CONNECT by PRIOR Clause

Similar Questions

• Hi all of that I'm using a connection by prior clause

  Hi all

  In my connect by prior clause, I have a request. I want to pass the values to the query of my outer query. Is this possible in Oracle.Can all help me in this. Also paste my request for your reference

  SELECT rct2.trx_number

  rctl2.customer_trx_line_id

  fifs.flex_value_set_id

  gcc.code_combination_id

  ffvc.child_flex_value

  ffvc.parent_flex_value

  gcc.segment4

  ffvc.description

  ffvc.lvl

  OF ra_customer_trx_all rct2

  ra_customer_trx_lines_all rctl2

  ra_cust_trx_line_gl_dist_all rctlgd

  gl_code_combinations gcc

  (SELECT ffvv.flex_value_set_id

  ffvv.flex_value child_flex_value

  , LEVEL lvl, ffvv.parent_flex_value

  ffvv.description

  OF fnd_flex_value_children_v ffvv

  WHERE 1 = 1

  START WITH ffvv.flex_value = (SELECT

  GCC.segment4

  OF ra_customer_trx_all rct1,.

  ra_customer_trx_lines_all rctl1,

  ra_cust_trx_line_gl_dist_all rctlgd,

  gl_code_combinations gcc

  WHERE rct1.customer_trx_id = rctl1.customer_trx_id

  AND rctl1.customer_trx_line_id = rctlgd.customer_trx_line_id

  AND rct1.customer_trx_id = rctlgd.customer_trx_id

  AND rctlgd.account_class = 'REV '.

  AND rctlgd.code_combination_id = gcc.code_combination_id

  AND rct1.trx_number =: trx_number - want to move here from the rct2.trx_number that comes from the outer query

  AND rctla.customer_trx_line_id =: trx_line_id)-to move rctl2.customer_trx_line_id here which comes from the outer query

  Ffvv.parent_flex_value CONNECT BY PRIOR = ffvv.flex_value) ffvc

  s fnd_flex_value_sets

  fnd_id_flex_segments FIFS

  gl_ledgers l

  WHERE rct2.customer_trx_id = rctl2.customer_trx_id

  AND rctl2.customer_trx_line_id = rctlgd.customer_trx_line_id

  AND rct2.customer_trx_id = rctlgd.customer_trx_id

  AND rctlgd.code_combination_id = gcc.code_combination_id

  AND fifs.flex_value_set_id = ffvc.flex_value_set_id

  AND fifs.application_id = 101

  AND fifs.flex_value_set_id = s.flex_value_set_id

  AND fifs.application_column_name = 'SEGMENT4. '

  AND fifs.id_flex_code = ' GL #

  AND l.chart_of_accounts_id = fifs.id_flex_num

  AND l.ledger_id = rct2.set_of_books_id

  AND rctlgd.account_class = 'REV '.

  AND rctl2.customer_trx_line_id =: trx_line_id - spend Trx id

  AND rct2.trx_number =: trx_number - number of trx Pass

  Hi all

  I have used a workaround to achieve this.

  I created a function and added the query above to function and the values passed as parameters in two places

  Thank you

• Clarification regarding the understanding of the ORACLE CONNECT BY PRIOR Clause

  Dear all,
  
  I'm trying to understand the ORACLE CONNECT BY CLAUSE, and I wrote a query to check my understanding.
  
  
  I wrote the below two queries.
  
  Select the level, lpad ('* ', 2 * (level - 1),'* '). t_ename Ename, empno, sys_connect_by_path(ename,'/') enames, Bishop of PEM
  -where empno = empno
  Start by empno = 7654
  -start with mgr is null
  Connect by prior empno = mgr;
  
  LEVEL T_ENAME ENAMES EMPNO, MGR
  ---------- -------------------- -------------------------------------------------- ---------- ----------
  1 MARTIN 7654 7698 /MARTIN
  
  Explanation: Start by EMPNO = 7654 here means, the root node is 7654. Connect BY PRIOR EMPNO = average MGR-> for the empno = 7654 are all the people under him. IE. Mgr with 7654.
  
  Select the level, lpad ('* ', 2 * (level - 1),'* '). t_ename Ename, empno, sys_connect_by_path(ename,'/') enames, Bishop of PEM
  -where empno = empno
  Start by empno = 7698
  -start with mgr is null
  Connect by prior empno = mgr
  
  LEVEL T_ENAME ENAMES EMPNO, MGR
  --------- -------------------- -------------------------------------------------- ---------- ----------
  1 BLAKE 7698 7839 /BLAKE
  2 * ALLEN/BLAKE/ALLEN 7499 7698
  2 * 7521 7698 WARD/BLAKE/WARD
  2 * MARTIN/BLAKE/MARTIN 7654 7698
  2 * TURNER, BLAKE/7844 7698 TURNER
  2 * JAMES BLAKE/JAMES 7900 7698
  
  Explanation: Start by EMPNO = 7698 here means, the root node is 7698. Connect BY PRIOR EMPNO = average MGR-> for the empno = 7698 are all the people under him. IE. Mgr with 7698.
  
