Other posts should answer most of your questions. Jared's answer assumes
only three levels and doesn't work when we add levels or branches to the
tree. ID is a system generated key and has no meaning. SQL only, no SQL*Plus
stuff. NODEORDER only refers to the sort within a node because you should
not have to resequence all the data just to add a node, change the sort
within a node or move a node to another parent.
-----Original Message-----
Sent: Tuesday, November 05, 2002 1:19 PM
To: Multiple recipients of list ORACLE-L
Okay, my answer was almost correct (almost correct = wrong). Jared's answer
is right on, given the current data set.
What happens when the data is changed? Does ID have meaning or is it the
sequence in which the row was added? NODEORDER is 'sequential', but the
starting values vary within the parent node (sometimes start with 1, others
with 0).
Old college trick, instead of answering the question, you challenge the
validity of the question.
Can you use sql only or assume sql*plus is available?
-----Original Message-----
Sent: Tuesday, November 05, 2002 10:24 AM
To: Multiple recipients of list ORACLE-L
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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