SQL Brain Teaser Challenge

From: Orr, Steve <sorr_at_rightnow.com>
Date: Tue, 05 Nov 2002 09:24:01 -0800
Message-ID: <F001.004FBDCA.20021105092401@fatcity.com>

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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Received on Tue Nov 05 2002 - 11:24:01 CST