left outer join, after summarizing

Similar Questions

• 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

• Bad result in a left outer join in 12.1.0.2

  Hallo,

  We discovered a strange behaviour in a query. The query provides values in a column of outer join where there is no corresponding value in the table is attached to the outside.

  When you expand this request by the "ORDER BY" then this query gives the correct result.

  Example:

  SQL > desc tb_a
  Name                                Null?    Typ
  -------------------------------------------- ----------------------------
  ID NOT NULL NUMBER (19)

  SQL > desc tb_b
  Name                                Null?    Typ
  -------------------------------------------- ----------------------------
  CLOSED NOT NULL NUMBER (1)
  ID NOT NULL NUMBER (19)

  CCS_APPLICATION@icw01> select * from tb_a where id in (4148,4141,4195);

  ID
  ----------
  4148
  4141
  4195

  CCS_APPLICATION@icw01> select * from tb_b where id in (4148,4141,4195);

  INTERNAL ID
  ---------- ----------
  4148 0

  CCS_APPLICATION@icw01> SELECT
  2      b.id                            AS b_id,
  3      a.id                            AS a_id,
  4 b.closed AS b_closed
  5
  6 tb_a a
  7 LEFT OUTER JOIN tb_b b ON a.id = b.id
  8 WHERE a.id IN (4148, 4195, 4141)
  9 ORDER BY ASC a.id
  10;

  B_ID ALLOCATION A_ID B_CLOSED
  ---------- ---------- ----------
  4141
  4148 4148 0
  4195

  CCS_APPLICATION@icw01> SELECT
  2      b.id                            AS b_id,
  3      a.id                            AS a_id,
  4 b.closed AS b_closed
  5
  6 tb_a a
  7 LEFT OUTER JOIN tb_b b ON a.id = b.id
  8 WHERE a.id IN (4148, 4195, 4141)
  9 - ORDER BY ASC a.id
  10;

  B_ID ALLOCATION A_ID B_CLOSED
  ---------- ---------- ----------
  4148 4148 0
  4141 4141
  4195 4195

  instance parameter:

  VALUE OF TYPE NAME

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

  compatible string 12.1.0.2.0

  optimizer_features_enable string 12.1.0.2

  After ""alter system set optimizer_features_enable = ' 11.2.0.4 ';"  the query provides the correct result in both cases (ordered and unordered).

  Now the final question: is this a bug?

  1480970 wrote:

  Hallo!  Yes, I searched the Support of Oracle. I found some similar entries, but not an exact match. To fix some issues

  with 12.1.0.2.

  There is another interesting clue when look you on the execution plan:

  Note

  -----

  -the dynamic statistics used: dynamic sampling (level = 2)

  - This is an adaptation plan

  We have disabled (= FALSE) optimizer_adaptive_features and the query provides the correct values.

  This could be a solution for us.

  Looks like a pretty tight match for bug 18430870, even if it affects the two 12.1.0.1 and 12.1.0.2, which contradicts the Martin trial against 12.1.0.1.

  The description of the bug mentions disabling "_projection_pushdown" (set to false) should also be a viable solution, perhaps if you want to give that a try and see if it is a different bug or not.

  There are also a number of one-time fixes already available for download, maybe your version / platform is already covered, if the bug applies.

  Randolf

• Oracle: Use LEFT OUTER JOIN, but convert the data to an external list

  Hello, all,.

  I know it can be done; I just don't remember how I got it done, oh there are so many years.

  Assumes that the tables exist for groups and individuals.  People can belong to several groups.

  SELECT g.groupName, p.lastName || ', ' || p.firstName as fullName
  FROM groups g LEFT OUTER JOIN groupPersonAssociation gpa ON gpa.groupID = g.groupID
                LEFT OUTER JOIN person p ON p.personID = gpa.personID
  ORDER BY g.groupName, fullName
  

  This gives us:

  Group One          Alpha, Daniel
  Group One          Bravo, Charles
  Group One          Charlie, Chuck
  Group Two          Beta, Alpha
  Group Two          Delta, Bonnie
  Group Three        Echo, Bunny
  Group Three        Golf, Samuel
  Group Three        November, Stan
  

  How word the SQL to get the data as:

  Group One          Alpha, Daniel | Bravo, Charles | Charlie, Chuck
  Group Two          Beta, Alpha | Delta, Bonnie
  Group Three        Echo, Bunny | Golf, Samuel | November, Stan
  

  V/r,

  ^_^

  I finally thought to it.  I was using incorrect keywords on Google.

