Outer join complete Multi Table

What is the syntax to make a full outer join on three tables. All the examples I can find use using the clause, but in my case, the column names are not the same in the three tables

Hello

Use IT instead of USE.

Assuming that all the 3 tables will have a common value in a column (whatever it is called):

...
FROM           table_x  x
FULL OUTER JOIN      table_y  y     ON     y.idy     = x.idx
FULL OUTER JOIN      table_z  z     ON     z.idz     = COALESCE (x.idx, y.idy)

When comes the time to join table_z, you won't know which of the previous tables match, so you must use COALESCE to try them all.

If you need to join several tables in this way, the expression of COALESCE for the nth table will have (n - 1) arguments.

Tags: Database

Similar Questions

  • 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

  • What is the Assembly of 4 or 5 tables FULL OUTER JOIN logic?

    the query is as below:

    as you can see, I need FULL OUTER JOIN, these 5 tables and get a column from each table that is based on three common columns (ORG_iD, CUST_ID, CURRENT_DT).

    And I wonder what is the logic of the 5 FULL OUTER JOIN table?
    It will return if there is a folder that exists in the single table COL1? or table, COL2 and COL3 and so on. Basically any combination of these 5 tables.

        SELECT (CASE WHEN COL1.ORG_ID IS NOT NULL THEN COL1.ORG_ID
                 WHEN COL2.ORG_ID IS NOT NULL THEN COL2.ORG_ID
                 WHEN COL3.ORG_ID IS NOT NULL THEN COL3.ORG_ID
                 WHEN COL4.ORG_ID IS NOT NULL THEN COL4.ORG_ID
                 ELSE COL5.ORG_ID 
            END) ORG_ID,
            (CASE WHEN COL1.CUST_ID IS NOT NULL THEN COL1.CUST_ID
                 WHEN COL2.CUST_ID IS NOT NULL THEN COL2.CUST_ID
                 WHEN COL3.CUST_ID IS NOT NULL THEN COL3.CUST_ID
                 WHEN COL4.CUST_ID IS NOT NULL THEN COL4.CUST_ID
                 ELSE COL5.CUST_ID
            END) CUST_ID, 
            (CASE WHEN COL1.CURRENT_DT IS NOT NULL THEN COL1.CURRENT_DT
                 WHEN COL2.CURRENT_DT IS NOT NULL THEN COL2.CURRENT_DT
                 WHEN COL3.CURRENT_DT IS NOT NULL THEN COL3.CURRENT_DT
                 WHEN COL4.CURRENT_DT IS NOT NULL THEN COL4.CURRENT_DT
                 ELSE COL5.CURRENT_DT 
            END) CURRENT_DT, 
          'VENDORS' as ITEM_NAME,
          V_AGE_IND as ACCT_AGE_IND,  
          COL1.AMT_ZSHT,  
          COL2.AMT_YFYS,  
          COL3.AMT_YFYS, 
          COL4.AMT_YFYS,  
          COL5.AMT_BZL,  
          'NAV',
          sysdate
    FROM COL1  
              FULL OUTER JOIN COL2 ON (COL1.ORG_ID=COL2.ORG_ID AND COL1.CURRENT_DT=COL2.CURRENT_DT AND COL1.CUST_ID=COL2.CUST_ID)  
              FULL OUTER JOIN COL3 ON (COL2.ORG_ID=COL3.ORG_ID AND COL2.CURRENT_DT=COL3.CURRENT_DT AND COL2.CUST_ID=COL3.CUST_ID)  
              FULL OUTER JOIN COL4 ON (COL3.ORG_ID=COL4.ORG_ID AND COL3.CURRENT_DT=COL4.CURRENT_DT AND COL3.CUST_ID=COL4.CUST_ID)  
              FULL OUTER JOIN COL5 ON (COL4.ORG_ID=COL5.ORG_ID AND COL4.CURRENT_DT=COL5.CURRENT_DT AND COL4.CUST_ID=COL5.CUST_ID) 
    Any ideas?

