Subquery factoring and DB Link

Similar Questions

• Subquery factoring and materialize Hint

  WITH t AS
          (SELECT MAX (lDATE) tidate
             FROM rate_Master
            WHERE     Code = 'G'
                  AND orno > 0
                  AND TYPE = 'L'
                  AND lDATE <= ':entereddate')
  SELECT DECODE (:p1,  'B', RateB,  'S', RateS,  Rate)
    FROM rate_Master, t
   WHERE     Code = 'G'
         AND orno > 0
         AND TYPE = 'L'
         AND NVL (lDATE, SYSDATE) = tidate;

  In the example given the sub query returns just one line because of the max aggregate function. This by making a With clause that will be of any benefit? Also I assume / understand that factoring of subquery would be really useful when we try to do a sub query that returns multiple rows in a clause. Is my right to intrepration?
  
  Then add the / * + Materialize * / reference to a query with is required or the optimizer itself will do and perform a transformation of the temporary table. In my example, I have to give the hint in the query. Please discuss and help
  
  Thanks in advance.

  ramarun wrote:

  WITH t AS
  (SELECT MAX (lDATE) tidate
  FROM rate_Master
  WHERE     Code = 'G'
  AND orno > 0
  AND TYPE = 'L'
  AND lDATE <= ':entereddate')
  SELECT DECODE (:p1,  'B', RateB,  'S', RateS,  Rate)
  FROM rate_Master, t
  WHERE     Code = 'G'
  AND orno > 0
  AND TYPE = 'L'
  AND NVL (lDATE, SYSDATE) = tidate;
  

  In the example given the sub query returns just one line because of the max aggregate function. This by making a With clause that will be of any benefit? Also I assume / understand that factoring of subquery would be really useful when we try to do a sub query that returns multiple rows in a clause. Is my right to intrepration?

  Not quite.
  The subquery factoring should be used when you want to use the subquery results more than once in your query. So if you write a regular SQL statement, but find that it is necessary to write the same subquery more than once inside, then you can factor on this subquery using the WITH clause, so that it is executed once, and the results may then be referenced several times in the main query. This is what gives a performance advantage in many cases.

  Then add the / * + Materialize * / reference to a query with is required or the optimizer itself will do and perform a transformation of the temporary table. In my example, I have to give the hint in the query. Please discuss and help

  As mentioned the suspicion of materialization is not documented so should not be used in production code. Personally, I found that it can add significant performance gain of a weighted subquery where this subquery causes a large amount of data. Don't know why the optimizer is not always materialize subqueries by default but... not really looked inside a lot.

• Subquery factoring clause and the temporary table

  Is it possible (probably a hint) to specify the Oracle to create a temporary table in subquery factoring (with...) clause?
  
  So if I have a query

  with t1 as (select ...)
  select ...

  How can I do Oracle to generate a plan that creates a temporary table for t1 (and not using t1 as inline view)?

  Hello

  use the indicator to materialize:

  with t1 as (select /*+ materialize */ ...)
  select ...
  

  Herald tiomela
  http://htendam.WordPress.com

• Identify cycles recursive subquery factoring (RSF/CTE)

  It is possible to detect cycles AFTER executing a recursive factoring subquery with the cycle_clause.

  I need to detect already in the query, similar to the virtual CONNECT_BY_ISCYCLE in a hierarchical query.

  Here is a test case of the use of factoring of the recursive subquery (RSF) to the algorithm of Dijkstra shortest Implement path?.

  

  DROP TABLE edges;

  CREATE TABLE edges (char (1) src, dst char (1), distance, NUMBER (3, 0));

  DROP TABLE nodes.

  CREATE TABLE nodes (nodes CHAR (1));

  -INSERTION in the edges

  -normal direction

  INSERT INTO edges (SRC, DST, DISTANCE) VALUES ('A', '' B, 2');

  INSERT INTO edges (SRC, DST, DISTANCE) VALUES ('A', 'C', '4');

  INSERT INTO edges (SRC, DST, DISTANCE) VALUES ('A' 'd', '3' ");

  INSERT INTO edges (SRC, DST, DISTANCE) VALUES (' B', 'E', 7' ");

  INSERT INTO edges (SRC, DST, DISTANCE) VALUES ("C", "E", "3");

  INSERT INTO edges (SRC, DST, DISTANCE) VALUES (', 'E', '4');