  SELECT g.groupName, LISTAGG(p.lastName || ', ' || p.firstName,' | ') WITHIN GROUP (ORDER BY g.groupName) "fullName"
  FROM groups g LEFT OUTER JOIN groupPersonAssociation gpa ON ggpa.groupID = g.groupID
                LEFT OUTER JOIN person p ON p.personID = gpa.personID
  GROUP BY g.groupName
  ORDER BY g.groupName, fullName  
  

  Just in case someone else is going through this same desire.

  HTH,

  ^_^

• doubt left outer join

  Hi all

  I use under version

  Connected to Oracle Database 11g Express Edition Release 11.2.0.2.0

  SQL > SELECT E.ENAME,.

  2 D.DEPTNO,

  3 D.LOC

  4. TO EMP E,.

  DEPT 5 D

  6. WHERE = E.DEPTNO D.DEPTNO (+);

  ENAME, DEPTNO LOC

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

  DALLAS SMITH 20

  ALLEN 30 CHICAGO

  WARD 30 CHICAGO

  20 DALLAS JONES

  MARTIN 30 CHICAGO

  BLAKE 30 CHICAGO

  CLARK 10 NEW YORK

  SCOTT 20 DALLAS

  THE 10 NEW YORK KING

  TURNER 30 CHICAGO

  20 DALLAS ADAMS

  JAMES 30 CHICAGO

  FORD 20 DALLAS

  MILLER 10 NEW YORK

  40 BOSTON

  15 selected lines

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

  SQL > SELECT E.ENAME,.

  2 D.DEPTNO,

  3 D.LOC

  4. TO EMP E

  5 LEFT OUTER JOIN

  D 6 DEPT

  7. THE E.DEPTNO = D.DEPTNO;

  ENAME, DEPTNO LOC

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

  MILLER 10 NEW YORK

  THE 10 NEW YORK KING

  CLARK 10 NEW YORK

  FORD 20 DALLAS

  20 DALLAS ADAMS

  SCOTT 20 DALLAS

  20 DALLAS JONES

  DALLAS SMITH 20

  JAMES 30 CHICAGO

  TURNER 30 CHICAGO

  BLAKE 30 CHICAGO

  MARTIN 30 CHICAGO

  WARD 30 CHICAGO

  ALLEN 30 CHICAGO

  14 selected lines

  My doubt is both are same query is the same, is in ansi format and is in the format of the Oracle,.

  but the results are different.

  For the first query null is coming for unmatched records in the dept table

  but in the second query, it does not come

  Thank you

  Hello

  2947022 wrote:

  Hi all

  I use under version

  Connected to Oracle Database 11g Express Edition Release 11.2.0.2.0

  SQL > SELECT E.ENAME,.

  2 D.DEPTNO,

  3 D.LOC

  4. TO EMP E,.

  DEPT 5 D

  6. WHERE = E.DEPTNO D.DEPTNO (+);

  ENAME, DEPTNO LOC

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

  DALLAS SMITH 20

  ALLEN 30 CHICAGO

  WARD 30 CHICAGO

  20 DALLAS JONES

  MARTIN 30 CHICAGO

  BLAKE 30 CHICAGO

  CLARK 10 NEW YORK

  SCOTT 20 DALLAS

  THE 10 NEW YORK KING

  TURNER 30 CHICAGO

  20 DALLAS ADAMS

  JAMES 30 CHICAGO

  FORD 20 DALLAS

  MILLER 10 NEW YORK

  40 BOSTON

  15 selected lines

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

  SQL > SELECT E.ENAME,.

  2 D.DEPTNO,

  3 D.LOC

  4. TO EMP E

  5 LEFT OUTER JOIN

  D 6 DEPT

  7. THE E.DEPTNO = D.DEPTNO;

  ENAME, DEPTNO LOC

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

  MILLER 10 NEW YORK

  THE 10 NEW YORK KING

  CLARK 10 NEW YORK

  FORD 20 DALLAS

  20 DALLAS ADAMS

  SCOTT 20 DALLAS

  20 DALLAS JONES

  DALLAS SMITH 20

  JAMES 30 CHICAGO

  TURNER 30 CHICAGO

  BLAKE 30 CHICAGO

  MARTIN 30 CHICAGO

  WARD 30 CHICAGO

  ALLEN 30 CHICAGO

  14 selected lines

  My doubt is both are same query is the same, is in ansi format and is in the format of the Oracle,.

  but the results are different.

