Problem with XMLTABLE and LEFT OUTER JOIN

Similar Questions

• Problem with the implementation of Outer JOIN

  Hello Guru,
  
  Here's my scenario. I know that this can be achieved in several ways. But I need this in a single in aid Of JOINS EXTERNALrequest.
  
  I have two tables namely masters and transactions table. I need to join two columns (trx_name, trx_type) of the two tables and the need for all transactions. Part the trickiest is off 2 join columns, one column can be NULL in the main table.
  
  Script TABLES:
  
  create master table (mstr_no number, trx_name varchar2 (100), trx_type varchar2 (100), trx_module varchar2 (100));
  
  insert into masters values (1, 'yyyy', 'simple', 'cont');
  insert into masters values (2, 'bbbb', < NULL >, "cust");
  insert into masters values (3, 'yyyy', 'triple', "cont1");
  
  create table transaction (trx_no number, trx_name varchar2 (100), trx_type varchar2 (100));
  
  Insert in transaction values (1, 'yyyy', 'simple');
  Insert the transaction values (2, 'bbbb', 'double');
  Insert in transaction values (3, 'yyyy', 'triple');
  Insert in transaction values (4, 'cccc', 'purple');
  commit;
  
  I need output like below.
  
  1, 'yyyy', 'simple', 'cont '.
  2, 'bbbb', < NULL >, * "cust."
  3, 'yyyy', 'triple', "cont1".
  4, 'cccc', 'purple', < NULL >
  
  I need to join trx_name & trx_type in the two tables. Even if one of the column is null in the main table (mstr_no = 2), I have need of the corresponding value of trx_module since trx_name made match.
  
  
  My query:
  Select a.*, b.trx_module of transaction a, master b
  where a.trx_name = b.trx_name (+)
  and a.trx_type = NVL (b.trx_type (+), 'NULL')
  order of a.trx_no
  
  Appreciate your help.
  
  
  Concerning
  MN

  Maybe I'm wrong, but I don't think you can do this with a single outer join:

  with
  full_matches as (
  select *
    from master
   where trx_type is not null)
  ,
  partial_matches as (
  select *
    from master
   where trx_type is null)
  select a.*
       , coalesce(b.trx_module, c.trx_module) trx_module
    from transaction a
    left outer join full_matches b
      on (a.trx_name = b.trx_name and a.trx_type = b.trx_type)
    left outer join partial_matches c
      on (a.trx_name = c.trx_name)
   ;
  
  TRX_NO TRX_NAME             TRX_TYPE             TRX_MODULE
  ------ -------------------- -------------------- ------------
       1 aaaa                 single               cont
       3 aaaa                 triple               cont1
       2 bbbb                 double               cust
       4 cccc                 purple
  

  Re-reading my answer, I see that a short explanation might be useful: the query uses two outer joins: first masters lines with the name and type are joined, and then type the lines with missing. Him coalesce work to choose the most relevant information of module.

  Published by: Martin Preiss on June 7, 2013 20:24

• What is the difference between NOT IN and LEFT OUTER JOIN

  Hello
  
  I searched the difference between everywhere. But his powerlessness.
  Please tell me the differences.
  
  Thanks in advance
  KVB

  It's like comparing apples and oranges.

  NOT IN - exclude all lines matching the condition NOT IN (beware of NULL values).

  JOIN EXTERNAL - return of rows from the inner table even if there is no corresponding row in the outer table.

  SQL> -- NOT IN
  SQL> with x as
    2  (select 1 col1 from dual union all
    3   select 2 col1 from dual union all
    4   select 3 col1 from dual)
    5  ,    y as
    6  (select 1 col1 from dual)
    7  select *
    8  from   x
    9  where  col1 not in (1,2);
  
        COL1
  ----------
           3
  
  SQL> -- NOT IN (subquery)
  SQL>
  SQL> with x as
    2  (select 1 col1 from dual union all
    3   select 2 col1 from dual union all
    4   select 3 col1 from dual)
    5  ,    y as
    6  (select 1 col1 from dual)
    7  select *
    8  from   x
    9  where  col1 not in (select col1 from y);
  
        COL1
  ----------
           2
           3
  
  SQL> -- OUTER JOIN
  SQL> with x as
    2  (select 1 col1 from dual union all
    3   select 2 col1 from dual union all
    4   select 3 col1 from dual)
    5  ,    y as
    6  (select 1 col1 from dual)
    7  select *
    8  from   x
    9  left outer join
   10         y
   11  on x.col1 = y.col1;
  
        COL1       COL1
  ---------- ----------
           1          1
           3
           2
  
  SQL> -- Maybe it helps to contrast LOJ with just JOIN?
  SQL> with x as
    2  (select 1 col1 from dual union all
    3   select 2 col1 from dual union all
    4   select 3 col1 from dual)
    5  ,    y as
    6  (select 1 col1 from dual)
    7  select *
    8  from   x
    9  join   y
   10  on x.col1 = y.col1;
  
