On 10/26/07, Michael Glaesemann <grzm@xxxxxxxxxxxxxxx> wrote: > > On Oct 26, 2007, at 10:56 , Pat Maddox wrote: > > > A bunch of options so far...but there's really no way to do this with > > standard SQL? > > What do you mean by "standard SQL"? Trees aren't inherently relational. Right now my table looks like this: posts id body parent_id root_id created_at so if I've got the records (1, 'post 1', NULL, 1, '4pm') (2, 'post 2', NULL, 2, '8pm') (3, 'post 3', 1, 1, '6pm') (4, 'post 4', 1, 1, '5pm') (5, 'post 5', 4, 1, '6pm') (6, 'post 6', NULL, 1, '5pm') I'd like to do a select and get them all in this order: (1, 'post 1', NULL, 1, '4pm') (4, 'post 4', 1, 1, '5pm') (5, 'post 5', 4, 1, '6pm') (3, 'post 3', 1, 1, '6pm') (6, 'post 6', NULL, 1, '5pm') (2, 'post 2', NULL, 2, '8pm') And reverse sorted would be: (2, 'post 2', NULL, 2, '8pm') (6, 'post 6', NULL, 1, '5pm') (1, 'post 1', NULL, 1, '4pm') (3, 'post 3', 1, 1, '6pm') (4, 'post 4', 1, 1, '5pm') (5, 'post 5', 4, 1, '6pm') Does that make sense? Pat ---------------------------(end of broadcast)--------------------------- TIP 9: In versions below 8.0, the planner will ignore your desire to choose an index scan if your joining column's datatypes do not match