  For the first query null is coming for unmatched records in the dept table

  but in the second query, it does not come

  Thank you

  In fact, these requests are not the same.

  The first is to find all the lines of the Department, with the corresponding lines of PGE (when there are).  This is equivalent to «FROM dept LEFT OUTER JOIN emp...» ».

  The second is to find all the rows in the emp of the lines of the Department (when there are any).  This is equivalent to «...» WHERE e.deptno = d.deptno (+).

• Left Outer Join DR.

  Hi Experts,

  I have a requirement that says - see the chart for the past 10 days, regardless the presence data table in fact.
  Lets consider an example - Time_dim product, are my dimension tables, Purchase_Order is my fact table.

  I did it for external Purchase_Order in left RPD with TIME_DIM and inner join with the PRODUCT table.  and execution of query of exit-
  Select T.Date, P.item, count (distinct PO.order_no)
  TIME_DIM t, PRODUCT P, PURCHASE_ORDER PO
  where T.date_key = PO.date_key
  and P.item = in. agenda
  and P.item = 'laptop ';

  The query generated by OBIEE left outer join, but when the condition P.item = "Notebook" included in the query, and if there are no orders for this product in one of the date, that date will not come in the result set.

  the query to be generated by the OBIEE is-

  
  Select T.Date, PO.item, count (distinct PO.order_no)
  TIME_DIM t,.
  (SELECT P.ITEM, IN. ORDER_NO
  PRODUCT P, PO PURCHASE_ORDER
  WHERE P.item = in. agenda
  and P.item = 'Laptop') IN.
  WHERE T.date_key = PO.date_key (+);

  How to design the RPD to achieve this. All pray to advise on this. Thanks in advance.

  Thank you
  Chantal

  Hello

  You are on 11.1.1.7?

  I would say that your condition can be made without using external and maintenance of product and the standard between the FACT dimension, time inner join join.

  If you enable your property analysis OBIEE "Include Null values" will automatically return all the elements of time and product matching your filter (so you'll need to add a filter on 'Date' to limit it to the last 10 days or you will have a unique day of your time dimension).

  If you filter then on "Laptop", even if there is not a single value in order for "Laptop" in the last 10 days, he will be there on the screen.

  Easy, clean and you keep your inner join between the facts and Dimensions.

  Take a look at this example, I just did on SampleApp 406:

  

  Selection of 12 months (year 2010) and a customer (id = 89) and income. The model has only an inner join. I activate the option "Include Null values" and here is the result.

  

  A line with cells only empty because there is not a single revenue for customer 89 in 2010. This is exactly your condition.

  Honestly, do not touch your model using the outer join, you will have more side effects than benefits. Every single scan will do the outer join and you'll have a lot of data 'empty' return of the DB (more large data set containing just the null values) and probably you need the outer join in 15 to 25% of your analysis.

  Keep things simple, it will be faster and easier to maintain.

• 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.

• Help for a LEFT OUTER JOIN query

  Hello, all,.

  I'm having some trouble setting up an Oracle 11 g Server SQL query, and I could use some help.

  Let's say tableA is blogs; tableC is comments for blog entries; tableB is the associative array:

  tableA
  blogID        blogTitle       blogBody      dateEntered
  1             This is a test  More text...  2016-05-20 11:11:11
  2             More testing    Still more!   2016-05-19 10:10:10
  3             Third charm!!   Blah, blah.   2016-05-18 09:09:09
  

  tableC
  commID        userID          userText      dateEntered
  10            Bravo           I like it!    2016-05-20 11:21:31
  11            Charlie         I don't!      2016-05-20 11:31:51
  12            Alpha           Do it again!  2016-05-19 10:20:30
  13            Bravo           Still more?   2016-05-19 10:30:50
  14            Charlie         So, what?     2016-05-19 10:35:45
  15            Bravo           Blah, what?   2016-05-18 09:10:11
  16            Alpha           Magic number! 2016-05-18 09:11:13
  

  tableB
  blogID        commID
  1             10
  1             11
  1             12
  2             13
  2             14
  3             15
  3             16
  
  
  
  

  I'm trying to get blogID, blogTitle, blogBody and the number of comments for each blog entry.  But, since I'm on to_char() for date and COUNT (commID) for the total number of comments, I am not "a group by expression.

  Here is an example of pseudo-SQL of what I'm trying.

