SELECT LEVEL, treenode.*
FROM treenode
START WITH parentid=0
CONNECT BY PRIOR ID = parentid
ORDER SIBLINGS BY PARENTid , nodeorder
Raj
______________________________________________________
Rajendra Jamadagni MIS, ESPN Inc.
Rajendra dot Jamadagni at ESPN dot com
Any opinion expressed here is personal and doesn't reflect that of ESPN Inc.
QOTD: Any clod can have facts, but having an opinion is an art!
-----Original Message-----
From: Orr, Steve [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, November 05, 2002 12:24 PM
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: Orr, Steve
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