The recursive subquery factoring vs hierarchical query

Similar Questions

• ORA-30654: 'missing DEFAULT keyword' in the recursive subquery factoring

  The subquery recursive factoring works without clause CYCLE and cycle only.

  When I have a cycle (uncomment SELECT 5, 2 double UNON ALL) ORACLE detects the cycle properly, but when I include the clause CYCLE I get ORA-30654.

  I'm pretty shure that this statement already worked in my DB (11.2.0.3.0 - 64 bit)

  CREATE TABLE temp_rsq (a,b)
  AS
  SELECT 1, 2 FROM dual UNION ALL
  SELECT 2, 3 FROM dual UNION ALL
  SELECT 3, 4 FROM dual UNION ALL
  SELECT 4, 5 FROM dual UNION ALL
  --SELECT 5, 2 FROM dual UNION ALL
  SELECT 5, 6 FROM dual UNION ALL
  SELECT 6, 7 FROM dual;
  --
  --
  
  -- b-> new A
  WITH cte (  pb, a,    b,      weg) AS
  (SELECT   NULL, a,    b,        a
     FROM temp_rsq
    WHERE a=1
  UNION ALL
  SELECT cte.b, n.a a, n.b, weg+n.a
     FROM temp_rsq n
     JOIN cte
       ON (cte.b=n.a))
  SEARCH depth FIRST BY a SET abst
  --CYCLE a SET is_cycle to '1' DEFAULT '0'
  SELECT * FROM cte;
  
  
  
  

  Found last al:

  It does not with cusor_sharing = FORCE, only with the cursor sharing = TRUE!

  For me, a very strange behavior.

• 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

• Hierarchical data, how to aggregate over levels in hierarchical query?

  Hello
  
  I hope someone can help me.
  
  I held in a data table ("" what part was built in what other part of when when? "')
  ID parent_id build_in build_out
  1 NULL NULL NULL
  2/1 2010 2012
  3 2 2011 2013
  4 2 2013 NULL
  
  What are the parts is stored in a separate table.
  
  Now I want to know when when which part was built in the first level, in the example, I want to know that
  1 was simply a part of 1
  2 was part of 1 between 2010 and 2012
  3 was part of 1 between 2011 and 2012
  4 has never been a part of 1
  
  I tried several approaches - if there is a fixed number of levels between the types, I am interested I can do using joins and more/less (similarly as some nvl). Unfortunately this is not always the case (some parts appear on several levels).
  If I'm only interested in the parts that are never deleted, I can get by using a style of connect request and get the current configuration. But I can't seem to understand the time connecting part to the high level in such a query, or even filtering absurd combinations (like "4 in 1" in the example above). I could deal with the data recovered outside the database, but I prefer not.
  
  Is there a way to obtain the hierarchical data with an aggregate (min, max) for all levels?

  What version of Oracle you are on?

  In 11.2.x, you can use the recursive subquery factoring. Something like

  with t (id, parent_id, build_in, build_out) as (
  select 1, null, null, null from dual union all
  select 2, 1, 2010, 2012 from dual union all
  select 3, 2, 2011, 2013 from dual union all
  select 4, 2, 2013, null from dual
  )
  , c1 (root_id, id, parent_id, build_in, build_out) as (
  select id, id, parent_id, 0, 9999
  from t
  where parent_id is null
  union all
  select root_id, t.id, t.parent_id
  , greatest(nvl(t.build_in,0), nvl(b.build_in,0))
  , least(nvl(t.build_out,9999), nvl(b.build_out,9999))
  from c1 b, t
  where b.id = t.parent_id
  )
  select * from c1
  where build_in < build_out
  ;
  ROOT_ID ID    PARENT_ID  BUILD_IN  BUILD_OUT
  ------- ----- ---------- --------- ----------
  1       1                0         9999
  1       2     1          2010      2012
  1       3     2          2011      2012      
  

  With a hierarchical query by using the syntax connection, you could do something like

  select * from (
  select connect_by_root id as root, id
  , greatest(nvl(build_in,0), nvl(prior build_in,0)) as max_in, least(nvl(build_out,9999), nvl(prior build_out,9999)) as min_out
  from t
  start with parent_id is null
  connect by parent_id = prior id
  )
  where max_in < min_out
  ;
  

  but it is not powerful enough. This version compares the dates between a current and previous levels, but the recursive subquery is to compare the dates in the current level for the winners of the comparisons to the previous level. Not sure if it's an important distinction for your needs, however.