  SELECT a.blogID, a.blogTitle, a.blogBody, to_char(a.dateEntered,'YYYY-MM-DD HH24:MI:SS') as dateEntered, COUNT(c.commID) as total
  FROM tableA a LEFT OUTER JOIN tableB b ON b.blog_ID = a.blog_ID
                LEFT OUTER JOIN tableC c ON c.commID = b.commID
  WHERE a.blogID = '1'
  GROUP BY blogID, blogTitle, blogBody
  ORDER BY to_date(dateEntered,'MM-DD-YYYY HH24:MI:SS') desc
  

  I'm sure it's something simple, but I just DO NOT see it.  Can you help me?

  V/r,

  ^_^

  Try:

  GROUP BY a.blogID, a.blogTitle, a.blogBody, to_char(a.dateEntered,'YYYY-MM-DD HH24:MI:SS')
  

  See you soon

  Eddie

• 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

• 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

• 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.

• Left outer join error 2

  Hi all

  11.2.0.1

  I have two 2 left outer join in my application, but I got errror

  Ths is supported?

  It is something like:

  Select tab1, b.col1, a.col1, c.col1 one

  LEFT OUTER JOIN tab2 b

  ON a.id = b.id;

  LEFT OUTER JOIN tab 3 c

  ON a.id = c.id;

  My syntax is correct? I got the error ora-942.

  Help, please...

  Thank you

  Then use script below

  SELECT A.ACRNUMBER AS ACR_NO, A.LASTNAME, A.GIVENNAME AS FIRSTNAME, A.MIDDLENAME, A.BIRTHDATE, A.GENDER,NLV2(F.COUNTRYID3,'XXX'), A.PROBSTAYLENGTH AS PROB_STAY_LENGHT,A.SECTIONISSUED AS SECTION_ISSUED, A.RESIDENCECERTIFICATENUMBER AS RESIDENCE_CERTIFICATE_NO, A.ACTIVESTATUS AS STATUS, B.CLIP
  IMAGE AS PHOTO, C.SIGBMP AS SIGNATURE,D.CARDISSUErNUMBER AS CARD_NO, D.CARDEXPIRYDATE AS CARD_VALIDITY, D.CARDSERIALNUMBER AS CARD_SERIAL_NO, D.CARDISSUEDATE AS CARD_DATE_ISSUED, nvl2( d.CARDSERIALNUMBER,'0','1') AS CARD_STATUS
         From   Acrmaster  A Inner Join  Pictures  B  On A.Acrnumber=B.Acrnumber
                         Inner Join  Signature  C On B.Acrnumber=C.Acrnumber
                         Inner Join  Acrcarddetails D On   C.Acrnumber=D.Acrnumber
                         Left Outer Join Acrblockcarddetails E On (D.Acrnumber=E.Acrnumber And D.Cardserialnumber=E.Cardserialnumber)
                         Left Outer Join Countrymaster_Lk F ON  A.NATIONALITY=F.ID
  

  You have error here:

  left outer join ACRBLOCKCARDDETAILS E

  Left Outer join Countrymaster_Lk F on d.acrnumber = e.acrnumber and d.CARDSERIALNUMBER = e.CARDSERIALNUMBER and A.NATIONALITY = F.ID

  Concerning

  Mr. Mahir Quluzade

• 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')
  

• 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 without data in a table

  I have two tables defined (see below). Emp table has given, and there is still no data in the table of Emp_Type! Now it is empty.
  
  I want to write a query that returns the data from the tables, even if there is no data in the Emp_type table. I use a left outer join but it return nothing. Can anyone help?
  

  create table EMP
  (
    EMPID   NUMBER(10,2),
    EMPNAME VARCHAR2(100)
  );
  
  INSERT INTO emp (empid,empname) values (1, 'Mark');
  INSERT INTO emp (empid,empname) values (2, 'Jason');
  INSERT INTO emp (empid,empname) values (3, 'Kevin');
  INSERT INTO emp (empid,empname) values (4, 'Drew');
  INSERT INTO emp (empid,empname) values (5, 'Jessica');
  INSERT INTO emp (empid,empname) values (6, 'Pena');
  INSERT INTO emp (empid,empname) values (7, 'Roxanne');
  
  create table EMP_Type
  (
    ID          NUMBER(10,2),
    EMPID       NUMBER(10,2),
    TYPE_ID     NUMBER(10,2),
    END_DATE    DATE
  );
  
  No data
  
  select *
    from emp e
    left outer join emp_Type t
      on e.empid = t.empid
   WHERE t.type_id = 1
     and t.end_date is null;

  WHERE t.type_id = 1

  will be ever true when table has no rows (or t.type_id is NULL?)