    Thank you

    Hello

    When you perform a JOIN EXTERNAL COMPLETE multi - build that way, whenever you add a new table, you can just join the previous table because this table could be all NULL values for a given row. You can reach each new table to the first table, either, for the same reason. You must enclose each new table with all previous tables, like this:

    FULL OUTER JOIN COL2 ON  COL2.ORG_ID     =           COL1.ORG_ID
               AND COL2.CURRENT_DT =           COL1.CURRENT_DT
               AND COL2.CUST_ID    =           COL1.CUST_ID
FULL OUTER JOIN COL3 ON  COL3.ORG_ID     = COLAESCE (COL1.ORG_ID,     COL2.ORG_ID)
               AND COL3.CURRENT_DT = COALESCE (COL1.CURRENT_DT, COL2.CURRENT_DT)
               AND COL3.CUST_ID    = COALESCE (COL1.CUST_ID,    COL2.CUST_ID)
FULL OUTER JOIN COL3 ON  COL4.ORG_ID     = COLAESCE (COL1.ORG_ID,     COL2.ORG_ID,     COL3.ORG_ID)
               AND COL4.CURRENT_DT = COALESCE (COL1.CURRENT_DT, COL2.CURRENT_DT, COL3.CURRENT_DT)
               AND COL4.CUST_ID    = COALESCE (COL1.CUST_ID,    COL2.CUST_ID,    COL3.CUST_ID)
FULL OUTER JOIN COL3 ON  COL5.ORG_ID     = COLAESCE (COL1.ORG_ID,     COL2.ORG_ID,     COL3.ORG_ID,     COL4.ORG_ID)
               AND COL5.CURRENT_DT = COALESCE (COL1.CURRENT_DT, COL2.CURRENT_DT, COL3.CURRENT_DT, COL4.CURRENT_DT)
               AND COL5.CUST_ID    = COALESCE (COL1.CUST_ID,    COL2.CUST_ID,    COL3.CUST_ID,    COL4.CUST_ID)

    I hope that answers your question.
    If not, post a small example data (CREATE TABLE and only relevant columns, INSERT statements), and the results you want from this data.
    Explain how you get these results from these data.
    Always tell what version of Oracle you are using.

  • Join between the tables with Order By

    Hello

    Sorry if my question has already been postponed. I Googled, but I can't find any solution.

    This is the version of my DB: database Oracle 10 g Enterprise Edition Release 10.2.0.3.0.

    I created 3 tables. They contain up to 4000 lines.

    When I join without the order of closed I got a very quick response (less than 1 second), when I simply add Order By it gets more than 30 seconds!

    The order is on a Date field from the first table.

    Here's my query:

    Select *.

    tableOne tone

    two tableTwo full outer join on tone.message_id = ttwo.message_id

    outer join complete tableThree on two.field_id = tthree.field_id

    tone.ts desc order;

    I have some difficulties to export the plan to explain it...

    Should I create any clue? where? on the orders of field? on the join field?

    Any tips? Thank you.

    NO, it would not be

    Oracle does not work like that.

    ORDER BY is always, really always done LAST.

    ALWAYS.

    The second statement is a JOIN EXTERNAL LEFT and not a FULL OUTER JOIN, so the results will be different.

    Why "assume you' yours 'assumptions' are always right?

    They are not. You were probably in your dipers when I was already performance in Oracle 6.0.33

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

    Sybrand Bakker

    Senior Oracle DBA

  • Issue of OUTER JOIN

    So, for the following data:
    CREATE TABLE MOVIE
  (TK         NUMBER(10,0) primary key,
   TITLE      VARCHAR2(40),
   CODE_GENRE VARCHAR2(2));
/
CREATE TABLE ACTOR
  (TK NUMBER(10,0) primary key, 
   NAME VARCHAR2(40));
/
CREATE TABLE MOVIE_ACTOR_XREF
  (TK_MOVIE NUMBER(10,0) references movie (tk),
   TK_ACTOR NUMBER(10,0) references actor (tk));
/
INSERT INTO MOVIE (TK,TITLE,CODE_GENRE) VALUES (1,'MAN ON FIRE','D');
INSERT INTO MOVIE (TK,TITLE,CODE_GENRE) VALUES (2,'THE MATRIX','SF');
INSERT INTO MOVIE (TK,TITLE,CODE_GENRE) VALUES (3,'PEE WEE''S BIG ADVENTURE','C');
INSERT INTO MOVIE (TK,TITLE,CODE_GENRE) VALUES (4,'NACHO LIBRE','C');
INSERT INTO MOVIE (TK,TITLE,CODE_GENRE) VALUES (5,'UP','CH');
INSERT INTO MOVIE (TK,TITLE,CODE_GENRE) VALUES (6,'HELLRAISER','H');
INSERT INTO MOVIE (TK,TITLE,CODE_GENRE) VALUES (7,'BILL AND TED''S EXCELLENT ADVENTURE','C');
INSERT INTO MOVIE (TK,TITLE,CODE_GENRE) VALUES (8,'MEMENTO','M');
INSERT INTO MOVIE (TK,TITLE,CODE_GENRE) VALUES (9,'GHOST BUSTERS','C');
INSERT INTO MOVIE (TK,TITLE,CODE_GENRE) VALUES (10,'MUMFORD','C');