  
  I UNDERSTAND TO CONNECT BY PRIOR IS CORRECT?
  
  Please correct me if I'm wrong.
  
  Thank you
  MK.

  Yes, you're right

  Kind regards
  Sayan M.

• Rebuild the parent/child (connect by prior and where clause)

  Hi guys, you have a quick question for you - code as follows:
  

  
  

  with t1 as)

  Select 1 id, 'Name1' some_name, null, 'Y' parent_id of double some_flag

  Union select 2 id, 'Name2', 1 parent_id, 'Y' some_flag of the double

  Union select 3 id, "Name3", 2 parent_id, "n" of the double some_flag

  Union select id 4, 'Name4.1', 3 parent_id, 'Y' some_flag of the double

  Union select id 5, 'Name4.2', 3 parent_id, 'Y' some_flag of the double

  Union select id 6, 'Name4.3', 3 parent_id, 'Y' some_flag of the double

  Union select id 7, "Name5", 6 parent_id, 'Y' some_flag of the double

  )

  SELECT id, the_name, parent_id, null new_parent_id

  de)

  SELECT id, lpad (' ', (level 1) * 3) | some_name the_name, parent_id

  from t1

  where some_flag = 'Y '.

  Start with id = 1

  connect by parent_id = prior id

  brothers and sisters of order by some_name

  )

  Output:

  ID, THE_NAME, PARENT_ID, NEW_PARENT_ID

  1, Name1,

  2, name2, 1,.

  4, Name4.1,3,

  5, Name4.2,3.

  6, Name4.3, 3,.

  7, name5, 6,.

  As you can see, id = 3 is not displayed (because of where some_flag = 'Y'), but parent_id for id in (4,5,6) shows 3.

  -Only by using SQL - display 'current parent in valid results' as new_parent_id?

  In this example that would be showing new_parent_id = 2 for the id in (4,5,6)

  Currently on Oracle 11 g 1 material.

  Similar theme (manipulate path), but seems to work with the brothers and SŒURS of ORDER BY.

  SELECT id, the_name, parent_id, some_flag, new_parent_id

  FROM (SELECT id, lpad (' ', (LEVEL - 1) * 3): some_name the_name,)

  some_flag, parent_id,

  REGEXP_SUBSTR)

  REGEXP_SUBSTR)

  SYS_CONNECT_BY_PATH)

  DECODE (some_flag, 'Y', id), ' / ').

  '[0-9]+[/]+[0-9]+$'),

  (0-9] +') new_parent_id

  FROM t1

  WHERE some_flag = 'Y '.

  START WITH id = 1

  CONNECT BY PRIOR ID = parent_id

  ORDER OF brothers and SŒURS some_name);

• Connect by PRIOR vrs Connect by LEVEL

  Hello

  I'm trying to understand the difference between CONNECT BY PRIOR and CONNECT BY LEVEL

  I know that CONNECT BY PRIOR is basically specifying a hierarchy and which column / field is the parent (using the PRIOR)

  for example

  SELECT employe_id, employee_name, level

  Employees

  CONNECT BY PRIOR employee_id = manager_id;

  
  Q1. What is CONNECT BY LEVEL (for example as shown below) practice? What type of relationship normally described? (below, it is just producucing a series of integers)

  SELECT the level

  OF the double

  CONNECT BY LEVEL < = 10;

  T2. Use LEVEL with CONNECT BY in this way, a subversive use of the CONNECT IN knowledge is really intended to be used this way?

  any advice appreciated,

  Jim

  Hi, Jim,.

  Jimbo wrote:

  Hello

  I'm trying to understand the difference between CONNECT BY PRIOR and CONNECT BY LEVEL

  I know that CONNECT BY PRIOR is basically specifying a hierarchy and which column / field is the parent (using the PRIOR)

  for example

  SELECT employe_id, employee_name, level

  Employees

  CONNECT BY PRIOR employee_id = manager_id;

  Sometimes, CONNECT BY using hierarchical data, such as a tree.  In these cases, it may be useful to use the terns as "parent" and "child."  The query above is one of these cases.

  In more general terms, CONNECT BY returns the union (UNION ALL, specifically) of lines found by the START WITH clause (LEVEL = 1) and lines found by the clause CONNECT BY (LEVEL > 1).  If there is no START WITH clause, every row in the table will appear on LEVEL = 1.

  If there is no line on LEVEL = N, then Oracle research lines that meet the CONNECT BY clause, to search for the lines that are connected to each line level = N, for put the LEVEL N + 1.  If the CONNECT BY clause is set to TRUE for a given row, while the rank will appear on level N + 1 =.  Often (but not always) the CONNECT BY clause uses the PRIOR operator.  After PRIOR consent refers to something = N level.

  Q1. What is CONNECT BY LEVEL (for example as shown below) practice? What type of relationship normally described? (below, it is just producucing a series of integers)

  SELECT the level

  OF the double

  CONNECT BY LEVEL<>

  Since there is no START WITH clause, each line of the double table will be LEVEL = 1.