  If you are on 11.2 I advise to use the recursive subquery factoring. If this isn't the case, you can try the link by version.

  Kind regards
  Bob

• Getting the line without the use of hierarchical query values

  Hi all
  
  
  I want to know the hierarchical values without using a hierarchical query. I have two tables EMP, DEPT
  
  EMP table is to have two columns (empid, mgrid)
  DEPT table is to have two columns (deptid, empid)
  
  Data of the EMP
  
  1,
  2, 1
  3, 2
  4, 3
  
  Data DEPT
  
  10, 1
  
  Each time, I gave deptid = 10, I need to know the this deptid empid then who is this empid (child levels as well). In this case, the output should be
  1
  2
  3
  4
  
  I don't want to use hierarchical query.
  
  Thanks in advance.
  
  
  Thank you
  PAL

  You can use the RECURSIVE subquery, if you are 11 GR 2, like this

  SQL> with EMP (empid,mgrid) as
    2  (
    3  select 1,null from dual union all
    4  select 2, 1 from dual union all
    5  select 3, 2 from dual union all
    6  select 4, 3 from dual
    7  ),
    8  dept(deptid,empid) as
    9  (
   10  select 10, 1  from dual
   11  ),
   12  t(empid,mgrid) as
   13  (
   14  select empid,mgrid
   15  from emp e
   16  where empid in (
   17      select d.empid
   18      from dept d
   19      where deptid = 10
   20                  )
   21  union all
   22  select e.empid,e.mgrid
   23  from emp e, t
   24  where e.mgrid = t.empid
   25  )
   26  select *
   27  from t;
  
       EMPID      MGRID
  ---------- ----------
           1
           2          1
           3          2
           4          3
  

  Yet, the question is valid - why can't use you a hierarchical query?

  Published by: JAC on May 29, 2013 12:37

• recursive subquery

  I have a table with ID of origin and destination.
  There may be a dynamic number of connections (wait is no more than 5) and the relationship is still one by one: A-> B-> C-> D
  A or B or C or D can only apear in a relationship, this average-> C does not already connected to B e because.
  
  I want to create one interviewed who receives the last values of destination and a column with the original values that are associated with this return value:
  for example:
  If my parameter value is D, my result will be a column of three lines: A, B, C
  If my parameter value is C, my result will be a column with two lines: A, B
  
  example:

  create table test_list as (
  select 32000 origin, 68200 destination from dual
  union all
  select 60000 origin, 168200 destination from dual
  union all
  select 8200 origin, 36600 destination from dual
  union all
  select 36600 origin, 8400 destination from dual
  union all
  select 8400 origin, 61800 destination from dual
  )

  I tried to conect by front and connect as root but could not achieve.
  found also some articles on 'Recursive subquery factoring' but it got even worse because I couldn't make it work.
  My database is 'Oracle Database 11g Enterprise Edition Release 11.2.0.2.0 - 64 bit Production'
  
  

  SELECT origin   from test_list  where 
      origin in (select *  FROM (
       SELECT CONNECT_BY_ROOT destination
        FROM test_list
      CONNECT BY PRIOR origin = destination and destination =68200 )
  )

  So in my list:
  32000 connects to 68200
  60000 connects to 168200
  8200 connects to 36600
  36600 connects to 8400
  8400 connects to 61800
  
  My results are expected:
  If the parameter 'VALUE_TO_SEARCH' = 68200 I wait only: 32000
  If the parameter 'VALUE_TO_SEARCH' = 168200 I wait only: 60000
  If the parameter 'VALUE_TO_SEARCH' = 61800 I expect: 8400,36600,8200
  If the "VALUE_TO_SEARCH" parameter = 32000 I expect any results.
  