  INSERT INTO edges (SRC, DST, DISTANCE) VALUES (' B', 'F', 4' ");

  INSERT INTO edges (SRC, DST, DISTANCE) VALUES ('C', 'F', "2");

  INSERT INTO edges (SRC, DST, DISTANCE) VALUES (', 'F', '1');

  INSERT INTO edges (SRC, DST, DISTANCE) VALUES (' B', 'G', 6' ");

  INSERT INTO edges (SRC, DST, DISTANCE) VALUES ('C', 'G', "4");

  INSERT INTO edges (SRC, DST, DISTANCE) VALUES (', 'G', '5');

  INSERT INTO edges (SRC, DST, DISTANCE) VALUES ('E', 'H', '1');

  INSERT INTO edges (SRC, DST, DISTANCE) VALUES ('F', 'H', '6');

  INSERT INTO edges (SRC, DST, DISTANCE) VALUES ('G', 'H', '3');

  INSERT INTO edges (SRC, DST, DISTANCE) VALUES ('E', 'I', '4');

  INSERT INTO edges (SRC, DST, DISTANCE) VALUES ('F', 'I', '3');

  INSERT INTO edges (SRC, DST, DISTANCE) VALUES ('G', 'I', '3');

  INSERT INTO edges (SRC, DST, DISTANCE) VALUES ('H', 'J', '3');

  INSERT INTO edges (SRC, DST, DISTANCE) VALUES ('I', 'J', '4');

  -inversion

  INSERT INTO edges (DST, SRC, DISTANCE) VALUES ('A', '' B, 2');

  INSERT INTO edges (DST, SRC, DISTANCE) VALUES ('A', 'C', '4');

  INSERT INTO edges (DST, SRC, DISTANCE) VALUES ('A' 'd', '3' ");

  INSERT INTO edges (DST, SRC, DISTANCE) VALUES (' B', 'E', 7' ");

  INSERT INTO edges (DST, SRC, DISTANCE) VALUES ("C", "E", "3");

  INSERT INTO edges (DST, SRC, DISTANCE) VALUES (', 'E', '4');

  INSERT INTO edges (DST, SRC, DISTANCE) VALUES (' B', 'F', 4' ");

  INSERT INTO edges (DST, SRC, DISTANCE) VALUES ('C', 'F', "2");

  INSERT INTO edges (DST, SRC, DISTANCE) VALUES (', 'F', '1');

  INSERT INTO edges (DST, SRC, DISTANCE) VALUES (' B', 'G', 6' ");

  INSERT INTO edges (DST, SRC, DISTANCE) VALUES ('C', 'G', "4");

  INSERT INTO edges (DST, SRC, DISTANCE) VALUES (', 'G', '5');

  INSERT INTO edges (DST, SRC, DISTANCE) VALUES ('E', 'H', '1');

  INSERT INTO edges (DST, SRC, DISTANCE) VALUES ('F', 'H', '6');

  INSERT INTO edges (DST, SRC, DISTANCE) VALUES ('G', 'H', '3');

  INSERT INTO edges (DST, SRC, DISTANCE) VALUES ('E', 'I', '4');

  INSERT INTO edges (DST, SRC, DISTANCE) VALUES ('F', 'I', '3');

  INSERT INTO edges (DST, SRC, DISTANCE) VALUES ('G', 'I', '3');

  INSERT INTO edges (DST, SRC, DISTANCE) VALUES ('H', 'J', '3');

  INSERT INTO edges (DST, SRC, DISTANCE) VALUES ('I', 'J', '4');

  -SELECT the recursive subquery factoring (RSF) / Common Table Expressions (CTE)

  -INCLUDE the starting point of Sub

  WITH the railways (root, src, dst, path, distance, cost, lev, iscyc) AS

  (SELECT 'A', NULL, 'A', 'A', 0, 0, 1, 0

  OF the double

  UNION ALL

  SELECT p.root,

  e.SRC,

  e.DST,

  p.Path | ',' || e.DST,

  e.distance,

  p.cost + e.distance,

  p.Lev + 1,

  -good idea?

  p.iscyc

  RAILWAYS p

  JOIN the edges e WE (e.src = p.dst)

  AND lev + 1 < = 3

  )

  Lev RESEARCH FIRST WIDTH, cost line_no SET

  CYCLE of dst SET is_cycle to 1 by DEFAULT 0

  SELECT *.