  The CONNECT BY clause means that CONNECT BY always means: If there is a line at = N, then it will be connected to any line that fills the CONNECT BY conditions.

  So that the lines will be added to the level = 2?  All the lines where condition 2<= 10="" is="" true,="" so="" every="" row="" in="" the="" dual="" table="" will="" be="" on="" level="">

  What lines will be added to the level = 3?  All the lines where condition 3<= 10="" is="" true,="" so="" every="" row="" in="" the="" dual="" table="" will="" be="" on="" level="">

  ...

  What lines will be added to the level = 10?  All the lines where the condition of 10<= 10="" is="" true,="" so="" every="" row="" in="" the="" dual="" table="" will="" be="" on="" level="10." 10="" is="" true,="" so="" every="" row="" in="" the="" dual="" table="" will="" be="" on="" level="">

  What lines will be added to the level = 11?  All the lines where condition 11<= 10="" is="" true,="" so="" no="" row="" in="" the="" dual="" table="" will="" be="" on="" level="">

  Since no rows are added to the LEVEL = 11, CONNECT BY query stop right there.

  Q2. Use LEVEL with CONNECT BY in this way, a subversive use of the CONNECT IN knowledge is really intended to be used this way?

  You can think of it as being subversive.  It was certainly not how planners and the deveopers at Oracle for CONNECT BY work in versions 2 to 8; you got an error if you tried in these versions.  It is also an exception to the rule that no line can be his own ancestor.

• START WITH and CONNECT BY PRIOR

  Hello

  Database: Oracle 11g

  1.

  SELECT ename, empno, mgr

  WCP

  START WITH empno = 7839

  CONNECT BY PRIOR MGR = EMPNO

  Result set:

  EMPNO, ENAME MGR

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

  KING 7839

  2.

  SELECT empno.ename, Bishop

  WCP

  START WITH mgr = 7839

  CONNECT BY PRIOR MGR = EMPNO

  Result set:

  EMPNO, ENAME MGR

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

  7566 JONES 7839

  7698 BLAKE 7839

  7782 CLARK 7839

  KING 7839

  KING 7839

  KING 7839

  My questions are:

  Q1. What is actually happening in the result defines two queries. I'm not able to grasp the difference when I use START WITH empno = 7839 and START WITH mgr =. 7839

  Q2. What is the difference between

  CONNECTION BY MGR PRIOR = EMPNO and

  CONNECT BY PRIOR EMPNO = MGR?

  can someone please help me here?

  Thank you

  Hello

  A CONNECT BY query looks like an operation UNION ALL of the data of different levels, numbered 1, 2, 3 and more.

  Level 1 is filled with the START WITH clause.

  If there are data on level N, then N + 1 level is filled, using the CONNECT BY clause, which generally refers to something on the N level via the PRIOR operator.  Another way to put it is that level N + 1 is filled by a self-join with lines that have already chosen the level N.

  If there is no data on the level of N, the query stops.

  Let's see how this applies to your queries.

  Level being such an important concept in CONNECT BY queries, you might want to see in all your CONNECT BY queries all test and debug the.

  1 query, including the level are:

  SELECT ename, empno, mgr

  LEVEL

  FROM scott.emp

  START WITH empno = 7839

  Empno = mgr PRIOR CONNECTION

  ;

  You will notice that I have re-arranged the CONNECT BY clause.  I find it a little more clear medium.  Of course, it never changes the results just if you say "x = y" or "y = x.

  The results, including the level, are:

  LEVEL OF ARCHBISHOP EMPNO, ENAME

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

  7839 KING 1

  What happened to produce these results?

  First level 1 has been met using the START WITH clause.  Level 1 is identical to the results of

  SELECT ename, empno, mgr

  AS LEVEL 1

  FROM scott.emp

  WHERE empno = 7839 - same as your START WITH clause

  ;

  It happens to be only 1 row in the table scott.emp who met the empno = 7839 condition, and we show a few columns of this line.

  That's all that need the level 1.  Something has been on level 1, so we're trying now to complete level 2, using the CONNECT BY condition.

  Any line that is included in the level 2 meets the empno = mgr PREREQUISITE condition, where the PREVIOUS operator refers to a line of level 1.  In this case, there is only 1 row at level 1, this line gets to have a NULL mgr.  Given that PRIOR mgr is NULL in this case, the condition to connect BY

  EmpNo = mgr BEFORE equals

  EmpNo = NULL and who obviously won't be true for any line, so nothing is added to level 2, and ends the query.

  Now let's look at application 2.  I'll add another column of debugging, called path, which I'll describe later:

  SELECT ename, empno, mgr

  LEVEL

  , SYS_CONNECT_BY_PATH (empno, "/") AS path

  FROM scott.emp

  START WITH mgr = 7839

  CONNECT BY PRIOR Mgr = empno

  LEVEL CONTROL

  path

  ;

  Output:

  EMPNO, ENAME MGR LEVEL PATH

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

  7566 7839 1 7566 JONES

  7698 7839 1 7698 BLAKE

  7782 7839 1 7782 CLARK

  7839 KING 2/7566/7839

  7839 KING 2/7698/7839

  7839 KING 2/7782/7839

  Again, we'll study how people got 1 level.  It happens to be 3 scott.emp lines that meet this condition START WITH, so there are 3 lines in the game at level 1.

