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

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• Full table on the right outer join scan

  Hello, I need help. Here's my query-
  
  SELECT / * + first_rows (100) * /.
  a.custom_gr_id, a.custom_gr_name AS customgroupname,
  a.custom_gr_type_id AS customtypeid, b.workspace_id, b.NAME, a.itime,
  a.is_shared_org AS est_partagee
  OF custom_group_pa one
  RIGHT OUTER JOIN
  account_workspace_pa b ON (a.workspace_id = b.workspace_id
  AND (a.delete_flag IS NULL or a.delete_flag <>1)
  AND a.owner_id = '123')
  WHERE b.workspace_id <>- 9999
  AND b.workspace_type_id = 1
  AND b.delete_flag = 0
  AND EXISTS (SELECT 1
  Of account_workspace_type c
  WHERE b.workspace_type_id = c.workspace_type_id)
  ITime to ORDER BY DESC;
  
  Account_workspace_pa almost 1 500 483 lines and custom_group_pa had 200 000 lines and account_workspace_type is a very small table of 10 lines.
  
  I always get a full on table account_workspace_pa table scan, while I have clues about workspace_id AND I index on workspace_id, workspace_type_id and delete_flag. The request is expected to produce less than 500 lines.
  
  Here's the plan I make-
  
  Plan
  COUNCIL of the SELECT STATEMENT: FIRST_ROWSCost: 3 931 bytes: 405 108 cardinality: 7 502
  8 SORTING ORDER BY ORDER BY a.ITIME DESCCost: 3 931 bytes: 405 108 cardinality: 7 502
  7 LOOPS IMBRIQUEES EXTERNAL (ACCOUNT_WORKSPACE_TYPE at ACCOUNT_WORKSPACE_PA) in CUSTOM_GROUP_PACost: 3 930 bytes: 405 108 cardinality: 7 502
  3 LOOPS IMBRIQUEES ACCOUNT_WORKSPACE_TYPE to ACCOUNT_WORKSPACE_PACost: 2 280 bytes: 210 056 cardinality: 7 502
  1 INDEX UNIQUE INDEX (SINGLE) ADR_DEV SCAN. SYS_C0025442 aliased as cCost: cardinality of 0 bytes: 6: 1
  TABLE 2 ACCESS FULL TABLE ADR_DEV. ACCOUNT_WORKSPACE_PA alias ADR_DEV_DATA Tablespaceb.WORKSPACE_ID b! = - 9999, b.WORKSPACE_TYPE_ID = 1 b.DELETE_FLAG = 0Cost: 2 280 bytes: 165 044 cardinality: 7 502
  VIEW 6 (Embedded SQL) cost: 0 bytes: 26 cardinality: 1
  TABLE 5 ACCESS BY INDEX ROWID TABLE ADR_DEV. Aliased as one on ADR_DEV_DATA TablespaceCost CUSTOM_GROUP_PA: 04:00 cardinality: 1
  4 INDEX RANGE SCAN INDEX ADR_DEV. IDX_CG_WSID_DF_OWID aliased as aa.delete_flag (+) is NULL, a.delete_flag (+)! = 1, a.owner_id (+) = '123', a.workspace_id (+) = b.workspace_idCost: cardinality 3: 1
  
  
  If someone can advise, why I always make a full table scan on table account_workspace_pa and what I can do to have a systematic index scan?
  
  Thank you.

  If you need all the lines then that would be a clue do for you?

  You don't necessarily need a full table scan. But if you have need of all ranks, it is certainly more effective to do a full scan (read through all of the blocks table in bulk, once) rather than navigate throu an index and jump from block to block to block (probably one at a time, probably having to hit most of them more than once until the work is done) after the ROWID as they occur in the index.

  Also, fully entered null will not appear in the index, so (depending on your data, NOT NULL constraints, etc.) it may be possible that navigate through an index will cause lines to be ignored.

  I think you look at that and see a problem, but the answer is: it's really the best way to accomplish what you're asking.

• Right outer join

  There are four tables:
  
  Family
  Parent
  Child
  Grandchild
  
  Ideally, they have all documents, such as
  
  F1 - P1 - C1 - G1
  F2 - P2 - C2 - G2
  F3 - P3 - C3 - G3
  
  But sometimes,.
  