INSERT INTO ACTOR (TK,NAME) VALUES (1,'DENZEL WASHINGTON');
INSERT INTO ACTOR (TK,NAME) VALUES (2,'DAKOTA FANNING');
INSERT INTO ACTOR (TK,NAME) VALUES (3,'KEANU REAVES');
INSERT INTO ACTOR (TK,NAME) VALUES (4,'PEE WEE HERMAN');
INSERT INTO ACTOR (TK,NAME) VALUES (5,'JACK BLACK');
INSERT INTO ACTOR (TK,NAME) VALUES (6,'ED ASNER');
INSERT INTO ACTOR (TK,NAME) VALUES (7,'PINHEAD');
INSERT INTO ACTOR (TK,NAME) VALUES (8,'GUY PIERCE');
INSERT INTO ACTOR (TK,NAME) VALUES (9,'BILL MURRAY');
INSERT INTO ACTOR (TK,NAME) VALUES (10,'DAN AKROYD');
INSERT INTO ACTOR (TK,NAME) VALUES (11,'JASON LEE');

INSERT INTO MOVIE_ACTOR_XREF (TK_MOVIE,TK_ACTOR) VALUES (1,1);
INSERT INTO MOVIE_ACTOR_XREF (TK_MOVIE,TK_ACTOR) VALUES (1,2);
INSERT INTO MOVIE_ACTOR_XREF (TK_MOVIE,TK_ACTOR) VALUES (2,3);
INSERT INTO MOVIE_ACTOR_XREF (TK_MOVIE,TK_ACTOR) VALUES (3,4);
INSERT INTO MOVIE_ACTOR_XREF (TK_MOVIE,TK_ACTOR) VALUES (4,5);
INSERT INTO MOVIE_ACTOR_XREF (TK_MOVIE,TK_ACTOR) VALUES (5,6);
INSERT INTO MOVIE_ACTOR_XREF (TK_MOVIE,TK_ACTOR) VALUES (6,7);
INSERT INTO MOVIE_ACTOR_XREF (TK_MOVIE,TK_ACTOR) VALUES (7,3);
INSERT INTO MOVIE_ACTOR_XREF (TK_MOVIE,TK_ACTOR) VALUES (8,8);
INSERT INTO MOVIE_ACTOR_XREF (TK_MOVIE,TK_ACTOR) VALUES (9,9);
INSERT INTO MOVIE_ACTOR_XREF (TK_MOVIE,TK_ACTOR) VALUES (9,10);
INSERT INTO MOVIE_ACTOR_XREF (TK_MOVIE,TK_ACTOR) VALUES (10,11);
    Why the following 2 queries do not produce the same results? The FILM table is the table of LEFTermost in outer joins, so I thought I could place criteria against it in the FROM clause and ending up with the same result:
    select m.title, m.code_genre, a.name
from movie m
     left outer 
       join movie_actor_xref x on (m.tk = x.tk_movie
                              and m.code_genre = 'C')
     left outer
       join actor a on (x.tk_actor = a.tk);
       
select m.title, m.code_genre, a.name
from movie m
     left outer 
       join movie_actor_xref x on (m.tk = x.tk_movie)
     left outer
       join actor a on (x.tk_actor = a.tk)
where m.code_genre = 'C';
    Thank you
    -= Chuck

    Hi, Chuck,

    chuckers wrote:
    So, for the following data:

    CREATE TABLE MOVIE
(TK         NUMBER(10,0) primary key,
TITLE      VARCHAR2(40),
CODE_GENRE VARCHAR2(2));
/
CREATE TABLE ACTOR
(TK NUMBER(10,0) primary key,
NAME VARCHAR2(40));
/
CREATE TABLE MOVIE_ACTOR_XREF
(TK_MOVIE NUMBER(10,0) references movie (tk),
TK_ACTOR NUMBER(10,0) references actor (tk));
/
INSERT INTO MOVIE (TK,TITLE,CODE_GENRE) VALUES (1,'MAN ON FIRE','D');
INSERT INTO MOVIE (TK,TITLE,CODE_GENRE) VALUES (2,'THE MATRIX','SF');
INSERT INTO MOVIE (TK,TITLE,CODE_GENRE) VALUES (3,'PEE WEE''S BIG ADVENTURE','C');
INSERT INTO MOVIE (TK,TITLE,CODE_GENRE) VALUES (4,'NACHO LIBRE','C');
INSERT INTO MOVIE (TK,TITLE,CODE_GENRE) VALUES (5,'UP','CH');
INSERT INTO MOVIE (TK,TITLE,CODE_GENRE) VALUES (6,'HELLRAISER','H');
INSERT INTO MOVIE (TK,TITLE,CODE_GENRE) VALUES (7,'BILL AND TED''S EXCELLENT ADVENTURE','C');
INSERT INTO MOVIE (TK,TITLE,CODE_GENRE) VALUES (8,'MEMENTO','M');
INSERT INTO MOVIE (TK,TITLE,CODE_GENRE) VALUES (9,'GHOST BUSTERS','C');
INSERT INTO MOVIE (TK,TITLE,CODE_GENRE) VALUES (10,'MUMFORD','C');

INSERT INTO ACTOR (TK,NAME) VALUES (1,'DENZEL WASHINGTON');
INSERT INTO ACTOR (TK,NAME) VALUES (2,'DAKOTA FANNING');
INSERT INTO ACTOR (TK,NAME) VALUES (3,'KEANU REAVES');
INSERT INTO ACTOR (TK,NAME) VALUES (4,'PEE WEE HERMAN');
INSERT INTO ACTOR (TK,NAME) VALUES (5,'JACK BLACK');
INSERT INTO ACTOR (TK,NAME) VALUES (6,'ED ASNER');
INSERT INTO ACTOR (TK,NAME) VALUES (7,'PINHEAD');
INSERT INTO ACTOR (TK,NAME) VALUES (8,'GUY PIERCE');
INSERT INTO ACTOR (TK,NAME) VALUES (9,'BILL MURRAY');
INSERT INTO ACTOR (TK,NAME) VALUES (10,'DAN AKROYD');
INSERT INTO ACTOR (TK,NAME) VALUES (11,'JASON LEE');

INSERT INTO MOVIE_ACTOR_XREF (TK_MOVIE,TK_ACTOR) VALUES (1,1);
INSERT INTO MOVIE_ACTOR_XREF (TK_MOVIE,TK_ACTOR) VALUES (1,2);
INSERT INTO MOVIE_ACTOR_XREF (TK_MOVIE,TK_ACTOR) VALUES (2,3);
INSERT INTO MOVIE_ACTOR_XREF (TK_MOVIE,TK_ACTOR) VALUES (3,4);
INSERT INTO MOVIE_ACTOR_XREF (TK_MOVIE,TK_ACTOR) VALUES (4,5);
INSERT INTO MOVIE_ACTOR_XREF (TK_MOVIE,TK_ACTOR) VALUES (5,6);
INSERT INTO MOVIE_ACTOR_XREF (TK_MOVIE,TK_ACTOR) VALUES (6,7);
INSERT INTO MOVIE_ACTOR_XREF (TK_MOVIE,TK_ACTOR) VALUES (7,3);
INSERT INTO MOVIE_ACTOR_XREF (TK_MOVIE,TK_ACTOR) VALUES (8,8);
INSERT INTO MOVIE_ACTOR_XREF (TK_MOVIE,TK_ACTOR) VALUES (9,9);
INSERT INTO MOVIE_ACTOR_XREF (TK_MOVIE,TK_ACTOR) VALUES (9,10);
INSERT INTO MOVIE_ACTOR_XREF (TK_MOVIE,TK_ACTOR) VALUES (10,11);

    Thanks for posting the CREATE TABLE and INSERT! It is very useful.

    Why the following 2 queries do not produce the same results? The FILM table is the table of LEFTermost in outer joins, so I thought I could place criteria against it in the FROM clause and ending up with the same result:

    We will call this first request 'Query 1.

    select m.title, m.code_genre, a.name
from movie m
left outer
join movie_actor_xref x on (m.tk = x.tk_movie
and m.code_genre = 'C')
left outer
join actor a on (x.tk_actor = a.tk);

    Left outer joins ensure that each row of the table to move will appear in the output.
    If movie.code_genre = 'C', then he might have a match in movie_actor_xref.
    If movie.code_genre! = 'C' (or NULL) then it will not be considered to match any line in movie_actor_xref, but since it is an outer join, that the rank of the film appears in the result set.