  What would be the best method to use for optimal performance and cash is with the CONNECT_BY_ROOT and "Se CONNECT BY FRONT" that I am I did wrong here?
  
  Best regards
  Ricardo Tomas

  Hello

  With the help of a WITH recursive clause:

  VARIABLE  value_to_search     NUMBER
  EXEC     :value_to_search := 168200;
  EXEC     :value_to_search := 61800;
  -- Only the last one above matters
  
  WITH     tree_results (origin, given_destination)     AS
  (
       SELECT  origin, destination
       FROM     test_list
       WHERE     destination     = :value_to_search
     UNION ALL
          SELECT  t.origin
       ,     r.given_destination
       FROM     test_list     t
       JOIN     tree_results  r  ON  r.origin  = t.destination
  )
  SELECT     *
  FROM     tree_results
  ;
  

• Recursive subquery in 11 GR 2

  I'm trying to learn more about the recursive subquery, but I'm stuck with this one. For each word, recursively remove 1, 2, 3... n characters of the word for all possible combinations. The word length is variable. For example, would be the expected result for the words 'abcd' and '123'. A PLSQL solution would be better?

  ABCD:

  1 deleted character: bcd, CDA, abd, abc

  2 deleted characters: cd, bd, BC., ad, ac, ab

  3 characters deleted: d, c, b, a

  4 removed characters: null

  123:

  1 deleted character: 12, 13 and 23

  2 deleted characters: 1, 2, and 3

  3 deleted characters: null

  Hello

  Here's a solution that makes everything you asked for, including characters removed (in order or appearance).

  As I used SYS_CONNECT_BY_PATH for a concatenated string of all of the included characters, so I used SYS_CONNECT_BY_PATH for a concatenated string of all characters between those included.

  WITH single_characters (c, n word) AS

  (

  SELECT Word

  ,       NULL                      AS c

  ,       1                         AS n

  The CBC

  UNION ALL

  SELECT Word

  SUBSTR (word, n - 1) c

  ,       n + 1                     AS n

  OF single_characters

  WHERE n<= length="">

  )

  SELECT Word

  REPLACE (SYS_CONNECT_BY_PATH (c, "ab")

  , 'ab '.

  )                       AS comb

  LENGTH (word) + 1 - LEVEL AS num_removed

  REPLACE (SYS_CONNECT_BY_PATH (SUBSTR (Word

  N ADVANCE

  , n - (n + 1 PREREQUISITE)

  )

  , "ba".

  )

  , "ba".

  ) || SUBSTR (Word, n) AS deleted

  OF single_characters

  START WITH n = 1

  CONNECT BY NOCYCLE Word = Word PREREQUISITE

  AND n > PREREQUISITE n

  AND LEVEL<= length="">

  ORDER BY word

  LEVEL DESC

  deleted

  ;

  Output:

  NUM_REMOVED DELETED WORD COMB

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

  123        23                   1 1

  123        13                   1 2

  123        12                   1 3

  123        3                    2 12

  123        2                    2 13

  123        1                    2 23

  123                             3 123

  ABCD bcd 1A

  ABCD DCO 1 b

  Abu abd 1 c

  ABCD abc 1 d

  AB cd 2 ABCD

  ABCD 2 ac comics

  ABCD BC 2

  ABCD ad 2 BC.

  ABCD ac 2 bd

  ABCD ab 2 cd

  ABCD d 3 abc

  ABCD c 3 abd

  ABCD b 3 CDA

  ABCD has 3 bcd

  ABCD ABCD 4

  lee        le                   1 e

  lee        le                   1 e

  lee        ee                   1 l

  lee        l                    2 ee

  lee        e                    2 le

  lee        e                    2 le

  lee                             3 lee

  With the exception of the order of the rows in the result, it's exactly what you asked in response #2.