  Given that the data on the level 1, the test of the query to complete level 2, referring to some PRIOR line on level 1.  Any line that meets the condition to connect BY, with a line any level 1 in the PREVIOUS line, will appear at level 2.

  Let's look at the line at level 1 where ename = 'JONES '.  Are there lines in sccott.emp that met the empno = mgr PREREQUISITE condition, where mgr PREREQUISITE is the column of Archbishop of the line with "JONES"?  Yes, there are one, the line with ename = 'KING', so that the rank is included at level 2.

  Let's look at the line at level 1 where ename = 'BLAKE '.  Are there lines in sccott.emp that met the empno = mgr PREREQUISITE condition, where mgr PREREQUISITE is the column of Archbishop of the line with "BLAKE"?  Yes, there are one, the line with ename = 'KING', so that the rank is included at level 2.

  Let's look at the line at level 1 where ename = 'CLARK '.  Are there lines in sccott.emp that met the empno = mgr PREREQUISITE condition, where mgr PREREQUISITE is the column of Archbishop of the line with 'CLARK '?  Yes, there are one, the line with ename = 'KING', so that the rank is included at level 2.

  There are thus 3 rows at level 2.  They happen to all be on the same line of the table emp; It is correct.  Remember, CONNECT BY is like a UNION ALL (not just a UNION).  It is a UNION of

  lines that are at level 1, because him meets the condition to BEGIN WITH, and

  lines that are at level 2 because puts it CONNECT BY condition regarding the 'JONES', and

  lines that are at level 2, because they meet the condition to connect BY regarding the "BLAKE", and

  lines that are at level 2, because they meet the condition to connect BY regarding the "CLARK".

  SYS_CONNECT_BY_PATH can enlighten us on that.  SYS_CONNECT_BY_PATH (empno, ' / ') shows the empno of each level that caused this line appears in the result set.  It's a delimited list /, where the nth element (i.e. the empno after bar oblique nth) is the empno who found the N level.

  Since there were data at level 2, the quert now trying to complete level 3.

  Is there all the rows in the table that satisfy the CONNECT BY condition (mgr PRIOR = empno) with respect to any line level 2?  No, Bishop is be NULL on all lines of level 2, so no line can satisfy this condition CONNECT BY, no lines are added at level 3, and ends the query.

  I hope that answers the question:

  Q1. What is actually happening in the result defines two queries. I'm not able to grasp the difference when I use START WITH empno = 7839 and START WITH mgr =. 7839

  I'll try to not be so detailed answering

  Q2. What is the difference between

  CONNECTION BY MGR PRIOR = EMPNO and

  CONNECT BY PRIOR EMPNO = MGR?

  These 2 CONNECT BY conditions are different where you put the PRIOR operator.  The operator PRIOR to switching as it change the direction, upward or downward, which move you through the tree you get from level to level.

  Bishop PRÉALABLE = empno means the employee on level N + 1 will be the same as the Manager of level N.  This means that higher level numbers will be the most senior people in the hierarchy.  This is called a query from the bottom up.  (Both of the queries that you have posted this same CONNECT BY exact state; both are requests from bottom to top).

  Mgr = empno PREREQUISITE or equivalent

  PRIOR empno = mgr means exactly the opposite.  When you move from level N to level N + 1 in the query, you will move to an older person, to a junior position in the hierarchy.  This is called a query from top to bottom.  The employee level N will be the Manager of wover is a level N + 1.

• Join the two trees connect by prior Start With and return only common records?

  Oracle 10g Release 2 (10.2)
  
  I have two tables which have structured data. The results, when running queries individually are correct, but I need to join tree a tree two to get only the common records between them.
  
  -Trees a
  SELECT ip_entity_name, entity_code, hier_level, entity_parent
  Of ip_hierarchy
  WHERE hier_level > = 3
  CONNECT BY PRIOR Entity_code = entity_parent
  START WITH entity_code = "MEWWD";
  
  -The two tree
  SELECT ip_entity_name, entity_code, hier_level, entity_parent
  Of ipt_hierarchy
  WHERE hier_level > = 3
  CONNECT BY PRIOR Entity_code = entity_parent
  START WITH entity_code = "IPNAM";
  
  
  If I understand correctly, the joints can not work with CONNECT BY / START WITH queries?
  
  A WITH clause is an option?
  
  If possible, I don't want to put a selection in a database to display object and join against other queries.
  
  Thank you.

  Hello

  jtp51 wrote:
  Oracle 10g Release 2 (10.2)
  ...
  If I understand correctly, the joints can not work with CONNECT BY / START WITH queries?

  Before Oracle 9 it was true. Since you're using Oracle 10, you can if you wish; but I'm guessing that you don't want in this case.

  A WITH clause is an option?

  If possible, I don't want to put a selection in a database to display object and join against other queries.

