I wondered if someone would try that. :-)
The link you provided is interesting but I don't think it fits the needs as
regards being able to add and rearrange nodes. (Maybe I got confused with
the little worms.) This is because the right column has to be twice the
number of rows and this number would need to be updated every time you added
or removed a row.
-----Original Message-----
Sent: Tuesday, November 05, 2002 12:45 PM
To: Multiple recipients of list ORACLE-L
and i got this to work:
SQL> select * from treenode;
ID PARENTID NODEORDER DESCRIPTION
---------- ---------- ---------- --------------------
1 0 0 top folder
9 1 0 1st subfolder
2 1 1 2nd subfolder
4 2 1 folder 2 item 1
3 2 2 folder 2 item 2
6 2 3 folder 2 item 3
7 1 2 3rd subfolder
5 7 0 folder 3 item 1
8 7 1 folder 3 item 2
9 rows selected.
</cheating with the order of the insert statements> ;-)
i had always thought the parentid was the way to go when representing trees
in the database, but Celko describes the "Nested-Set Model of Trees"
alternative in this ancient article:
http://www.dbmsmag.com/9603d06.html
makes your problem trivial.
anyway, thanks for the post Steve. learned something today - i'm going
home.
-----Original Message-----
Sent: Tuesday, November 05, 2002 1:15 PM
To: Multiple recipients of list ORACLE-L
Steve,
This works for me.
Jared
====================================================================
col nodelevel noprint
col parent noprint
col child noprint
select
a.nodelevel
, a.id id
, a.parentid
, a.nodeorder
, a.description
, decode(c.children,null,'N','Y') parent
, decode(p.children,null,'Y','N') child
from (
select level nodelevel, id, parentid, nodeorder, description
from treenode
start with parentid=0
connect by prior id = parentid
) a,
(
select parentid, count(*) children
from treenode b
group by parentid
) c,
(
select parentid, count(*) children
from treenode d
group by parentid
) p
where a.id = c.parentid(+)
and a.id = p.parentid(+)
order by
decode(parent
-- is a parent
, 'Y', nodelevel * id
-- is a child
, 'N', nodelevel * parentid + nodeorder
)
/
"Orr, Steve" <[EMAIL PROTECTED]>
Sent by: [EMAIL PROTECTED]
11/05/2002 09:24 AM
Please respond to ORACLE-L
To: Multiple recipients of list ORACLE-L <[EMAIL PROTECTED]>
cc:
Subject: SQL Brain Teaser Challenge
Challenge: present SQL results hierarchically and sort the nodes. Use sort
column without changing data. Here's the DDL/DML to start:
create table treenode (
id number not null
constraint pk_treenode
primary key,
parentid number not null,
nodeorder number not null,
description varchar2(20) null);
insert into treenode values(1,0,0,'top folder');
insert into treenode values(9,1,0,'1st subfolder');
insert into treenode values(7,1,2,'3rd subfolder');
insert into treenode values(2,1,1,'2nd subfolder');
insert into treenode values(8,7,1,'folder 3 item 2');
insert into treenode values(6,2,3,'folder 2 item 3');
insert into treenode values(5,7,0,'folder 3 item 1');
insert into treenode values(3,2,2,'folder 2 item 2');
insert into treenode values(4,2,1,'folder 2 item 1');
-----------------------------------------------------
Here's the data presented hierachically without the desired sort:
select * from treenode
start with parentid=0 connect by prior id = parentid;
ID PARENTID NODEORDER DESCRIPTION
---------- ---------- ---------- --------------------
1 0 0 top folder
9 1 0 1st subfolder
7 1 2 3rd subfolder
8 7 1 folder 3 item 2
5 7 0 folder 3 item 1
2 1 1 2nd subfolder
6 2 3 folder 2 item 3
3 2 2 folder 2 item 2
4 2 1 folder 2 item 1
-----------------------------------------------------
Desired SQL statement results:
ID PARENTID NODEORDER DESCRIPTION
---------- ---------- ---------- --------------------
1 0 0 top folder
9 1 0 1st subfolder
2 1 1 2nd subfolder
4 2 1 folder 2 item 1
3 2 2 folder 2 item 2
6 2 3 folder 2 item 3
7 1 2 3rd subfolder
5 7 0 folder 3 item 1
8 7 1 folder 3 item 2
-----------------------------------------------------
Kudos to anyone who can figure out how to do this via SQL.
Steve Orr
Bozeman, Montana
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