  If the order of the rows is important, I'm sure we can get it exactly as you wish; just explain exactly how it needs to be sorted.

• How to write a hierarchical query so that only the child nodes are displayed?

  Hi all

  I have a hierarchical query that I use in an area of tree demand APEX and there are nodes that have no children and I am trying to find a way to not display these nodes. Essentially if the user does not have to develop a lot of knots to know that nothing exists at the most detailed level.

  The data are based on the Oracle Fusion FND tables but for example purposes here is enough data to illustrate my question:

  
  

  create table APPL_TAXONOMY_HIERARCHY (SOURCE_MODULE_ID varchar2(30), TARGET_MODULE_ID varchar2(30));
  create table APPL_TAXONOMY_TL (module_id varchar2(30), description varchar2(100), user_module_name varchar2(30), language varchar2(5));
  create table APPL_TAXONOMY (MODULE_ID    varchar2(30),    MODULE_NAME    varchar2(30), MODULE_TYPE varchar2(10),    MODULE_KEY varchar2(30));
  create table TABLES (table_name varchar2(30), module_key varchar2(30));
  
  

  insert into APPL_TAXONOMY_TL values ('1', null, 'Oracle Fusion', 'US' );
  insert into APPL_TAXONOMY_TL values ('2', null, 'Financials', 'US' );
  insert into APPL_TAXONOMY_TL values ('3', null, 'Human Resources', 'US' );
  insert into APPL_TAXONOMY_TL values ('20', null, 'Accounts Payable', 'US' );
  insert into APPL_TAXONOMY_TL values ('10', null, 'General Ledger', 'US' );
  
  insert into APPL_TAXONOMY_HIERARCHY values ('1', 'DDDDDDDD');
  insert into APPL_TAXONOMY_HIERARCHY values ('2', '1');
  insert into APPL_TAXONOMY_HIERARCHY values ('3', '1');
  insert into APPL_TAXONOMY_HIERARCHY values ('4', '1');
  insert into APPL_TAXONOMY_HIERARCHY values ('10', '2');
  insert into APPL_TAXONOMY_HIERARCHY values ('20', '2');
  
  insert into APPL_TAXONOMY values ('1', 'Fusion', 'PROD', 'Fusion');
  insert into APPL_TAXONOMY values ('2', 'Financials', 'FAMILY', 'FIN');
  insert into APPL_TAXONOMY values ('10', 'GL', 'APP', 'GL');
  insert into APPL_TAXONOMY values ('3', 'Human Resources', 'FAMILY', 'HR');
  insert into APPL_TAXONOMY values ('20', 'AP', 'APP', 'AP');
  
  insert into tables values ('GL_JE_SOURCES_TL','GL');
  insert into tables values ('GL_JE_CATEGORIES','GL');
  

  My hierarchical query is as follows:

  with MODULES as
  (
  SELECT h.source_module_id, b.user_module_name, h.target_module_id
        FROM APPL_TAXONOMY_HIERARCHY H,
        APPL_TAXONOMY_TL B,
  APPL_TAXONOMY VL
  where H.source_module_id = b.module_id
  and b.module_id = vl.module_id
  and vl.module_type not in ('PAGE', 'LBA')
  UNION ALL
  select distinct table_name, table_name,  regexp_substr(table_name,'[^_]+',1,1) 
  from TABLES --needed as a link between TABLES and APPL_TAXONOMY
  union all
  select module_key as source_module_id, module_name as user_module_name, module_id as target_module_id  from appl_taxonomy VL where VL.module_type = 'APP')
  SELECT  case when connect_by_isleaf = 1 then 0
              when level = 1             then 1
              else                           -1
         end as status,
  LEVEL,
  user_module_name as title,
  null as icon,
  ltrim(user_module_name, ' ') as value,
  null as tooltip,
  null as link
        FROM MODULES
        START WITH source_module_id = '1'
        CONNECT BY PRIOR source_module_id = target_module_id
  ORDER SIBLINGS BY user_module_name;
  

