we get a single column.
  
  When I join all 3 tables as it appears in the query runs for 6 hours. (The
  
  SELECT / * + PARALLEL_INDEX(R,SAA_REP_PRODUCT_ACTIVITY_BI,12)
  PARALLEL(SUB,24)
  PARALLEL_INDEX(RR,SAA_REP_SUBSCRIBER_ACTIVITY_I1,8)
  
  */
  R.*, sub.mrc, sub. RATEPLANCODE, rr.measure_group_id
  Saa_hist.saa_rep_product_activity r
  JOIN scmsa_hist.scmsa_sub_feature_trueup sub
  ON sub.job_log_id = r.job_log_id
  AND sub.service_number = r.msisdn
  AND sub.ban = r.account_number
  AND sub.soc_code = r.product_id_code
  LEFT OUTER JOIN saa_hist.saa_rep_subscriber_activity rr
  ON rr.quantity = 1
  AND rr.service_universal_id = r.service_universal_id
  AND rr.msisdn = r.msisdn
  AND rr.account_number = r.account_number
  WHERE r.billing_subsystem_id = 1
  AND 1 = 1
  AND r.batch_key BETWEEN 674 AND 675;
  
  The explain command plan looks good for this one and the cost is also okay(222K). I don't see any S-> data flow in the Section of IN-OUT of the explain plan P.
  
  -----------------------------------------------------------------------------------------------------------------
  | ID | Operation | Name | TQ | IN-OUT | PQ Distrib.
  -----------------------------------------------------------------------------------------------------------------
  | 0 | SELECT STATEMENT |
  | 1. COORDINATOR PX |
  | 2. PX SEND QC (RANDOM). : TQ10001 | Q1, 01 | P > S | QC (RAND) |
  | 3. NESTED EXTERNAL LOOPS | Q1, 01 | SVCP |
  | 4. HASH JOIN | Q1, 01 | SVCP |
  | 5. JOIN FILTER PART CREATE | : BF0000 | Q1, 01 | SVCP |
  | RECEIVE 8 2 PX | Q1, 01 | SVCP |
  | 7. PX SEND LOCAL BROADCAST | : TQ10000 | Q1 00 | P > P | BCST LOCAL |
  | 8. PX HASH PARTITION ALL | Q1 00 | ISSUE |
  | 9. TABLE ACCESS BY LOCAL INDEX ROWID | SAA_REP_PRODUCT_ACTIVITY | Q1 00 | SVCP |
  | 10. CONVERSION OF BITMAP IN ROWID | Q1 00 | SVCP |
  | 11. BITMAP INDEX RANGE SCAN | SAA_REP_PRODUCT_ACTIVITY_BI | Q1 00 | SVCP |
  | 12. FILTER-HASH PARTITION PX JOIN | Q1, 01 | ISSUE |
  | 13. TABLE ACCESS FULL | SCMSA_SUB_FEATURE_TRUEUP | Q1, 01 | SVCP |
  | 14. RANGE OF PARTITION ALL THE | Q1, 01 | SVCP |
  | 15. HASH PARTITION ALL | Q1, 01 | SVCP |
  | 16. TABLE ACCESS BY LOCAL INDEX ROWID | SAA_REP_SUBSCRIBER_ACTIVITY | Q1, 01 | SVCP |
  | 17. INDEX RANGE SCAN | SAA_REP_SUBSCRIBER_ACTIVITY_I1 | Q1, 01 | SVCP |
  -----------------------------------------------------------------------------------------------------------------
  
  But when I joined the first two tables and run as shown below, the request is get executed in 13 Minutes.
  
  SELECT / * + PARALLEL_INDEX(R,SAA_REP_PRODUCT_ACTIVITY_BI,12)
  PARALLEL(SUB,24)
  */
  R.*, sub.mrc, sub. RATEPLANCODE, rr.measure_group_id
  Saa_hist.saa_rep_product_activity r
  JOIN scmsa_hist.scmsa_sub_feature_trueup sub
  ON sub.job_log_id = r.job_log_id
  AND sub.service_number = r.msisdn
  AND sub.ban = r.account_number
  AND sub.soc_code = r.product_id_code;
  
  Therefore, I created a temporary table with this data and made a left outer join with the other 3 table. It runs a few seconds and I can say all the data I am able to take less than 15 minutes to 6 hours.
  
  So my question is that why Oracle is a strange behavior.
  
  I tried all approaches to avoid this kind of behavior methods but unable to produce the same thing as how I did in the second approach.
  
  For example, I joined the first two tables and use NO_MERGE hint with the third table and always NO_MERGE, PUSH_PRED's not going anywhere. I tried Inline view and it doesn't give me any improvement in performance.
  I'm tempted to do it using WITH Clause with two tables and then joined the data set with the third table. But still does not.
  I tried hint ORDINATE and yet so big difference.
  I even tried to force some clues which are presentm, but the 50Million cost ranges from 200K.
  I think I'm exhausted all the possibilities.
  
  I'd appreciate any Oracle guru can help me what is the issue.
  
  Thank you
  NKM

  You could start trying to impose (with extensive predicates) what you discover or believe work better.
  In this case I usually found the ansi syntax not useful and often the optimizer ignore my intentions:(mais il n'y a pas de mal essayer)

  select x.*,
         rr.measure_group_id
    from (select r.*
                 sub.mrc,
                 sub.rateplancode,
            from (select *
                    from saa_hist.saa_rep_product_activity
                   where billing_subsystem_id = 1
                     and batch_key between 674 and 675
                 ) r,
                 scmsa_hist.scmsa_sub_feature_trueup sub
           where sub.job_log_id = r.job_log_id
             and sub.service_number = r.msisdn
             and sub.ban = r.account_number
             and sub.soc_code = r.product_id_code
         ) x,
         (select service_universal_id,
                 account_number,
                 msisdn,
                 measure_group_id
            from saa_hist.saa_rep_subscriber_activity
           where quantity = 1
         ) rr
   where x.service_universal_id = rr.service_universal_id(+)
     and x.msisdn = rr.msisdn(+)
     and x.account_number = rr.account_number(+)
  

  Concerning

  Etbin