  TRAIL pr

  WHERE 1 = 1

  - AND is_cycle = 0

  - AND DST = 'J '.

  ORDER BY lev;

  
  

  DISTANCE FROM ROOT SRC DST LEV ISCY LINE_NO IS_CYCLE COST PATH

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

  A           A     A         0     0     1     0       1       0

  A     A     D     A,D       3     3     2     0       3       0

  A     A     B     A,B       2     2     2     0       2       0

  A     A     C     A,C       4     4     2     0       4       0

  A     C     F     A,C,F     2     6     3     0       9       0

  A E C E A, C, 3 7 3 0 10 0

  AN E D A, D, E 4 7 3 0 11 0

  B G G, B, 6 8 3 0 12 0

  A C AN A, C, 4 8 3 0 13 1

  A D G, D, G 5 8 3 0 14 0

  G C G, C, 4 8 3 0 15 0

  B E A, B, E 7 9 3 0 16 0

  A     B     F     A,B,F     4     6     3     0       7       0

  A     B     A     A,B,A     2     4     3     0       6       1

  A     D     F     A,D,F     1     4     3     0       5       0

  A     D     A     A,D,A     3     6     3     0       8       1

  There is a powerful way to fill the column ISCY, while the recursion like the IS_CYCLE column?

  I tried the table nested in CTE before but had a few questions. The code below works in 11g but not 12 c:

  create or replace type nt_char in the table to the varchar2 (20);

  /

  WITH the railways (root, src, dst, path, distance, cost, lev, iscyc) AS

  (SELECT 'A', NULL, 'A', CAST (nt_char ('A') AS nt_char), 0, 0, 1, 0)

  OF the double

  UNION ALL

  SELECT p.root,

  e.SRC,

  e.DST,

  nt_char (e.dst) MULTISET UNION ALL p.Path,

  e.distance,

  p.cost + e.distance,

  p.Lev + 1,

  CASE WHEN e.dst MEMBER OF p.path, 1 ELSE 0 END AS iscyc

  RAILWAYS p

  JOIN the edges e WE (e.src = p.dst)

  AND lev + 1<>

  )

  Lev RESEARCH FIRST WIDTH, cost line_no SET

  CYCLE of dst SET is_cycle to 1 by DEFAULT 0

  SELECT *.

  TRAIL pr

  WHERE 1 = 1

  - AND is_cycle = 0

  - AND DST = 'J '.

  ORDER BY lev;

• Recursive subquery factoring: calculate aggregates

  Table T represents a tree. Each record is a node, and each node has only one parent. This query calculates the SUM() of each branch for each node.

  WITH T AS
          (SELECT  1 ID, NULL parent_id, NULL VALUE FROM dual UNION ALL
           SELECT 10 ID,    1 parent_id, 1000 VALUE FROM dual UNION ALL
           SELECT 20 ID,    1 parent_id, 2000 VALUE FROM dual UNION ALL
           SELECT 30 ID,   10 parent_id, 3000 VALUE FROM dual UNION ALL
           SELECT 40 ID,   10 parent_id, 4000 VALUE FROM dual UNION ALL
           SELECT 50 ID,   20 parent_id, 5000 VALUE FROM dual UNION ALL
           SELECT 60 ID,    1 parent_id, 6000 VALUE FROM dual UNION ALL
           SELECT 70 ID,   60 parent_id, 7000 VALUE FROM dual UNION ALL
           SELECT 80 ID,   70 parent_id, 8000 VALUE FROM dual
      ) SELECT CAST(LPAD(' ', (LEVEL-1)*4) || ID AS VARCHAR2(20))  id
          ,VALUE                                                   self_value
        , (SELECT SUM (VALUE)
           FROM   T t2
           CONNECT BY 
             PRIOR t2.ID = t2.parent_id
             START WITH ID = T.ID)                                 branch_value
      FROM   T
      CONNECT BY PRIOR t.id = t.parent_id
      START WITH t.parent_id IS NULL
      ORDER SIBLINGS BY t.id;
  
  ID                   SELF_VALUE BRANCH_VALUE
  -------------------- ---------- ------------
  1                                      36000
      10                     1000         8000
          30                 3000         3000
          40                 4000         4000
      20                     2000         7000
          50                 5000         5000
      60                     6000        21000
          70                 7000        15000
              80             8000         8000
  
  9 rows selected.
  