        COL1       COL1
  ---------- ----------
           1          1
  
  SQL> 
  

• Problems with XMLTABLE and XMLNAMESPACES

  Hello

  I have an XMLTYPE table. When I now want to query the data I get error ORA-19228: XPST0008 - undeclared identifier: local name prefix "sfa" "sfa:Contact"

  XML data:

  <? XML version = "1.0" encoding = "ISO-8859-15? >

  < version ftc: FATCA_OECD = "Chain" xsi: schemaLocation = "urn: oecd:ties:fatca:v1 " http://xmlns.Oracle.com/xdb/documentation/FatcaXML_v1.1.xsd "" xmlns: xsi = " http://www.w3.org/2001/XMLSchema-instance " xmlns:ftc = "urn: oecd:ties:fatca:v1" xmlns:sfa = "urn: oecd:ties:stffatcatypes:v1" > ""

  < ftc: MessageSpec >

  < sfa:TransmittingCountry > LI < / sfa:TransmittingCountry >

  < sfa:ReceivingCountry > U.S. < / sfa:ReceivingCountry >

  < sfa:MessageType > FATCA < / sfa:MessageType >

  < sfa:Contact > Alexander Zelzer < / sfa:Contact >

  < sfa:MessageRefId > 123456 < / sfa:MessageRefId >

  < sfa:CorrMessageRefId / >

  < sfa:ReportingPeriod > 2014 - 12 - 31 < / sfa:ReportingPeriod >

  < sfa:Timestamp > 2015-02-18 T 16: 38:02.000000 < / sfa:Timestamp >

  < / ftc: MessageSpec >

  < / ftc: FATCA_OECD >

  SQL query:

  SELECT msg.contact

  Of itx_ftc_fatca_oecd

  , XMLTABLE (xmlnamespaces ("urn: oecd:ties:fatca:v1' AS"ftc"," urn: oecd:ties:stffatcatypes:v1' AS "sfa"), ' / ftc: FATCA_OECD')

  PASSAGE object_value

  Messagespec COLUMNS XMLTYPE PATH ' / ftc: FATCA_OECD / ftc: MessageSpec') level 1

  , XMLTABLE ('/ ')

  PASSAGE level1.messagespec

  Msg of COLUMNS contact VARCHAR2 (38) PATH "sfa:Contact")

  ORDER BY 1;

  Has anyone an idea what is the problem?

  Thanks in advance.

  Kind regards

  Martin

  The result still contains namespace references while you are passing it in the second section XMLTABLE...

  WITH data
  AS ( SELECT XMLTYPE('
  
    
      LI
      US
      FATCA
      Alexander Zelzer
      123456
      
      2014-12-31
      2015-02-18T16:38:02.000000
    
  ' ) as CONTENT
      FROM dual
    )
  SELECT msg.contact
    FROM data tl
     , XMLTABLE(xmlnamespaces('urn:oecd:ties:fatca:v1' AS "ftc", 'urn:oecd:ties:stffatcatypes:v1' AS "sfa"),
                '/ftc:FATCA_OECD'
                PASSING tl.CONTENT
                COLUMNS
                  messagespec XMLTYPE PATH '/ftc:FATCA_OECD/ftc:MessageSpec'
                ) level1
     , XMLTABLE(xmlnamespaces('urn:oecd:ties:fatca:v1' AS "ftc", 'urn:oecd:ties:stffatcatypes:v1' AS "sfa"),
                '/'
                PASSING level1.messagespec
                COLUMNS
                  contact     VARCHAR2(38) PATH 'sfa:Contact'
               ) msg
  ORDER BY 1
  ;
  
  CONTACT
  --------------------------------------
  Alexander Zelzer
  

• Problem with LEFT OUTER JOIN

  Hello
  
  I am in charge of the migration of a SQL Server 2000 database to Oracle 11 g, under what I also migrate some predefined queries, that my client has. However I can't seem to get the syntax right and it keeps failing. Could you please help me? Thank you.
  
  Query:

  SELECT *,(select r.recsolins from gx.repara r where r.percod=c.percod and r.concod=c.concod and r.rectpo='I' and r.recsts='F' and r.grppercod=10 and r.recnro=(select max(t.recnro) from gx.repara t where t.percod=c.percod and t.concod=c.concod and t.rectpo='I' andt.recsts='F' ) ) as NROID
  FROM gx.CONABO c, gx.abonad a  
  LEFT OUTER JOIN gx.CALLES y ON  y.dptocod=10 and y.ciucod=524 and y.CALCOD=A.AboCalEsq1, 
  LEFT OUTER JOIN gx.CALLES Z ON  z.dptocod=10 and z.ciucod=524 and z.CALCOD=A.AboCalEsq2 
  ,gx.calles x WHERE c.PERCOD in (10,60) and CONSTSHAB in ('C','D','P')
  and a.percod=c.percod and a.abocod=c.abocod and
  x.dptocod=10 and x.ciucod=524
  and x.calcod=abocal
  order by c.percod,c.concod;

  The fields are correct, but I get not found when expected in FROM clause.
  
