Hierarchical Query

Question

Hi all,

I hope I'm able to explain the problem that is puzzeling me. I have the following hierarchical data set (this is just subset of 34K records)

PARENT_ID   CHILD_ID          EXAM
TUDA12802   TUDA12982         N 
TUDA12982   TUDA12984         J    
TUDA12984   TUDA999           J
TUDA12982   TUDA12983         N
TUDA12983   TUDA15322         J
TUDA12983   TUDA15323         J

This is a representation of the tree

TUDA12982 N
- TUDA12984 J
--  TUDA999 J
- TUDA12983 N
--  TUDA15322 J
--  TUDA15323 J

What I need is a list of all records with exam=N and the underlying exam = 'J' records, which can be nested.

select *
from test1 
connect by prior child_id = parent_id
start with child_id = 'TUDA12982'
order siblings by child_id;

Gives me

PARENT_ID      CHILD_ID          EXAM
TUDA12802   TUDA12982         N 
TUDA12982   TUDA12984         J    
TUDA12984   TUDA999           J
TUDA12982   TUDA12983         N
TUDA12983   TUDA15323         J
TUDA12983   TUDA15322         J

But what I need is

TUDA12802   TUDA12982         N 
TUDA12982   TUDA12984         J 
TUDA12984   TUDA999           J

The traversing needs to stop when I encounter a EXAM = 'N' record.

I need something like a 'stop with' clause.

select *
from test1 
connect by prior child_id = parent_id
start with child_id = 'TUDA12982'
stop with exam = 'N'
order siblings by child_id;

How can this be done?

Answer 1

Sounds like a simple query that gets the requested item and it's 'J' children is what you want, so wouldn't this work:

select *
from test1 
where child_id = 'TUDA12982'
or exam = 'J'
connect by prior child_id = parent_id
start with child_id = 'TUDA12982'
order siblings by child_id;

I don't have Oracle so I can't test if that works, but from what I understand of the syntax and what I just Googled it looks like it would work.

Answer 2

Robert,

You can do this by adding "exam = 'J'" to the connect by clause:

SQL> create table test1(parent_id,child_id,exam)
  2  as
  3  select 'TUDA12802', 'TUDA12982', 'N' from dual union all
  4  select 'TUDA12982', 'TUDA12984', 'J' from dual union all
  5  select 'TUDA12984', 'TUDA999', 'J' from dual union all
  6  select 'TUDA12982', 'TUDA12983', 'N' from dual union all
  7  select 'TUDA12983', 'TUDA15322', 'J' from dual union all
  8  select 'TUDA12983', 'TUDA15323', 'J' from dual
  9  /

Tabel is aangemaakt.

SQL>  select parent_id
  2        , child_id
  3        , exam
  4        , level
  5        , lpad(' ',2*level) || sys_connect_by_path(parent_id||'-'||child_id,'/') scbp
  6     from test1
  7    start with exam = 'N'
  8  connect by prior child_id = parent_id
  9      and exam = 'J'
 10  /

PARENT_ID CHILD_ID  E  LEVEL SCBP
--------- --------- - ------ ----------------------------------------------------------------------
TUDA12802 TUDA12982 N      1   /TUDA12802-TUDA12982
TUDA12982 TUDA12984 J      2     /TUDA12802-TUDA12982/TUDA12982-TUDA12984
TUDA12984 TUDA999   J      3       /TUDA12802-TUDA12982/TUDA12982-TUDA12984/TUDA12984-TUDA999
TUDA12982 TUDA12983 N      1   /TUDA12982-TUDA12983
TUDA12983 TUDA15322 J      2     /TUDA12982-TUDA12983/TUDA12983-TUDA15322
TUDA12983 TUDA15323 J      2     /TUDA12982-TUDA12983/TUDA12983-TUDA15323

6 rijen zijn geselecteerd.

Regards, Rob.