  In Oracle APEX, this gives a tree with the appropriate data, you can see here:

  https://Apex.Oracle.com/pls/Apex/f?p=32581:29 (username: guest, pw: app_1000);

  The SQL of the query results are:

  STATUS

  TITLE LEVEL

  VALUE

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

  	1	1 oracle Fusion	Oracle Fusion
  	-1	2 financial tables	Financials
  	-1	3 accounts payable	Accounts payable
  	0	4 AP	AP
  	-1	General Accounting 3	General Accounting
  	-1	4 GL	GL
  	0	5 GL_JE_CATEGORIES	GL_JE_CATEGORIES
  	0	5 GL_JE_SOURCES_TL	GL_JE_SOURCES_TL
  	0	2 human resources	Human resources

  The lowest level is the name of the table to level 5. HR is not any level under level 2, in the same way, "AP" (level 4) has nothing below, i.e. no level 5 and that's why I don't want to show these nodes. Is this possible with the above query?

  Thanks in advance for your suggestions!

  John

  Hello

  The following query will include only the nodes of level = 5 and their ancestors (or descendants):

  WITH modules LIKE

  (

  SELECT h.source_module_id

  b.user_module_name AS the title

  h.target_module_id

  To appl_taxonomy_hierarchy:

  appl_taxonomy_tl b

  appl_taxonomy vl

  WHERE h.source_module_id = b.module_id

  AND b.module_id = vl.module_id

  AND vl.module_type NOT IN ('PAGE', "LBA")

  UNION ALL

  SELECT DISTINCT

  table-name

  table_name

  , REGEXP_SUBSTR (table_name, ' [^ _] +')

  From the tables - required as a link between the TABLES and APPL_TAXONOMY

  UNION ALL

  SELECT module_key AS source_module_id

  AS user_module_name module_name

  module_id AS target_module_id

  Of appl_taxonomy vl

  WHERE vl.module_type = 'APP '.

  )

  connect_by_results AS

  (

  SELECT THE CHECK BOX

  WHEN CONNECT_BY_ISLEAF = 1 THEN 0

  WHEN LEVEL = 1 THEN 1

  OF ANOTHER-1

  The END as status

  LEVEL AS lvl

  title

  -, NULL AS icon

  , LTRIM (title, "") AS the value

  -, NULL as ToolTip

  -, Link AS NULL

  source_module_id

  SYS_CONNECT_BY_PATH (source_module_id - or something unique

  , ' ~' - or anything else that may occur in the unique key

  ) || ' ~' AS the path

  ROWNUM AS sort_key

  Modules

  START WITH source_module_id = '1'

  CONNECT BY PRIOR Source_module_id = target_module_id

  Brothers and SŒURS of ORDER BY title

  )

  SELECT the status, lvl, title, value

  -, icon, tooltip, link

  OF connect_by_results m

  WHEN THERE IS)

  SELECT 1

  OF connect_by_results

  WHERE the lvl = 5

  AND the path AS ' % ~' | m.source_module_id

  || '~%'

  )

  ORDER BY sort_key

  ;

  You may notice that subqueries modules and the connect_by_results are essentially what you've posted originally.  What was the main request is now called connect_by_results, and it has a couple of additional columns that are necessary in the new main request or the EXISTS subquery.

  However, I am suspicious of the 'magic number' 5.  Could you have a situation where the sheets you are interested in can be a different levels (for example, some level = 5 and then some, into another branch of the tree, at the LEVEL = 6, or 7 or 4)?  If so, post an example.  You have need of a Query of Yo-Yo, where you do a bottom-up CONNECT BY query to get the universe of interest, and then make a descendant CONNECT BY query on this set of results.

