Hierarchy problem once again

Dear All;
I have the following sample data shown below
create table t1_concept
  list_id varchar2(1000),
  list_model varchar2(400),
  list_num varchar2(1000)
insert into t1_concept
  (list_id, list_model, list_num)
values
  ('A11-11', '1', 'A22-2');
insert into t1_concept
  (list_id, list_model, list_num)
values
  ('A11-11', '1', 'A33-3'); 
insert into t1_concept
  (list_id, list_model, list_num)
values
  ('A11-11', '2', 'A44-4');
insert into t1_concept
  (list_id, list_model, list_num)
values
  ('A11-11', '1', 'A55-5');
insert into t1_concept
  (list_id, list_model, list_num)
values
  ('A44-4', '13', 'B11-1');
insert into t1_concept
  (list_id, list_model, list_num)
values
  ('A44-4', '8', 'B22-2');
insert into t1_concept
  (list_id, list_model, list_num)
values
  ('A44-4', '8', 'B33-3');
insert into t1_concept
  (list_id, list_model, list_num)
values
  ('A44-4', '8', 'B44-4');
insert into t1_concept
  (list_id, list_model, list_num)
values
  ('A55-5', '9', 'C11-1');
insert into t1_concept
  (list_id, list_model, list_num)
values
  ('A55-5', '9', 'C22-2');
create table t2_concept
  list_num varchar2(1000),
  max_list_model varchar2(1000)
insert into t2_concept
  (list_num, max_list_model)
values
  ('A44-4', 'A');   --changed to show that we can have letters
insert into t2_concept
  (list_num, max_list_model)
values
  ('A55-5', '9');
insert into t2_concept
  (list_num, max_list_model)
values
  ('A66-6', '10');I know things would have been easier with a good normalized table design. however unfortunately, I am taking over another programmer works and I am trying to look for a good solution.
I am here trying to create a hierarchy structure. the idea behind it is as following
if we start-off with List_ID = A11-1 and List_model = 1, then we are going to have the have the following children associated with it
A22-2, A33-3 and A55-5
now, these children have children of their own and the idea is to look in the t2_concept table to determine the latest list_model and utilize it in the recursion, so in this case if you look at t2_concept for A44-4 it has a max_list_model of A and we are then going to go back to t1_concept and determine all the associated children for A44-4 which are B22-2, B33-3 and B44-4 and we continue with this approach till the end.
this is output I want by the way
LEVEL      List_ID          
   1          A22-2
   1          A33-3
   1          A44-4
   2          B22-2
   2          B33-3
   2          B44-4
   1          A55-5
   2          C11-1
   2          C22-2I know things would have been easier with a good normalized table design. however unfortunately, I am taking over another programmer works and I am trying to look for a good solution.
All help and questions are appreciated.
Edited by: user13328581 on Aug 16, 2011 10:25 AM
Edited by: user13328581 on Aug 16, 2011 12:00 PM

Hi,
If you can't normalize the tables in reality, you can normalize them on the fly, like this:
WITH     joined_data      AS
     SELECT     t1.*
     ,     t2.max_list_model
     FROM           t1_concept     t1
     LEFT OUTER JOIN      t2_concept     t2  ON    t1.list_num     = t2.list_num
--     WHERE        ...      -- If you have other filtering, do as much as possible here
SELECT       LEVEL
,       list_num
FROM       joined_data          j
START WITH        list_id     IN ('A11-11')     -- Or any given values
CONNECT BY        list_id     = PRIOR list_num
     AND        list_model     >= PRIOR max_list_model
ORDER SIBLINGS BY  list_num
;You may not need a sub-query; joins are done before START WITH and CONNECT BY, but it's often faster if you do the joins in a separate sub-query.
Is the problem here that, for purposes of this query, you don't want to consider 'B11-1' to be a child of 'A44-4' because list_model > list_num? That's what the 2nd condition in the CONNECT BY clause is doing. The query above produces the results you want, but the condition I posted looks backwards to me.
Edited by: Frank Kulash on Aug 16, 2011 2:37 PM
Mystery solved: list_model and max_list_model are strings, so '8' is greater than '13'.
The following works for the sample data already given:
WITH     joined_data      AS
     SELECT     t1.*
     ,     t2.max_list_model
     FROM           t1_concept     t1
     LEFT OUTER JOIN      t2_concept     t2  ON    t1.list_num     = t2.list_num
--     WHERE        ...      -- If you have other filtering, do as much as possible here
SELECT       LEVEL
,       list_num
FROM       joined_data          j
START WITH        list_id     IN ('A11-11')     -- Or any given values
CONNECT BY        list_id     = PRIOR list_num
     AND        LPAD ( list_model                    -- This condition has been changed
                  , 1000
               )     <= PRIOR LPAD (max_list_model, 1000)
ORDER SIBLINGS BY  list_num
;and it may even work in your real data. (I have no way of knowing yet.)

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    I am still using 10.3.9, but I dont want to buy Tiger before I am not absolutly sure, that this will solve the problem. Every other program is up to date.
    Another Question: where does IDVD save the rendered files? I am slowly running out of HD space. (still having 23 GB, but every new try costs 3 GB)
    I have to say, I am very disapointed, I did not imaging, apple come with such problems in ILife 06.
    thanks and greetings everyone,
    Kristian
      Mac OS X (10.3.9)  

    Well, thats just great!
    Just as I was about to sove my problems with the DVD burning, my DSL modem quit its job. Now I've been offline a week and it was just like ****.
    A world without internet...
    Sorry, back to my burning issue.
    Yes, I tryed all the steps written in the link above, but it didnt work out.
    I deletet al the dots in fonts book, but the appeared back again. Why do they do this?
    I had also an other idea: what if i install IDVD 5 on my extern HD and try to burn the proojects with this version?
    But, since I installed IDVD 6 a month ago, all my IDVD projects (witch were originally created in IDVD 5) have been converted to be compatibel with IDVD 6.
    Is IDVD 5 able to read this files now?
    Thank you for answering, guys.
    Kristian
    Powerbook G4 1,33GHz   Mac OS X (10.3.9)  

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