  Yes, a WITH clause that is an option. Viewed online, as Zhxiang has shown, are another option. Either way gives you a regular display effect without creating a database object.

  You did not show a sample and the results, so no one can tell if a join is really what you want. Other possibilities include INTERSECT or an IN subquery.

• question of the practical exam 1Z0-047 - CONNECT BY PRIOR

  Hello
  Please see the question by clicking on the link:
  
  http://S10.postimg.org/58ph01n5l/1z0_047_connectby.PNG
  
  I échinés on it (well, for a few minutes or more) because I don't think that the answers are correct.
  Option 3 is almost correct, except that in my mind the CONNECT BY PRIOR is backwards.
  Option 4 is almost correct, except it takes LEVEL < = 5 rather than LEVEL < 5.
  
  In the end, I chose option 3 because it is more certain that the LEVEL should be < = 5 I had CONNECT it BEFORE the wrong way round.
  
  Now, it so is that I chose the "correct" answer but I am mistaken on option 3? Option 3 is really correct?
  My reasoning is this: we want the ancestors of Peter. This means that Peter's parents. CONNECTION BY ParentID = PRIOR PersonID will begin at the root (Peter Clark), and the next record found (using this SQL) will have ParentID = PersonId of pre-registration, namely, of Peter Clark PersonID. In other words, Peter Clark will be parent of the next record. But we are looking for ancestors of Peter Clark; Ancestors of Peter Clark are the parents of Peter Clark. Peter Clark cannot be the parent of the ancestor (or ancestors).
  
  I'm here or not?
  
  Thank you
  Jason

  I agree with you. The connection is evil around all 4 options.
  I set up a test with grandparents etc and connect ParentId = PersonId prior only got the start record.
  Connect prior ParentId = PersonId gives the correct descent.

  create table gr (parentid integer, personID integer, name varchar2(30));
  insert into gr values (null,1,'GGG');
  insert into gr values (1,2,'GGF');
  insert into gr values (null,3,'GGM');
  insert into gr values (2,4,'GF');
  insert into gr values (4,5,'F');
  insert into gr values (5,6,'Peter');
  select * from gr order by parentid, personid;
  
    PARENTID   PERSONID NAME
  ---------- ---------- ------------------------------
           1          2 GGF
           2          4 GF
           3          4 GF
           4          5 F
           5          6 Peter
                      1 GGG
                      3 GGM
  
  select level, gr.* from GR where level <= 5
  start with name = 'Peter'
  connect by ParentId = Prior PersonId
  Order Siblings by name;
  

  Output has only Peter:

       LEVEL   PARENTID   PERSONID NAME
  ---------- ---------- ---------- ------------------------------
           1          5          6 Peter
  
  1 row selected.
  

  The reverse

  select level, gr.* from GR where level <= 5
  start with name = 'Peter'
  connect by Prior ParentId = PersonId
  Order Siblings by name;
  
       LEVEL   PARENTID   PERSONID NAME
  ---------- ---------- ---------- ------------------------------
           1          5          6 Peter
           2          4          5 F
           3          2          4 GF
           4          1          2 GGF
           5                     1 GGG
           3          3          4 GF
           4                     3 GGM
  
  7 rows selected.
  

• Connect by prior and leaves...

  Hi gurus...
  I have real difficulty understanding connect by prior and leaves...
  Is there a good example that you can suggest?

  Hello

  Tina wrote:
  Hi gurus...
  I have real difficulty understanding connect by prior and leaves...

  What is "connect per sheet? Post an example.
  Did you mean SHEETISCONNECT_BY_? It is a pseudo-column, documented in the manual of the SQL language:
  http://download.Oracle.com/docs/CD/B28359_01/server.111/b28286/pseudocolumns001.htm#sthref796

  Is there a good example that you can suggest?

  Make your choice:

  http://www.adp-GmbH.ch/ora/SQL/connect_by.html
  http://www.oradev.com/connect_by.jsp
  http://www.oraclepassport.com/connect%20BY%20PRIOR.html

• Connect by prior request

  I know that the query to run the child in the product, but I want to get the products given the child node parent...
  
  can someone request for...
  
  for example
  
  Select
  LPAD (' ', (level - 1) * 2) | product_name as product_name,
  product_id,
  parent_id,
  level
  products
  Connect prior product_id = parent_id
  Start by product_id = 11123;
  
  the output is:
  11123
  -24545
  -67676
  but my requirement is if I give
  24545
  output must be
  11123
  24545
  
  entry = 67676
  
  ouput
  11123
  -24545
  -67676

  select
  lpad(' ', (level - 1) * 2) || product_name as product_name,
  product_id ,
  parent_id ,
  level
  from products
  connect by prior parent_id = product_id
  start with product_id = 11123;
  

  SY.