  F1 - P1 - C1 - null
  F2 - P2 - null - null
  F3 - P3 - null - null
  
  For the case of the latter, maybe I need to right outer join. If it's between two tables, the right outer join is easier. But among the four tables, inner family join parent, child of the outer join, then the big kid outer join. Maybe even this has been done? If an outer join in this case is not relevant, what other options are available?
  
  
  Thank you
  
  (ORACLE 11.2)

  Hello

  As Salomon, said

  FROM  p  LEFT  OUTER JOIN c
  

  means exactly the same thing that

  FROM  c  RIGHT OUTER JOIN p
  

  Everything you do with LEFT OUTER JOIN you could also do with RIGHT OUTER JOIN, and vice versa. If either one did not exist, you may do whatever you want with the other. In practice, it's just what's happening: most of the people always use LEFT OUTER JOIN and never use a RIGHT OUTER JOIN.

  You can have a series of outer joins cascading. If I understand your problem, a particular family may or may not have parents that belongs to him. If there are relatives, then you want to show the family with his parents to realteaed, but if there are no parents, so you want to show the family anyway. Similarly, you want to show parents whether or not they have children, and the children or not all my grandchildren are related to them. In general, which is written like this:

  ...
  FROM           family          f
  LEFT OUTER JOIN      parent          p    ON  p.family_id  = f.family_id
  LEFT OUTER JOIN      child          c    ON  c.parent_id  = p.parent_id
  LEFT OUTER JOIN      grandchild  g       ON  g.child_id   = c.child_id
  

  You can have a situation where (for example) a child is related to a family, but the child has no parent? In this case, you can still use LEFT OUTER JOIN, but join conditions would be different.

• 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

• Looking for right outer join help

  Hello

  Can anyone help me re - write the query using right below outer join (join ansi) instead of using the (+) symbol.

  Select cd.sku_id waitm,

  Decode (cl.invn_lock_code, NULL, 'OH', cl.invn_lock_code) WLOCN,

  Sum (CD.actl_qty) WQOH

  cd, a.case_lock cl, a.case_hdr ch, a.item_master im a.case_dtl

  where cd.case_nbr = ch.case_nbr

  and cl.case_nbr (+) = ch.case_nbr

  and cd.sku_id = im.sku_id

  and ch.stat_code < = '64'

  Cd.sku_id group,

  IM.sku_brcd,

  Decode (cl.invn_lock_code, NULL, 'OH', cl.invn_lock_code)

  order of cd.sku_id

  Kind regards

  Gannu

  Maybe something like this:

  of a.case_dtl cd

  Join a.case_hdr ch on (cd.case_nbr = ch.case_nbr)

  Join a.item_master im on (cd.sku_id = im.sku_id)

  join a.case_lock cl left (cl.case_nbr = ch.case_nbr)

  where ch.stat_code<=>

  I know you asked for right outer join, but I tend to write like that.  You can certainly change if you wish.

• Reg: Diff between Right Join and Right Outer Join

  Dear all,

  Kindly help me to find the difference between the two queries below where the results are the same.

  SELECT * FROM emp RIGHT JOIN dept WE emp.deptno = dept.deptno;

  
  SELECT * FROM emp RIGHT OUTER JOIN dept WE emp.deptno = dept.deptno;

  Thank you

  Both are same

• Difference between Inner join and right outer join...

  Which is precisely the difference between an Inner join and right outer join...

  JOIN INTERNAL:-to return all rows from the two tables where there is a football game. That is to say. the table resulting from all the rows and columns will have values.

  AND

  RIGHT OUTER JOIN:-Returns all rows in the second table, even if there is no match in the first table.

• Motorcycle deny non-American owners the right to join the PNM

  Like other owners of non-us I HAV been denied the right to join the feedback of Motorola (MFN) network.

  This confirms the thought racist and discriminating motorcycle against the owners other than the United States.

  
  

• Get the error invalid identifier in the left outer join

  I wrote a query and get the invalid identifier error during execution:
  the exact error is: ORA-00904: "D". "" ACCT_NO ": invalid identifier
  This query is not able to access the parent table alias in the subquery of the left outer join.
  