  Published by: n on June 5, 2012 13:47
  

• 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.

• 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

• 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.

• 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

• I'm having a lot of problems with firefox and cannot figure out how to get help. It all started when I updated to 13. I get all kinds of advertising popups, I can't play a

  I'm having a lot of problems with firefox and cannot figure out how to get help. It all started when I updated to 13. I get all kinds of advertising popups, I can't play a game on FaceBook called Farm Town at all, and I get a popup of AVG on the cookies that I can't get rid of. These issues are causing me to use Chrome quite often, although I like Fox better. I've searched and searched how to get help and find nothing. How can I get personalized technical help? These problems will not occur in Chrome at all. Thank you.

  Do a check with some malware malware, analysis of programs on the Windows computer.
  
  You need to scan with all programs, because each program detects a different malicious program.
  
  Make sure that you update each program to get the latest version of their databases before scanning.

  • http://www.malwarebytes.org/mbam.php - Malwarebytes' Anti-Malware
  • http://www.superantispyware.com/ - SuperAntispyware
  • http://www.microsoft.com/security/scanner/en-us/default.aspx - scan Microsoft Security
  • http://www.microsoft.com/windows/products/winfamily/defender/default.mspx - Windows Defender: Home Page
  • http://www.safer-networking.org/en/index.html - Spybot Search & Destroy

  Alternatively, you can write a check for an infection rootkit TDSSKiller.

  • http://support.Kaspersky.com/viruses/solutions?QID=208280684

  See also:

  • "Spyware on Windows": http://kb.mozillazine.org/Popups_not_blocked

• left and right outer join

  Hi gurus,

  Left outer join:

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

  Select * from A LEFT OUTER JOIN B on A.col = B.col;

  by definition, a left outer join brings the rows that match the join condition and lines not corresponding to table A.

  My question here is "corresponding to B and no matching rows in A" is that nothing, but... SELECT * FROM A;

  can someone pls clarity...

  Thank you.

  Imagine that you had:

  TableA

  COLUMN1 COLUMN2

  'A'                1

  'B'                2

  'C'                3

  TABLEB

  COLUMN1 COLUMN2

  'A'                 'X1'

  'A'                 'X2'

  'B'                 'Y'

  'D'                 'Z'

  SELECT * FROM TABLEA;

  A 1

  B 2

  C 3

  Now, if you want to join (first column is either in table A or B, (deuxieme from tableA, third table B)

  SELECT * FROM TABLEA A JOIN TABLEB B ON (A.COLUMN1 = B.COLUMN1)

  A 1 X 1

  A 1 X 2

  B 2 Y

  SELECT * FROM TABLE LEFT B EXTERNAL ON (A.COLUMN1 = B.COLUMN1) JOIN TABLE

  A 1 X 1

  A 1 X 2

  B 2 Y

  C 3 {null}

  SELECT * FROM TABLE A TABLE RIGHT OUTER JOIN B (A.COLUMN1 = B.COLUMN1)

  A 1 X 1

  A 1 X 2

  B 2 Y

  D {null} Z

  SELECT * FROM TABLE A TABLE FULL OUTER JOIN B (A.COLUMN1 = B.COLUMN1)

  A 1 X 1

  A 1 X 2

  B 2 Y

  C 3 {null}

  D {null} Z

  HTH

• left outer join and the where clause for the table to the right

  I want to join two tables a and b, where a is a must and b is a result set in option. When I use a left outer join to a to b, I want to achieve:
  
  1. Select a single column, two columns of b (not the join columns)
  2 - even if theres no friendly on the join column does not return data from one.
  3. If there is a match applies when the criteria on column b (table in option)
  
  so, how can I avoid no_data_found in this case? When I apply where criteria for b, so it does not return the data from one, which is a must.

  Sounds like a regular outer join to me...

  select a.col1, b.col2, b.col2
  from   tableA a
         left outer join tableB b
         on (a.id = b.id and b.colX = 'X')
  

• LEFT OUTER JOIN, trigger after QUERY problem

  Hello
  
  Guide to please the following
  
  I wrote under query in QUERY after a BLOCK of TABULAR DATA, not as a single text element, called INVENTORY_ITEM
  
  Select c.cat |' '|| s.SubCat |' '|| L1.lvl1 POINT IN: DATABLOCK. INVENTORY_ITEM
  of itemcat c
  LEFT OUTER JOIN itemsubcat s on (c.catid = s.catid)
  LEFT OUTER JOIN lvl1 l1 on (s.subcatid = l1.subcatid)
  
  When I compile the module an error is generated
  
  «* ' Met the 'LEFT' symbol when waiting for one of the following values for the group with intersect less order start union where connect '.» *
  
  Top query works fine with ORACLE SQL DEVELOPER.
  
