Steve,

I'm not sure why as of yet, but I had some success by creating two segmented
indexes.  One on PARENTID, NODEORDER and the second on NODEORDER, PARENTID:

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

Seems strange, though, and I don't have the time to research it.  We had the
exact same problem for our BOM structures and ended up writing a recursive
PL/SQL procedure and cursors to do it.  Not nice.

Also, I'm using CBO on 8.1.7.4 and have analyzed the tables and indexes w/o
histograms using DBMS_STATS.  BTW, the explain plan is a helluva lot better
with the indexes...  :)

HTH!  GL!

Rich


Rich Jesse                           System/Database Administrator
[EMAIL PROTECTED]              Quad/Tech International, Sussex, WI USA


> -----Original Message-----
> From: Orr, Steve [mailto:sorr@;rightnow.com]
> Sent: Tuesday, November 05, 2002 11:24 AM
> To: Multiple recipients of list ORACLE-L
> 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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Author: Jesse, Rich
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