    Let's call the other "Application 2" quuery

    select m.title, m.code_genre, a.name
from movie m
left outer
join movie_actor_xref x on (m.tk = x.tk_movie)
left outer
join actor a on (x.tk_actor = a.tk)
where m.code_genre = 'C';

    When all the joins are finsihed, all lines of film will be in the result set, if it corresponds to what in the other tables.
    Unlike the query 1, 2 the query has a WHERE clause. You can think of the WHERE clause as being applied after all the joins are completed. The WHERE clause in the query 2 will remove all the lines where movie.code_genre! = 'C' (or NULL), so the final results will not include necessarily all movie lines.

    Let's look at a specific example.
    This line appears in the output of query 1:

    TITLE                CO NAME
-------------------- -- --------------------
UP                   CH

    but there is no line of corresoponding in query output 2. Why not? What is the difference between Query1 and Query2? The difference is the condition of wheher "where m.code_genre = 'C'" applies to one of the outer joins (as in the query 1) or the game (as in query 2) set of results.
    Try to run this query, I'll call the query 0 because it contains less query 1 or 2 of the query:

    select m.title, m.code_genre, a.name
from movie m
     left outer
       join movie_actor_xref x on (m.tk = x.tk_movie)
     left outer
       join actor a on (x.tk_actor = a.tk)
;

    Note that the query 0 does not speak code_genrel; It contains only the code that is used by query 1 and query 2. 0 query output includes this line:

    TITLE                CO NAME
-------------------- -- --------------------
UP                   CH ED ASNER

    Query 1 produces a line of title = 'UP' but that line had a NULL value in the name column. This is because the query 1 was a condition of additional join between the film and actor_movie_xref. In query 1, lines of these two tables were regarded as a match only if movie.cide_genre =' it. The line with title = 'UP' a 'CH' code_genre, so it is not considered a match with actor_movie_xref. Since it is an outer join, however, the line stays in the result set, but the columns that would be provided by actor_move_xref are all NULL. 1 query is an outer join to the table of the actor. "Join condition"x.tk_actor = a.tk"fails, because x.tk_actor is null (all x columns have NULL value) on the title line =" upward ", but, again, since it is an outer join, this line is in the result set.

    The difference between the 0 and 2 of the query request is 2 query has a WHERE clause. When the query 2 finishes to make all joins, but before she applies the WHERE clause, its result set is identical to the query result 0 set, including a line with title = 'UP '. Then the WHERE clause is applied, and any line that doesn't have a 'C' for his code_genre is eliminated.

    Published by: Frank Kulash, October 13, 2011 15:44
    Added example 'UP '.

  • using (+) or left outer join

    Dear Expert;

    I've been playing by using the two symbol... and realized that they do the same thing... Is it true...? or am I wrong.

    Thank you.

    Hello

    user13328581 wrote:
    Dear Expert;

    I've been playing by using the two symbol... and realized that they do the same thing... Is it true...? or am I wrong.

    They all have two outer joins. The + rating was the original way to do it in Oracle. LEFT, RIGHT, and FULL OUTER JOIN introduced in Oracle 9, but the old way is still supported.

    There are some situations (such as an outer join complete and outer-join a table to two different tables) that are better with the ANSI syntax (it's LEFT OUTER JOIN). It is possible to get the same results using +, but it must be combersome and/or inefficient workarounds. I suggest that you use always LEFT OUTER JOIN (or FULL OUTER JOIN, or, on occasions RIGHT OUTER JOIN). I think that you will find it easier and less error-prone.

  • [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.

  • Former Outer Join syntax on constants

    Hello

    Can someone please tell me why the below two queries are performing differently. How to join a constant by using the old syntax used.

    Also, how is 3 different query of query 2?

    CREATE TABLE T1 (X VARCHAR2 (10), B INTEGER);

    CREATE TABLE T2 (Y VARCHAR2 (10), B INTEGER);

    INSERT INTO T1 VALUES('XXX',10);
    INSERT INTO T1 VALUES('XXX',10);
    INSERT INTO T1 VALUES('XXX',10);
    INSERT INTO T1 VALUES('YYY',20);
    INSERT INTO T1 VALUES('YYY',20);
    INSERT INTO T1 VALUES('YYY',20);

    INSERT INTO T2 VALUES('AAA',10);
    INSERT INTO T2 VALUES('BBB',20);

    Query 1

    SELECT T1.*, T2.* FROM T1 LEFT JOIN T2

    ON T1. B = T2. B

    AND T1. B = 20

    -OUTPUT

    =========

    BBB YYY 20 20

    BBB YYY 20 20

    BBB YYY 20 20

    NULL NULL 10 XXX

    NULL NULL 10 XXX

    NULL NULL 10 XXX

    Query 2

    SELECT T1.*, T2.* FROM T1, T2

    ON T1. B (+) = T2. B

    AND T1. B ( + ) = 20;

