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

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  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.

• Hierarchical connect by and start with and joined?

  I have an Employees table and a table of identifiers. The table of identifiers is hierarchical, with parents and children. Each employee has one or more identifiers, but that a unique identifier is considered to be the "primary" or root for each employee identifier. Unfortunately, the employee table can point to one of the children identifier lines and not the root. I need a quick query to reach employees with their most recent ID (root).
  
  Here's the code to define the problem.

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  insert into employees (employeeid, fakeNationalID, empname) values (2,'002000002','James Jones');
  
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  insert into realids (realidkey, fakeNationalID, realNationalID, parent_realidkey) values
     (1,'001000001','111111111',3);
  insert into realids (realidkey, fakeNationalID, realNationalID, parent_realidkey) values
     (2,'002000002','222222222',null);
  insert into realids (realidkey, fakeNationalID, realNationalID, parent_realidkey) values
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           from (
             select realNationalID, fakeNationalID
                from realids
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                order by level desc)
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  It works, but it is NOT very effective:

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  employeeid   realid       empname
  ----------   -----------  ------------
  1            333333333     John Smith
  2            222222222     James Jones

  You can imagine what it would be like with 100K lines or more. It takes about 3 minutes to run.
  
  It seemed like a good way to do it, but with a sub query.

  select e.employeeid, e.fakenationalid, e.empname, sub.realnationalid
     from employees, 
        (select realidkey, fakenationalid, realnationalid, parent_realidkey
           from realids r
           start with r.fakenationalid = e.fakenationalid
           connect by prior r.parent_realidkey = r.realidkey) sub

  Unfortunately, it produces an invalid identifier on e.fakenationalid (in the beginning with the clause).
  
  Anyone has any ideas on how to get top most parent node of the realids for each row in the employees table? In real life, there are 6 or more employees tables across multiple remote instances of what children in the realids table and how much to the parents. We always want the highest parent of the page realid. Any help would be appreciated.

  Hello

  Thanks for posting the sample data in a convenient form!
  It is always useful to post your version of Oracle, too, especially when it comes with CONNECT BY queries.

  What follows is what you asked for in Oracle 10:

  WITH     got_roots   AS
  (
       SELECT     CONNECT_BY_ROOT     fakenationalid     AS leaf_id
       ,     realnationalid
       FROM     realids
       WHERE     CONNECT_BY_ISLEAF     = 1
       START WITH      fakenationalid IN ( SELECT  fakenationalid
                                            FROM    employees
                             )
       CONNECT BY     realidKEY     = PRIOR parent_realidkey
  )
  SELECT     e.employeeid
  ,     r.realnationalid
  ,     e.empname
  FROM     employees     e
  JOIN     got_roots     r     ON     r.leaf_id     = e.fakenationalid
  ;
  

  In any query, call a function defined by the user for each line is going to be slow. Fortunately, Oracle now has the built-in functions and operators that can take the place of get_parentid. The CONNECT_BY_ROOT operator, which was introduced in Oracle 10, is the key to the problem. In Oracle 9, you can get the same results using SYS_CONNECT_BY_PATH.

  It is generally faster to CONNECT BY query separately, and then join some other tables you need for results.

  You had a good idea in your last query. The problem was that void and employees were equal tables in the FROM clause, and you cannot establish a correlation between equals. You can only correlate a subquery to its Super application. You could make to this general idea work by changing void in a scalar sub-requete, which can be connected to the employees, but I think it would be much less effective than what I posted above.

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  =======================

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