  This is the query:
  SELECT D.PROD_DESC_TX, BASE. ASSET_NUM, BASE. PROD_ID, BASE.NAME
  OF TABLE1 D
  LEFT OUTER JOIN
  (
  SELECT ASSET_NUM, PROD_ID, B.SID
  FROM TABLE2 E
  JOIN IN-HOUSE TABLE3 HAS IT E.PROD_ID = A.ROW_ID
  JOIN INTERNAL TABLE 4 C ON A.PAR_PROD_INT_ID = C.ROW_ID
  INTERNAL TABLE5 JOIN B ON C.ROW_ID = B.PAR_ROW_ID
  AND B.TYPE = 'VALUE '.
  AND B.NAME = 'VALUE '.
  WHERE E.ASSET_NUM = ((CASE WHEN LTRIM (RTRIM (D.BANK_NO)) = '021 ' THEN '021' ELSE ' 020' END) |) LTRIM (RTRIM (D.APPL_CD)). LTRIM (RTRIM (D.ACCT_NO)))
  ) BASE ON ((CASE WHEN LTRIM (RTRIM (D.BANK_NO)) = '021 ' THEN '021' ELSE ' 020' END) |) LTRIM (RTRIM (D.APPL_CD)). LTRIM (RTRIM (D.ACCT_NO))) = BASE. ASSET_NUM
  WHERE D.BANK_NO = 'VALUE '.
  AND D.APPL_CD = 'VALUE '.
  AND D.ACCT_NO = 'VALUE '.
  
  Edited by: user648525 13 Sep, 2011 01:21

  I can easily look at your request at this time (using my iPhone on a train).

  But trying out this line in the WHERE (the one with the invalid reference) clause in the main query.

  Who is... in the view of inline, use only the logic of the join (the clauses).

  Overflow the filtering logic (WHERE clause) to the main request. You may need to select some additional inline view columns to make these references work.

• How to find the right server to use windowsmail to send and receive e-mails?

  How to find the right server to use windowsmail to send and receive e-mails?

  Who is your current e-mail server? (Which is after the @ in your address).
   
  Windows Mail: Setting up an account of end-to-end
  http://Windows.Microsoft.com/en-us/Windows-Vista/Windows-mail-setting-up-an-account-from-start-to-finish 

• Error: you have an error in your SQL syntax; consult the manual for your version of the MySQL server for the right syntax to use near ')' at line 5

  Original title: how to solve this problem:

  System error

  You have an error in your SQL syntax; consult the manual for your version of the MySQL server for the right syntax to use near ')' at line 5

  Hi Roger,

  It seems that you have problems with the SQL syntax. The question you have posted is related to encoding and it would be better suited to the MSDN Community. Please report it in the community below.

  http://social.msdn.Microsoft.com/forums/SQLServer/en-us/home

  Hope this is useful.

• Comparing the same outer join column.

  Hello
  
  I have the following query.
  