    OUTPUT

    =======

    BBB YYY 20 20

    BBB YYY 20 20

    BBB YYY 20 20

    Request 3

    SELECT T1.*, T2.* FROM T1, T2

    ON T1. B (+) = T2. B

    ET T2. B ( + ) = 20;

    -OUTPUT

    =========

    BBB YYY 20 20

    BBB YYY 20 20

    BBB YYY 20 20

    NULL NULL 10 XXX

    NULL NULL 10 XXX

    NULL NULL 10 XXX

    Thank you and best regards,

    Mathieu

    Hi, nada.

    967250 wrote:

    Hello

    My apologies for errors in queries.

    Please find the three queries

    QUERY1

    SELECT T1.*, T2.*
    T1 LEFT JOIN T2
    ON T1. B = T2. B
    AND T1. B = 20;

    QUERY2

    SELECT T1.*, T2.*
    FROM T1, T2
    WHERE T1. B = T2. B ( + )
    AND T1. B (+) = 20;

    QUERY3
    SELECT T1.*, T2.*
    FROM T1, T2
    WHERE T1. B = T2. B ( + )
    AND T2. B (+) = 20;

    Query 1 and 3 produce the same results. I want to know is what is the difference between Q2 and (T1 or T3)

    Thank you

    Mathieu

    In the query above 1, t1 is the required table (LEFT OUTER JOIN means the table on the LEFT is needed, and the other table is optional).

    Query 3 is the way to do the same in the old notation.

    If you use the old notation of outer join, the + sign is used by reference to the table as an option.  It should always apply for t1 or t2, not sometimes one and sometimes the other, which is what you do in the application 2.  In the State:

    T1. B = T2. B ( + )

    you say that t2 is optional; in the other condition

    T1. B (+) = 20

    you say that t1 is optional.  I don't know why that does not raise an error.  Apparently, it is just to ignore the sign in this last condition.

  • 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.

  • Outer join without outer join

    I'm trying to do is to learn how to do an outer join without specifying the outer join complete, right or left. Why? For most of the reasons for performance and gain a better understanding of the Oracle.
    Thank you

    Charles.

    Another option would be a scalar subquery in the select part, as:

    SQL> select * from t;

        ID DESCR
---------- ----------
         1 T One
         2 T Two
         3 T Three

SQL> select * from t1;

        ID DESCR
---------- ----------
         1 T1 One
         2 T1 Two

SQL> select t.*, t1.descr t1_descr
  2  from t
  3     left join t1
  4        on t.id = t1.id;

        ID DESCR      T1_DESCR
---------- ---------- ----------
         1 T One      T1 One
         2 T Two      T1 Two
         3 T Three

SQL> select t.*, (select t1.descr
  2               from t1
  3               where t.id = t1.id) t1_descr
  4  from t;

        ID DESCR      T1_DESCR
---------- ---------- ----------
         1 T One      T1 One
         2 T Two      T1 Two
         3 T Three

    This can be faster if the correlated T1 column is indexed, and there is a "large" number of records in t which do not correspond with t1.

    John

  • Join with conditions of outer join statement

    Hello
    The following was extracted from the 11 GR 2 document guide performance, which I am not quite understand.
    Can anyone provide some examples to clearly explain this. Thank you

    For join instructions examples with outer join conditions, the table with the outer join operator must come after the other table in the condition in the join order. The optimizer does not consider join orders that violate this rule. Similarly, when a subquery was transformed into an antijoin or a semi-join, arrays of the subquery must come after these tables in the outer query to which they were connected or block in correlation. However, semijoins and antijoins of hash are override this order condition in certain circumstances

  • Problem format (LEFT OUTER JOIN)?

    THE addresses of Mutiple of return as a single record + current addresses

    >
    Hi all

    I have to back student addresses home and dormitory under a single registration.

    He had to go something like this:
    LAST_NAME     FIRST_NAME     ADDY_TYPE   ADDRESS             ZIP
Smith                John             HOME         123 Awesome St     10003
Smith               John               DORM         Oak Quad             10013
    In this format the desired:
    LAST_NAME  FIRST_NAME     ADDY_TYPE  ADDRESS         ZIP        ADDY_TYPE     ADDRESS     ZIP
Smith            John      HOME        123 Awesome St  10003   DORM            Oak Quad     10013
    You also need to get their last addresses by date.
    The database hold records of all students have places moving from dorm to dorm
    and some permanent residence ("HOME") has changed as well.
    To return only a DORM address and only a HOME address
    for each student.

