based on the date it was written, that was before hints were available
for use. so it makes sense to include the dummy comparison to the
indexed column in the where clause as described in the note.
You mean Oracle sometimes thinks of these things BEFORE we do? nah!
- Jared Still <jkstill_at_cybcon.com> wrote:
>
> Note 1009345.6 on MetaLink provides a solution
> similar to the one proposed with an index.
>
> Jared
>
> On Tuesday 05 November 2002 09:24, Orr, Steve wrote:
> > 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
> --
> Please see the official ORACLE-L FAQ: http://www.orafaq.com
> --
> Author: Jared Still
> INET: jkstill_at_cybcon.com
>
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--
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--
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Received on Thu Nov 07 2002 - 07:18:25 CST