  I tried to reach the same result of this query using the new syntax for subquery factoring. Any help would be really appreciated!

  Hello

  I think it's one of those things that CONNECT BY is better.

  Here's a way to do it using a recursive clause (AND not CONNECT BY):

  WITH recursive_results (ancestor_id, descendant_id, value, lvl, lineage) AS

  (

  SELECT id AS ancestor_id

  id LIKE descendant_id

  value

  ,       1                     AS lvl

  , TO_CHAR (id, "9999") AS line

  T

  UNION ALL

  SELECT r.ancestor_id

  t.id AS descendant_id

  t.valeur

  r.lvl + 1 AS lvl

  r.lineage | ' /'

  || To_char (t.id, '9999') AS line

  T

  JOIN recursive_results r WE t.parent_id = r.descendant_id

  )

  SELECT LPAD (' ' ')

  , 4 * (

  SELECT MAX (lvl) - 1

  OF recursive_results

  WHERE descendant_id = m.ancestor_id

  )

  ) || ancestor_id AS indented_id

  SUM (CASE WHEN ancestor_id = descendant_id THEN value END) AS self_value

  The amount (value) AS branch_value

  OF recursive_results m

  GROUP BY ancestor_id

  ORDER BY)

  SELECT MAX (lineage) DUNGEON (DENSE_RANK LAST ORDER BY lvl)

  OF recursive_results

  WHERE descendant_id = m.ancestor_id

  )

  ;

  Output (even you have):

  INDENTED_ID SELF_VALUE BRANCH_VALUE

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

  1                                      36000

  10-1000-8000

  30 3000 3000

  40 4000 4000

  20-2000-7000

  50 5000 5000

  60 6000 21000

  70-7000-15000

  80-8000-8000

• The recursive subquery factoring vs hierarchical query

  Experts,
  
  Here it is two queries, I'm executing using hierarchical of recursive subquery and new classical approach. Problem came out was different for the recursive subquery factoring, as this is not displayed parent-child approach, while forwards connect shows parent-child mode. Query 1, I use hierarchical as show good output parent-child, is approaching the 2 query displays all of the nodes of level 1 and then level 2 nodes. I want the query 2 output to be the same as query 1. change query 2 as required.
  
  Note: the output of the two queries post as in sqlplus and toad. Please copy and use in your command prompt.
  
  QUERY 1:
  
  with the hand in the form (select 1 id, name 'John', null, mgrid Union double all the)
  Select 2 id, the name of 'michael', null, mgrid Union double all the
  Select 3 id, the name of "peter", null, mgrid Union double all the
  Select 4 id, the name of "henry", 1 mgrid Union double all the
  Select 5 id, "nickname", mgrid 2 Union double all the
  Select 6 id, "pao" name, mgrid 3 of all the double union
  Select 7 id, the name of 'kumar', mgrid 3 of all the double union
  Select 8 id, the name 'parker', mgrid 3 of all the double union
  Select 9 id, the name of "mike", 5 double mgrid),
  Select lpad (' ', 2 *(level-1)). name, key start with mgrid level is null connect by prior id = mgrid.
  
  OUTPUT:
  
  LEVEL NAME
  ------------------------------ ----------
  John 1
  Henry 2
  Michael 1
  Nick 2
  Mike 3
  Stone 1
  PAO 2
  Kumar 2
  Parker 2
  
  9 selected lines.
  
  QUERY 2:
  
  with the hand in the form (select 1 id, name 'John', null, mgrid Union double all the)
  Select 2 id, the name of 'michael', null, mgrid Union double all the
  Select 3 id, the name of "peter", null, mgrid Union double all the
  Select 4 id, the name of "henry", 1 mgrid Union double all the
  Select 5 id, "nickname", mgrid 2 Union double all the
  Select 6 id, "pao" name, mgrid 3 of all the double union
  Select 7 id, the name of 'kumar', mgrid 3 of all the double union
  Select 8 id, the name 'parker', mgrid 3 of all the double union
  Select 9 id, the name of "mike", 5 double mgrid),
  / * Select lpad (' ', 2 *(level-1)). name, key start with mgrid level is null connect by prior id = mgrid. * /
  secmain (id, name, mgrid, hierlevel) as (select id, name, mgrid, 1 hierlevel of the main where mgrid is null
  Union of all the
  Select m.id, $m.name, m.mgrid, sm.hierlevel + 1 in m main join secmain sm on(m.mgrid=sm.id))
  cycle is_cycle set id 1 default 0
  Select lpad (' ', 2 *(hierlevel-1)). name, secmain hierlevel.
  