• 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

• hierarchical query of VO in the ofa page

  Hello

  I try to use the hierarchical query in VO for the display of all levels of supervisors for employee and want to display in populist:

  This is the query

  REPLACE SELECT distinct (mgx.full_name, "',' ') supervisor_full_name;

  XPP. Employee_number

  OF per_assignments_x pax,.

  per_people_x ppx,

  per_people_x mgx,

  per_positions pp,

  per_jobs pj,

  per_position_definitions ppd

  WHERE ppx.person_id = pax.person_id

  AND ppx.current_employee_flag = 'Y '.

  AND mgx.person_id = pax.supervisor_id

  AND pp.position_id = pax.position_id

  AND pj.job_id = pax.job_id

  AND ppd.position_definition_id = pp.position_definition_id

  START WITH ppx.employee_number =: 1

  CONNECT BY PRIOR MGX.employee_number = ppx.employee_number AND LEVEL < 6

  I'm trying to get this VO implemented PR when I run the page in jdeveloper with parameter binding, it's getting error.

  oracle.apps.fnd.framework.OAException: oracle.jbo.SQLStmtException: 27122 Houston: SQL error in the preparation of the statement.

  I would like to delete the parameter binding and executing CO VO past these two statements may be:

  START WITH ppx.employee_number =: 1

  CONNECT BY PRIOR MGX.employee_number = ppx.employee_number AND LEVEL < 6

  How can I do?

  Thank you

  MK

  has worked?

• Summarize the costs with hierarchical query

  Hello Oracle experts! I need some advice because I am very new to the hierarchical queries really new to Oracle. I have the following table:

  CREATE TABLE routes
  (
       from  VARCHAR2(15), 
       to  VARCHAR2(15), 
       cost NUMBER
  ); 
  INSERT INTO routes VALUES('San Francisco', 'Denver', 1000); 
  INSERT INTO routes VALUES('San Francisco', 'Dallas', 10000); 
  INSERT INTO routes VALUES('Denver', 'Dallas', 500); 
  INSERT INTO routes VALUES('Denver', 'Chicago', 2000); 
  INSERT INTO routes VALUES('Dallas', 'Chicago', 600); 
  INSERT INTO routes VALUES('Dallas', 'New York', 2000); 
  INSERT INTO routes VALUES('Chicago', 'New York', 3000);
  INSERT INTO routes VALUES('Chicago', 'Denver', 2000);
  

  I want to calculate the costs through the hierarchy, to get the following result:

  FROM            TO             COST
  --------------- --------------- -----
  San Francisco Dallas   10000   //San Francisco -> Dallas
  San Francisco Denver 1000    //San Francisco -> Denver
  San Francisco Chicago   10600   //San Francisco -> Dallas -> Chicago (10000 + 600)
  San Francisco New York   12000   //San Francisco -> Dallas -> New York (10000 + 200)
  San Francisco Chicago   3000    //San Francisco -> Denver -> Chicago (1000 + 2000)
  San Francisco Dallas   1500    //San Francisco -> Denver -> Dallas (1000 + 500) 
  etc..
  
  

  I from imagined using the CONNECT BY PRIOR statement to get the hierarchy and have written a query that runs through him:

  SELECT CONNECT_BY_ROOT from, to
   FROM routes 
    CONNECT BY NOCYCLE PRIOR to = from;
  

  But I have absolutely no idea how summarize right costs, I could use some help.

  Thank you for your advice.

  Hello

  Here's a way

  SELECT DISTINCT

  CONNECT_BY_ROOT from_city AS from_city

  to_city

  , SYS_CONNECT_BY_PATH (to_city, '->') is ARRESTED - if wanted

  , XMLQUERY (SYS_CONNECT_BY_PATH (cost, '+'))

  BACK CONTENT

  ) .getnumberval () SUCH as total_cost

  CHANNELS

  WHERE CONNECT_BY_ROOT from_city <> to_city

  CONNECT BY NOCYCLE from_city = to_city PRIOR

  ORDER BY from_city

  to_city

  stops

  ;

  FROM and TO are the keywords of the Oracle, they really ugly column names.  I have changed the from_city and to_city.