    I'm looking at possibly a function BOX to put on a line/record and RANK BY() or DENSE_RANK to determine the last addresses.

    Hope I'm making some sense and as always very grateful for any help. Thank you.
    >

    The correct code provided by Frank Kulash here:
    WITH    got_rnum     AS
(
     SELECT     last_name
     ,     first_name
     ,     addy_type
     ,     address
     ,     zip
     ,     ROW_NUMBER () OVER ( PARTITION BY  last_name
                               ,                    first_name
                         ,             addy_type
                         ORDER BY        addy_date     DESC     NULLS LAST
                       ) AS rnum
     FROM    table_x
--     WHERE     ...          -- Any filtering goes here
)
SELECT    last_name
,       first_name
,       MIN (CASE WHEN addy_type = 'HOME' THEN address END)     AS home_address
,       MIN (CASE WHEN addy_type = 'HOME' THEN zip     END)     AS home_zip
,       MIN (CASE WHEN addy_type = 'DORM' THEN address END)     AS dorm_address
,       MIN (CASE WHEN addy_type = 'DORM' THEN zip     END)     AS dorm_zip
FROM       got_rnum
WHERE       rnum     = 1
GROUP BY  last_name
,            first_name
;
    I need to add a 'NATION' field, located on another table for the addresses of welcome for foreign students.
    I made a LEFT OUTER JOIN with the table of the NATION and the release came out like this:
    LAST_NAME  FIRST_NAME     ADDY_TYPE  ADDRESS   ZIP      NATION    ADDY_TYPE       ADDRESS     ZIP
Smith            John      HOME        Rue Henry M1V 4F4  CANADA      null              null             null
Smith            John      null        null      null     null        DORM               Oak Quad     10013
    My desired output would be like this:
    LAST_NAME  FIRST_NAME     ADDY_TYPE  ADDRESS   ZIP      NATION    ADDY_TYPE       ADDRESS     ZIP
Smith            John      HOME        Rue Henry M1V 4F4  CANADA    DORM              Oak Quad     10013
    Maybe it's something I'm not right. What is the way I'm joining tables?

    As always very grateful for your contributions. Thank you.

    Give a glance to your group by, you group on the nation, but it has two values (NULL and CANADA). Try this way:

    WITH    got_rnum     AS
(
     SELECT     STUINFO_id
  , STUINFO_last_name
     ,     STUINFO_first_name
     ,     ADDRESSLIST_atyp_code
     ,     ADDRESSLIST_street_line1
  , ADDRESSLIST_street_line2
  , ADDRESSLIST_city
  , ADDRESSLIST_stat_code
     ,     ADDRESSLIST_zip

, ADDRESSLIST_natn_code
, NATION_nation
     ,     ROW_NUMBER () OVER ( PARTITION BY  STUINFO_last_name
                               ,                    STUINFO_first_name
                         ,                         ADDRESSLIST_atyp_code

                         ORDER BY        ADDRESSLIST_from_date     DESC     NULLS LAST
                       ) AS rnum
     FROM STUINFO JOIN CLASSROSTER ON STUINFO_pidm = CLASSROSTER_pidm JOIN ADDRESSLIST ON ADDRESSLIST_pidm = STUINFO_pidm LEFT OUTER JOIN NATION ON ADDRESSLIST_NATN_CODE =NATION_CODE
-- The WHERE part determines if the student is currently enrolled in a class
-- ADDRESSLIST_to_date is the last date the student will be living in that residence
WHERE ADDRESSLIST_atyp_code IN ('PR', 'CA') and  STUINFO_change_ind IS NULL and STUINFO_last_name !='Registrar'
and CLASSROSTER_term_code='200909' and CLASSROSTER_PTRM_CODE IN ('D', 'D1', 'D2') and CLASSROSTER_CAMP_CODE='1'