  SELECT Replens.Date1               AS date1,
    Replens.USER_ID                  AS Operator1,
    Picks.USER_ID                    AS Operator1,
    Users.Time                       AS Logged_in_hours,
    Replens.Quantity                 AS Pallet_replened,
    SUM(Replens.Quantity/Users.Time) AS Replen_rate,
    Picks.Quantity                   AS Cases_picked,
    SUM(Picks.Quantity/Users.Time)   AS Pick_rate
  FROM
    (SELECT libclientreports.getshift(itl.dstamp) Date1,
      ITL.USER_ID,
      COUNT(ITL.tag_id) Quantity
    FROM INVENTORY_TRANSACTION ITL
    WHERE ITL.code                                     = 'Replenish'
    AND TO_CHAR(itl.dstamp, 'DD-Mon-YYYY HH24:MI:SS') >= '01-Mar-2010 06:00:00'
    AND TO_CHAR(itl.dstamp, 'DD-Mon-YYYY HH24:MI:SS') <= '01-Mar-2010 18:00:00'
    GROUP BY libclientreports.getshift(itl.dstamp),
      ITL.USER_ID
    ORDER BY ITL.USER_ID,
      libclientreports.getshift(itl.dstamp)
    ) Replens,
    (SELECT libclientreports.getshift(itl.dstamp) Date1,
      ITL.USER_ID,
      SUM(ITL.update_qty) Quantity
    FROM INVENTORY_TRANSACTION ITL
    WHERE ITL.to_loc_id                                = 'CONTAINER'
    AND TO_CHAR(itl.dstamp, 'DD-Mon-YYYY HH24:MI:SS') >= '01-Mar-2010 06:00:00'
    AND TO_CHAR(itl.dstamp, 'DD-Mon-YYYY HH24:MI:SS') <= '01-Mar-2010 18:00:00'
    GROUP BY libclientreports.getshift(itl.dstamp),
      ITL.USER_ID
    ORDER BY ITL.USER_ID,
      libclientreports.getshift(itl.dstamp)
    ) Picks,
    (SELECT USER_ID,
      SUM(LibClientReports.getLoggedInHours (USER_ID, to_date('01-Mar-2010 06:00:00', 'DD-Mon-YYYY HH24:MI:SS'), to_date('01-Mar-2010 18:00:00', 'DD-Mon-YYYY HH24:MI:SS')) )/60 TIME
    FROM APPLICATION_USER
    GROUP BY USER_ID
    ) Users
  WHERE Users.Time != '0'
  AND Users.USER_ID = Picks.USER_ID (+)
  AND Users.USER_ID = Replens.USER_ID (+)
  GROUP BY Replens.Date1,
    Replens.USER_ID,
    Picks.USER_ID,
    Users.Time,
    Picks.Quantity,
    Replens.Quantity
  ORDER BY Replens.USER_ID,
    Replens.Date1

  The problem I have is with the operator1 columns, I've done outer joins on the Users table that contains all users that will be returned in the tables of the peaks and Replens.
  However, I want to merge this operator1 into one column. For example, here is the current data, I have for the two columns of the operator 1;
  

  Operator1       Operator1_1
  
  D9AGY19             D9AGY19
  D9BEELB             D9BEELB
  D9FULOPS     (null)
  D9GAWKOWA     D9GAWKOWA
  D9LIDDING     D9LIDDING
  D9NOWAKS     D9NOWAKS
  (null)             D9AGY10
  (null)             D9AGY13
  (null)             D9AGY15
  (null)             D9AGY17

  And here is the data I want to see;
  

  Operator1
  
  D9AGY19
  D9BEELB
  D9FULOPS
  D9GAWKOWA
  D9LIDDING
  D9NOWAKS
  D9AGY10
  D9AGY13
  D9AGY15
  D9AGY17

  Notice the USER_ID to replace the values (null).
  
  Could someone please explain to me how to make this happen? If you want more information please let me know.
  
  
  Kind regards
  
  SM.

  This should do if Op1 and Op1_1 are always the same or one or the other is null. If both are filled and have different values, then it will not work:

  nvl(Replens.USER_ID, Picks.User_ID) as                  AS Operator1
  

• 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.

• My web site have a blank space in the right side when using chrome

  It seems perfectly on Safari, desktop and mobile, but in my android phone, I use chrome and nothing when I get this white ditch on the right. I even test in resizer to google to check it out and it seems fine. I only get this problem when using chrome. Has it to do with the fact that I'm the test catalyst for business?

  I think it's a bug. because when I delete some items, it works well and everything seems to be good.

  Please help me!

  Here is the link to the site

  http://bufeteostos02.BusinessCatalyst.com/index.html

  Thanks for sharing the file with me.

  You have created a black rectangle in the home page to give it a black background. This rectangle is the origin of the problem.

  I tried now to get rid of the question.

  1 remove the black rectangle of the homepage.

  2. go in the master page and fill color of the browser set to black.

  3 publish the file and it works perfectly.

  I am also you send the file with my changes.

  Kind regards

  Vivek

• How can I get the new tab to open all the way to the right when I use the link open in a new tab?

  How can I get the new tab to open all the way to the right of the tabs old when I use the link open in a new tab?

  Set the pref browser.tabs.insertRelatedAfterCurrent false on the subject: config page.

  To open the topic: config page, type Subject: config in the address bar (address) and press the 'Enter' key, as you type the url of a Web site to open a Web site.
  
  If you see a warning then you can confirm that you want to access this page.
  
  You can use the filter at the top bar of the on: page config to more easily spot a pref.