• connect by prior (hierarchical query)

  Hi all
  
  Some1 asked me a question that goes like this:
  
  Source
  
  emp_no dep_no
  10 110
  20 110
  30 110
  40 110
  10 120
  10 130
  130 20
  
  write a query to get the following output from the source above
  
  emp_no dept_no
  110 10203040
  10 120
  130 1020
  
  Now I stumbled upon solutions with the terms "connect by front" but I'm nt able to understand how oracle produces this result, could someone point me to a good article or a text that can explain this concept very well (I searched on google and have seen many articles, but I could not able to understand since these articles were not all explain)
  
  Concerning
  Rahul

  Hi, Rahul,

  Welcome to the forum!

  What you want to do is called chain aggregation . This page shows the different ways to do it:
  http://www.Oracle-base.com/articles/10G/StringAggregationTechniques.php

  SYS_CONNECT_BY_PATH (which requires a CONNECT BY query) is just a medium. (From Oracle 11.2, LISTAGG is clearly the best way to string aggregation. Before that, SYS_CONNECT_BY_PATH seems to be the most popular, especially if the results must be in order.)

  The following pages are introductions to CONNECT BY queries:
  http://www.adp-GmbH.ch/ora/SQL/connect_by.html
  http://www.oradev.com/connect_by.jsp

• Queries on hierarchical data

  
  Hi all

  I have it here is the scenerio

  The entry like:

  power outlet	new_outlet
  one	b
  b	c
  c	d
  d	one
  AA	BB
  BB	CC
  CC	AA
  l	m
  m	n
  x	There
  There	x

  expected results will be:

  one	d
  b	d
  c	d
  AA	CC
  BB	CC
  l	n
  m	n
  x	There
  There	x

  Please help me on this. I tried with you connect by prior clause but not able to get the exact output.

  Thank you very much

  I lost the create table statement, but if you create a table "T" with the data that you have posted:

  with case_2 as (
    select outlet, new_outlet, connect_by_root(new_outlet) root_outlet
    from t
    start with new_outlet not in (select outlet from t)
    connect by new_outlet = prior outlet
  ), case_3 as (
    select outlet, new_outlet root_outlet from t
    where outlet < new_outlet
    and (outlet, new_outlet) in (
      select new_outlet, outlet from t
    )
  ), case_1_start as (
    select outlet, new_outlet from t
    minus
    select outlet, new_outlet from case_2
    minus
    select * from case_3
    minus
    select root_outlet, outlet from case_3
  ), case_1 as (
    select outlet, min(connect_by_root(new_outlet)) root_outlet
    from case_1_start a
    start with outlet > new_outlet
    connect by nocycle new_outlet = prior outlet
    group by outlet, new_outlet
  )
  select outlet, root_outlet from case_1
  union all
  select outlet, root_outlet from case_2
  union all
  select outlet, root_outlet from case_3
  order by 2,1;
  

  OUTLET	ROOT_OUTLET
  one	one
  b	one
  c	one
  d	one
  AA	AA
  BB	AA
  CC	AA
  l	n
  m	n
  x	There

  Post edited by: StewAshton - PS. I removed the MAX()... GROUP BY, and then put it back with MIN() because it is necessary if you have cycles in cycles.

• Reason for ORA-00600: internal error code, arguments: [sorgetqb_1], [2],...

  Hello
  
  I have problems with a bellows to query:
  
  Select CUSTOMER_ID, DECODE (NVL (CURR_LEVEL_FUACC, ROOT_LEVEL_FUACC), 'A', 10', 20', 30', 40)
  in (select CUSTOMER_ID, POOLED_FU_ACCOUNT CURR_LEVEL_FUACC, CONNECT_BY_ROOT (POOLED_FU_ACCOUNT)
  (Select B.CUSTOMER_ID, CUSTOMER_ID_HIGH, POOLED_FU_ACCOUNT ROOT_LEVEL_FUACC
  CUSTOMER_ALL b, RLH_PROCESS_CUST CPP
  where (B.CUSTOMER_ID = CPP. CUSTOMER_ID and CPP. BILLCYCLE_GROUP = 3))
  First, CUSTOMER_ID_HIGH is nothing connect prior CUSTOMER_ID = CUSTOMER_ID_HIGH)
  
  What is the reason for this error?
  at or near line number: 0
  SQL return code:-600
  SQL error message:
  [ORA-00600: internal error code, arguments: [sorgetqb_1], [2] [] [] [], [], []]
  
  P.S.:
  Oracle Database 10 g Enterprise Edition release 10.2.0.3.0 - 64bi
  PL/SQL version 10.2.0.3.0 - Production
  CORE Production 10.2.0.3.0
  AMT for HP - UX: release 10.2.0.3.0 - Production
  NLSRTL Version 10.2.0.3.0 - Production
  
  
  Thank you
  
  Fernando

  Solutions are provide in the following note:

  Note: 415648.1 -ORA-600 [Sorgetqb_1] with connection by prior Clause *.

• CONNECTION BY level/prior/root

  Hi all
  
  I'm trying to CONNECT BY level/prior/root. But I can barely decipher what it actually does. I think it is a very useful method provided by oracle and can have several uses. can someone help me on this please.
  
  I googled on this topic but everywhere I could find only employee and manager problem that will be solved by query below.
  