and (ADDRESSLIST_to_date is NULL OR ADDRESSLIST_TO_DATE > SYSDATE)
)
SELECT    STUINFO_id
,   STUINFO_last_name
,       STUINFO_first_name
,       MIN (CASE WHEN ADDRESSLIST_atyp_code = 'PR' THEN ADDRESSLIST_STREET_LINE1  END)     AS PR_ADDRESSLIST_STREET_LINE1
,       MIN (CASE WHEN ADDRESSLIST_atyp_code = 'PR' THEN ADDRESSLIST_STREET_LINE2  END)     AS PR_ADDRESSLIST_STREET_LINE2
,       MIN (CASE WHEN ADDRESSLIST_atyp_code = 'PR' THEN ADDRESSLIST_city     END)     AS PR_ADDRESSLIST_city
,       MIN (CASE WHEN ADDRESSLIST_atyp_code = 'PR' THEN ADDRESSLIST_stat_code     END)     AS PR_ADDRESSLIST_stat_code
,       MIN (CASE WHEN ADDRESSLIST_atyp_code = 'PR' THEN ADDRESSLIST_zip     END)     AS PR_ADDRESSLIST_zip
,       MIN (CASE WHEN ADDRESSLIST_natn_code IS  NULL THEN ADDRESSLIST_natn_code     END)     AS PR_ADDRESSLIST_natn_code
, MIN(NATION_nation) NATION_nation
,       MIN (CASE WHEN ADDRESSLIST_atyp_code = 'CA' THEN ADDRESSLIST_STREET_LINE1  END)     AS CA_ADDRESSLIST_STREET_LINE1
,       MIN (CASE WHEN ADDRESSLIST_atyp_code = 'CA' THEN ADDRESSLIST_STREET_LINE2  END)     AS CA_ADDRESSLIST_STREET_LINE2
,       MIN (CASE WHEN ADDRESSLIST_atyp_code = 'CA' THEN ADDRESSLIST_city     END)     AS CA_ADDRESSLIST_city
,       MIN (CASE WHEN ADDRESSLIST_atyp_code = 'CA' THEN ADDRESSLIST_stat_code     END)     AS CA_ADDRESSLIST_stat_code
,       MIN (CASE WHEN ADDRESSLIST_atyp_code = 'CA' THEN ADDRESSLIST_zip     END)     AS CA_ADDRESSLIST_zip

FROM       got_rnum
WHERE       rnum     = 1
GROUP BY  STUINFO_last_name
,            STUINFO_first_name
,         STUINFO_id
ORDER BY STUINFO_last_name;

    Max

  • 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

  • 3 left outer join tables

    Hi all

    I have 3 tables A, B, C

    Create table a (varchar2 (100)) of the currency;

    insert into a values (GBp);

    insert into a values (GBP);

    insert into a values (GBX);

    Create table B (varchar2 (100) currency, number);

    insert into B values (GBP, 61.1);

    Create a table (minor_currency varchar2 (100), major_currency varchar2 (100));

    insert into values of C (GBp, GBP);

    insert into values of C (GBX, GBP);

    I need to get the rate table B by linking the A with the currency as a condition of joining.  (left outer join)

    For currencies which are not in table B, table should be attached with C minor currency-based

    and get the major_currency and join with table B

    Ex:

    something like this:

    Select B.rate from A, B, C

    WHERE (A.currency = B.currency or (A.currency = C.minor_currency and B.currency = C.major_currency)

    O/P: for GBp and GBX currency, I need to get the rate as 61.1 in table B, but B currency is GBP. So I need to get the major_currecny for GBp, GBX table C and join with the table B

    Thank you

    Sasi

    Hi all

    Thanks for your time. Its done

  • The table can be outer joined to a single other table

    Hello

    Using oracle 11.2.0.3

    Select *.

    from tablea

    tableb

    tablec

    where tablea.col = tableb.col (+)

    and tablea.col = tablec.col (+)

    throws error table can be attached externally to another table.

    What is the best/recommended way to allow more than one outer join table?

    Thank you

    Hello

    Use ANSI join rating:

    SELECT *.

    FROM tablea a

    LEFT OUTER JOIN tableb b ON a.col = b.col

    LEFT OUTER JOIN tablec c ON a.col = c.col

    ;

    Are you sure the query you posted is really what you're running?  The query you posted shoulddn can't cause the error, because the tableb and tablec are each being only outside attached to 1 other table.  You get the error if the + signs were reversed in the two conditions.

    In any case, I suggest to use the ANSI syntax for all joins, especially outer joins.  You will never get the error "Table can be external attached to a single other table", and which is one of the reasons why it is better.

    I hope that answers your question.

    If not, post a small example data (CREATE TABLE and only relevant columns, INSERT statements) for all of the tables involved and also publish outcomes from these data.

    Explain, using specific examples, how you get these results from these data.

    Always say what version of Oracle you are using (for example, 11.2.0.2.0).

    See the FAQ forum: https://forums.oracle.com/message/9362002

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