  Any solution please.
  
  Kind regards

  Difference

• Modeling of the left outer join

  Hello world

  I'm tender hand to you guys for a modeling help

  I have a FACT, the customers, the Dim_Date and CUST_ADDRESS of tables to model

  Fact and the client are joined through CUST_ID

  FACT and DATE are joined through DATE_ID

  CUST_ADDRESS must be attached to the top of the model through CUST_ID, DATE_ID and this join must be Left outer because sometimes the address does not exist or is not current, which means DATE_ID could be different between Dim_Date and CUST_ADDRESS

  If it were to join internal, model would have been easy, because of the outside left that I am unable to model, it's pretty good.

  Application under
  Select D.DATE, C.CUST_NAME, CA. ADDRESS, F.AMOUNT
  Of
  F FACT
  JOIN THE
  CUSTOMER C
  ON C.CUST_ID = F.CUST_ID
  JOIN THE
  DIM_DATE D
  ON F.DATE_ID = D.DATE_ID
  LEFT OUTER JOIN
  CUST_ADDRESS CA
  ON C.CUST_ID = CA. CUST_ID AND C.DATE_ID = D.DATE_ID

  Thanks in advance

  When I add the CUSTOMER and in FACT LTS CUST_ADDRESS

  Stop it!

  Don't add CUSTOMER and CUST_ADDRESS in the FACT of LTS. Why would add you to the LTS DO?

  You design a management model: CUSTOMER is a dimension and it has its own logical table this logic table join with a logical join in the activity diagram. Ditto for CUST_ADDRESS.

  So the change, I missed earlier is CUST_ADDRESS contains no Cust_ID (ACTUALLY existing), but contains a Cust_NO, and the table to translate Cust_NO in Cust_ID is CUSTOMER?

  No problem...

  Let's start with a new alias of CUSTOMER (to keep more simple to understand at the moment), call as you want, but this new alias will be the link between the FACT and CUST_ADDRESS.

  In LTS of the dimension 'Address', you have CUST_ADDRESS initially, add an inner join on the new alias that you created in the LTS of the CUSTOMER. So now your 'Address' logical dimension contains the Cust_NO and Cust_ID and this will make the join to FACT.

  Between CUST_ADDRESS and the CLIENT, you can keep an inner join, because the target is not not for get the address of the customer, but is having the Cust_ID in the address line.

  Give it a try at that.

  But do not add these tables in the LTS, they are logical dimensions.

• Left Outer Join help...

  Hello world

  I'm still trying to learn the SQL, and I can not specifically with the left outer join.  I normally join tables using equijoin, but I don't get the right data set returns, and designed with the help of a left or right outer join would solve the problem...

  Here is my SQL that works properly with 1 left outer join.  I build the slow query in the SQL following, you will see where I see the error.  I don't expect you to understand the data and the columns, I'm trying to join, I think the problems I encounter are related to syntax, and I hope that you can find where are my syntax errors.

  Select

  s.Name as "Pseudonym,"

  SV.view_name as "name of the view.

  s_view. Name

  Of

  s s_screen,

  s_screen_view sv

  outer join left s_view

  WE (sv.view_name = s_view.name)

  where

  SV.screen_id = ' 1-866 A - 1X3LU' and

  s.ROW_ID = sv.screen_id and

  s.repository_id = ' 866 A-1-1"and

  s_view.repository_id = ' A-1-1 866 ";

  Here is the SQL code where I encounter the following error...

  Error:

  ORA-00904: "SV". "" View_name ": invalid identifier

  00904, 00000 - '% s: invalid identifier '.

  * Cause:

  * Action:

  Error on line: column 14: 7

  Problematic SQL:

  Select

  s.Name as "Pseudonym,"

  SV.view_name as "name of the view.

  s_view. Name,

  s_applet. Name as Applet

  -b.SID like "name of the cmdlet.

  Of

  s s_screen,

  s_screen_view sv,

  wti s_view_wtmpl_it

  outer join left s_view

  WE (sv.view_name = s_view.name)

  outer join left s_applet

  WE (wti.name = s_applet.name)

  where

  SV.screen_id = ' 1-866 A - 1X3LU' and

  s.ROW_ID = sv.screen_id and

  s.repository_id = ' 866 A-1-1"and

  s_view.repository_id = ' A-1-1 866 ";

  Thanks in advance for your help.

  Chris

  > ORA-00904: "S_VIEW_WEB_TMPL." "" ROW_ID ": invalid identifier

  I don't see this table in your FROM clause.