   SELECT employee_id, last_name, manager_id
     FROM employees
     CONNECT BY PRIOR employee_id = manager_id;

  But if we just query as below, we can get the employee and his manage without CONNECT BY.

  SELECT employee_id, last_name, manager_id
     FROM employees;

  Other queries is obtained to connect by is below, but he could not also include:

   select SYSDATE-41 + level - 1 the_date
                              from dual
                            connect by level <= SYSDATE - SYSDATE-41 + 1

  Please help me.
  
  Published by: J2EE_Life on December 6, 2011 08:12

  J2EE_Life wrote:
  Hi all

  I'm trying to CONNECT BY level/prior/root. But I can barely decipher what it actually does. I think it is a very useful method provided by oracle and can have several uses. can someone help me on this please.

  I googled on this topic but everywhere I could find only employee and manager problem that will be solved by query below.

  SELECT employe_id, last_name, manager_id
  Employees
  CONNECT BY PRIOR employee_id = manager_id;

  But if we just query as below, we can get the employee and his manage without CONNECT BY.

  SELECT employe_id, last_name, manager_id
  Employees;

  In fact, those who are not the same.
  If you use the standard table of hr.employees, the CONNECT BY query above returns 315 lines, but the secoind query return lines only 107.

  CONNECTION is useful to show the recursive relationships. It can show you, for example, not only that reports directly to a data manager, but the whole hierarchy of persons under a given head. For example, the following shows the hierarchy starting with an employee named "Rooster":

  SELECT  employee_id, last_name, manager_id
  ,     LEVEL                         AS lvl
  ,     SYS_CONNECT_BY_PATH (last_name, '/')     AS path
  FROM    hr.employees
  START WITH      last_name          = 'De Haan'
  CONNECT BY       PRIOR employee_id      = manager_id;
  
  SELECT     SYSDATE - SYSDATE-41 + 1
  FROM DUAL;
  

  Output:

  EMPLOYEE_ID LAST_NAME  MANAGER_ID LVL PATH
  ----------- ---------- ---------- --- --------------------------
          102 De Haan           100   1 /De Haan
          103 Hunold            102   2 /De Haan/Hunold
          104 Ernst             103   3 /De Haan/Hunold/Ernst
          105 Austin            103   3 /De Haan/Hunold/Austin
          106 Pataballa         103   3 /De Haan/Hunold/Pataballa
          107 Lorentz           103   3 /De Haan/Hunold/Lorentz
  

  Very briefly, here's how it works.
  The result set of a query CONNECT BY is the UNION ALL of several queries, all sharing a common SELECT clause.
  Any line that meets the conditions of the START WITH clause is supposed to be at LEVEL 1. So you can think of the first branch of the UNION of ALL being (in this case):

  SELECT  employee_id, last_name, manager_id
  ,     LEVEL                         AS lvl
  ,     SYS_CONNECT_BY_PATH (last_name, '/')     AS path
  FROM    hr.employees
  WHERE   last_name     = 'De Haan'
  UNION ALL ...
  

  Other rows in the result set will be LEVEL = N if the conditions of the CONNECT BY clause, where the PRIOR operator refers to any line level = N - 1.
  For example, the line with name = "Hunold" satisfies the condition to connect BY when the FIRST line is the line where last_name = "Rooster", Hunold is in the game with the LEVEL of results = 2.
  Another example: the line with name = 'Lorentz' meets the condition to connect BY when the FIRST line is the line where last_name = "Hunold", Hunold is in the result set with LEVEL = 3.
  This is a classic example of a tree structure, where the rows in the table are connected to other lines in a parent-child relationship. In a tree, each row has 0 or 1 of the parents, but a line can have any humber children. Haan has 0 parent and 1 child (that is, Hunold). Hunold has 1 parent (De Haan) and 4 children. CONNECT BY queries are useful to deal with the trees like this, especially when the parent-child relationship may extend to a number any levels.

  Other queries is obtained to connect by is below, but he could not also include:

  Select the level - 1 the_date + SYSDATE-41
  of the double
  connect by level<= sysdate="" -="" sysdate-41="" +="">

  I don't understand whether you.
  SYSDATE-SYSDATE is just 0
  0 - 41 + 1 is just-40, so you might as well say:

  select SYSDATE-41 + level - 1 the_date
                              from dual
                            connect by level <= -40;
  

  In addition, LEVEL is always a positive integer, then the CONNECT BY State will never have the value TRUE, and if the CONNECT BY clause can never be TRUE, there is no reason to have a CONNECT BY clause.
  A more reasonable request is:

  SELECT     SYSDATE + LEVEL - 1     AS the_date
  FROM      dual
  CONNECT BY     LEVEL     <= 4     -- or any positive number
  ;
  

  which produces this output, the given number of days (4 in this example) from today:

  THE_DATE
  --------------------
  06-Dec-2011 11:08:24
  07-Dec-2011 11:08:24
  08-Dec-2011 11:08:24
  09-Dec-2011 11:08:24
  

  This is a very common way to generate a Counter of Table (a table, or, as in this case, a result set, that matters). For example, if you need for the aggregates per day in a table and you want to show 0 for days that never exist in the table, then you outer join in a table of counter like the one shown above.
  This is a common use of CONNECT BY, but this is an unusual exception to the rules that is prohibited govern CONNECT BY queries. I suggest you uderstand how the query on hr.employees, above, the work before you start trying to figure out how it works.