  OUTPUT:
  
  NAME HIERLEVEL
  ------------------------------ ----------
  John 1
  Michael 1
  Stone 1
  Henry 2
  Nick 2
  Parker 2
  Kumar 2
  PAO 2
  Mike 3
  
  9 selected lines.

  For example

  SQL> with main as(select 1 id,'john' name,null mgrid from dual union all
    2  select 2 id,'michael' name,null mgrid from dual union all
    3  select 3 id,'peter' name,null mgrid from dual union all
    4  select 4 id,'henry' name,1 mgrid from dual union all
    5  select 5 id,'nick' name,2 mgrid from dual union all
    6  select 6 id,'pao' name,3 mgrid from dual union all
    7  select 7 id,'kumar' name,3 mgrid from dual union all
    8  select 8 id,'parker' name,3 mgrid from dual union all
    9  select 9 id,'mike' name,5 mgrid from dual),
   10  secmain (id,name,mgrid,hierlevel) as
   11  (select id,name,mgrid,1 hierlevel from main
   12   where mgrid is null
   13   union all
   14   select m.id,m.name,m.mgrid,sm.hierlevel+1 from main m join secmain sm on(m.mgrid=sm.id))
   15  search depth first by name set seq
   16  cycle id set is_cycle to 1 default 0
   17  select lpad(' ',2*(hierlevel-1))||name name,hierlevel from secmain order by seq;
  
  NAME        HIERLEVEL
  ---------- ----------
  john                1
    henry             2
  michael             1
    nick              2
      mike            3
  peter               1
    kumar             2
    pao               2
    parker            2
  
  9 rows selected.
  
  SQL> 
  

  Published by: Dom Brooks on November 23, 2011 13:52
  Edited for the scope of the search and detection of cycle

• I get e-mails bounce, which I have not sent. These emails contain a message and a link. Could I have a virus?

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  I use OS Yosemite 10.10.5. Thanking you in advance for your time.

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  I use Windows 7 Enterprise.
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  Following some help online suggestions on selection of Firefox's default mail client:

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  Original title: windows help and support links does not not correctly__

  my colleague tried to 'fix' my computer while I was on vacation and it reindexed. Now when I try to load the links help and support it loads Word instead of Explorer. How can I solve this problem?

  Hey dhender,

  Looks like the file association is messed up. Change the file association for help and Support and see if it works.

  Here's how to fix the associated file.

  a. Click Start and type default programs in the start search field.

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  d. find the .hlp file.

  e. Select .hlp file and click Change program.

  f. Select Windows Winhlp32 Stub list.

  This should be it.

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  Change the programs that Windows uses by default

  http://Windows.Microsoft.com/en-us/Windows-Vista/change-which-programs-Windows-uses-by-default

  If you are interested, you can look in this article as well.

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  Original title: problem e mail

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  Suggestions;

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  2. make sure that your Internet Explorer, or your main browser has all its faults.

  You can do both of the above by following these steps...

  Start button > right column, click default programs > click Set Your Default programs...

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  Download exefix_cu.reg attached at the end of this article and save it on the desktop. Right click on the REG file and chooseMerge.

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• Compilation of FDI: mismatched CPU target between the compiler and the linker

  Hello

  When I compile my project, it gives an error of link:

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  Looking at orders of compilation that I see the files objects created with - V4.6.3, gcc_ntox86_cpp and the linker, try using - V4.6.3, gcc_ntoarmv7le_cpp

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  I tried "Restore default" and change options around, but it didn't fix the problem.   Is this a bug?   How can I fix the "-V4.6.3, gcc_ntox86_cpp" question?   Is there something that I don't see?  Do I have to manually edit the .project files to fix the problem?

  The version of the IDE is

  Version: 10.1.0
  Build id: v201302012246

  Thank you!

  Solved my problem.   Is go to the cpp files individual and selected "resource Configurations"-> reset by default.   Not sure why she chose x 86 front and the change of target does not spread.