• Get the value of the root of a hierarchical query

  I'm writing a hierarchical query for one of my paintings.
  
  I need to know the value of the root by entering a value of sheet of any level.
  
  for example: I enter a value which is at level 4. I need to find the value of the root without displaying the other levels.
  
  Help, please.
  
  Thank you.

  TRY THIS,

  SELECT     ename, PRIOR  ename MGRNAME
        FROM emp
       WHERE ename = 
  START WITH mgr IS NULL
  CONNECT BY PRIOR empno = mgr
  

  Published by: NSK2KSN on August 9, 2010 18:10

• Hierarchical query, differentiate between the leafs and nodes?

  Hello
  
  I started coding a small program PL/SQL (not finished) with a hierarchical query:
  

  DECLARE
  
  CURSOR cur IS
  select 
    op_id||' '||op_descript as description, level
  from operations
  connect by prior op_id = op_parent_op_id
  start with op_id = 0;
  
  BEGIN
  
  HTP.prn('<ul id="arbre2" class="filetree">');
    for c in cur loop
      HTP.prn('<li><span class="folder">'||c.description||'</span></li>');
    end loop;
  
    HTP.prn('</ul>');
  END;

  This program must return a hierarchical HTML code like this (he does not at the moment):
  

  <ul id="arbre" class="filetree">
   <li><span class="folder">Folder 1</span>
    <ul>
     <li><span class="file">Item 1.1</span></li>
    </ul>
   </li>
   <li><span class="folder">Folder 2</span>
    <ul>
     <li><span class="folder">Subfolder 2.1</span>
      <ul>
       <li><span class="file">File 2.1.1</span></li>
       <li><span class="file">File 2.1.2</span></li>
      </ul>
     </li>
     <li><span class="file">File 2.2</span></li>
    </ul>
   </li>
   <li class="closed"><span class="folder">Folder 3 (closed at start)</span>
    <ul>
     <li><span class="file">File 3.1</span></li>
    </ul>
   </li>
   <li><span class="file">File 4</span></li>
  </ul>

  then jQuery will make a tree.
  
  But I don't know how to differentiate leafs and nodes in the request, to allow me to specify a class = 'file' or class = 'file '.
  
  Does anyone have a solution?
  
  A more general question might be: how to generate a hierarchical HTML code exactly as it is above my table operations (op_id, op_parent_op_id, descript).
  
  Thank you.
  
  Yann.

  Discover the CONNECT_BY_ISLEAF pseudo-column. Returns 1 if it is a leaf and zero otherwise.

  HTH!

• How to convert the hierarchical query of SQL Server (CTE) to Oracle?

  How to convert the hierarchical query of SQL Server (CTE) to Oracle?
  
  WITH cte (col1, col2) AS
  (
  SELECT col1, col2
  FROM dbo. [tb1]
  WHERE col1 = 12
  UNION ALL
  SELECT c.col1, c.col2
  FROM dbo. [tb1] AS c INNER JOIN cte AS p ON c.col2 = p.col1
  )
  DELETE one
  FROM dbo. [tb1] AS an INNER JOIN b cte
  ON a.col1 = b.col1

  Hello
  Something like this maybe:

  DELETE FROM dbo.tb1 a
   WHERE EXISTS (
    SELECT 1
      FROM dbo.tb1 b
    WHERE a.co11 = b.col1
        AND a.col2 = b.col2
     START WITH b.col1 = 12
    CONNECT BY b.col2 = PRIOR b.col1)
  

  Although you need to do here is to check that CONNECT it BY SELECT, returns records you wait first, then the DELETION should work too.