  Published by: Frank Kulash, December 6, 2011 11:15

  The pages below are introductions to CONNECT BY:
  http://asktom.Oracle.com/pls/asktom/f?p=100:11:0:P11_QUESTION_ID:489772591421
  http://Philip.Greenspun.com/SQL/trees

• logical prior connection problem

  Hi all

  I have the scenario where I fill in amount of parent based on her children:

  create table xx_pack_content)

  child varchar2 (255),

  parent varchar2 (255),

  child_quantity NUMBER

  );

  insert into xx_pack_content values ('OBJ2', 'OBJ1', 50)

  insert into xx_pack_content values ("OBJ3", "OBJ2", 2)

  insert into xx_pack_content values ('OBJ4', 'OBJ2', 6)

  SELECT the level,

  x.Child,

  x.Parent,

  x.child_quantity

  OF xx_pack_content x

  START WITH x.parent =: p_parent_val

  CONNECT BY PRIOR x.child = x.parent

  Brothers and SŒURS ORDER BY level, x.child;

  Code:

  FUNCTION parent_qty (p_parent_val) RETURN VARCHAR2

  CURSOR lcu_cur

  IS to SELECT the level,

  x.Child,

  x.Parent,

  x.child_quantity

  OF xx_pack_content x

  START WITH x.parent = p_parent_val

  CONNECT BY PRIOR x.child = x.parent

  Brothers and SŒURS ORDER BY level, x.child;

  TYPE level_tbltype IS TABLE OF lcu_cur % ROWTYPE;

  level_tbl level_tbltype;

  lv_parent_content VARCHAR2 (30): = NULL;

  lv_level_low VARCHAR2 (30): = NULL;

  lv_level_high VARCHAR2 (30): = NULL;

  BEGIN

  OPEN lcu_cur (p_parent_val);

  Get the lcu_cur COLLECT in BULK IN level_tbl;

  CLOSE Lcu_cur;

  DBMS_OUTPUT. Put_line ("Count:" | ") level_tbl. (COUNT);

  lv_level_low: = level_tbl (level_tbl. PREMIER level);

  lv_parent_content: = level_tbl (level_tbl..) Quantity FIRST) | » x';

  BECAUSE me IN 2.level_tbl. LAST

  LOOP

  lv_level_high: = level_tbl (i) level;

  IF (lv_level_low <>lv_level_high) THEN

  lv_parent_content: = lv_parent_content | level_tbl (i) .child_quantity;

  lv_level_low: = level_tbl (i) level;

  ON THE OTHER

  lv_parent_content: = lv_parent_content | ' ('| level_tbl (i) .child_quantity |) » +';

  lv_level_low: = level_tbl (vpe_idx2) level.

  END IF;

  return (lv_parent_content);

  END LOOP;

  DBMS_OUTPUT. Put_line ('lv_parent_content: ' | lv_parent_content);

  END;

  expected results:

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

  I need to build a function that returns the total amount of parent based on the parameter

  When I pass "OBJ1" he shud be returned as 50 x (2 + 6) x'

  When I pass 'OBJ2' he shud be returned as '(2+6) x'

  If there is more than one child for the parent's should be like "(2 + 6 +..)" x'

  I wrote a function above, but need help to get a correct output.

  Don't know what you want to do when / if have brothers and sisters separated children but anyway...

  Oracle Database 11 g Enterprise Edition Release 11.2.0.3.0 - 64 bit Production
  With partitioning, OLAP, Data Mining and Real Application Testing options

  SQL > SELECT LISTAGG)
  2. DECODE (COUNT (*), 1,)
  MAX 3 (TO_CHAR (child_quantity)),
  4              '(' || LISTAGG (child_quantity, '+')
  5 THE GROUP)
  6 ORDER BY child) | ')') || 'x', ' ')
  7 IN GROUP)
  8 LEVEL CONTROL) Qty.
  9 FROM xx_pack_content
  10 START WITH PARENT = "OBJ1".
  11 children to connect BY PRIOR = PARENT
  12 GROUP BY LEVEL;

  QTY.
  --------------------------------------------------------------------------------
  50 x (2 + 6) x

  SQL > SELECT LISTAGG)
  2. DECODE (COUNT (*), 1,)
  MAX 3 (TO_CHAR (child_quantity)),
  4              '(' || LISTAGG (child_quantity, '+')
  5 THE GROUP)
  6 ORDER BY child) | ')') || 'x', ' ')
  7 IN GROUP)
  8 LEVEL CONTROL) Qty.
  9 FROM xx_pack_content
  10 START WITH PARENT = 'OBJ2 '.
  11 children to connect BY PRIOR = PARENT
  12 GROUP BY LEVEL;

  QTY.
  --------------------------------------------------------------------------------
  (2 + 6) x

  SQL >