subquery join order by
Hello,
(I googled and read docs before sending this e-mail).
Is it necessary to use order by twice (inside and outside) to get the
proper order if I have an ordered subqery in a join?
select * from (select distinct on (b_id) * from a order by b_id, id) sub
left join b on b.id = sub.b_id;
or
select * from (select distinct on (b_id) * from a order by b_id, id) sub
left join b on b.id = sub.b_id order by b_id;
It seems to me that it's enough to use 'order by' only inside wheter 'by
desc' or 'by asc' (b_id), however I'd like to be sure.
Thank you.
Mage
On 19 November 2010 01:36, Mage <mage@mage.hu> wrote:
Hello,
(I googled and read docs before sending this e-mail).
Is it necessary to use order by twice (inside and outside) to get the proper
order if I have an ordered subqery in a join?select * from (select distinct on (b_id) * from a order by b_id, id) sub
left join b on b.id = sub.b_id;or
select * from (select distinct on (b_id) * from a order by b_id, id) sub
left join b on b.id = sub.b_id order by b_id;It seems to me that it's enough to use 'order by' only inside wheter 'by
desc' or 'by asc' (b_id), however I'd like to be sure.Thank you.
Mage
You should always use ORDER BY on the outer-most part of the query
since that's what will be finally returning your data. Don't bother
with ordering sub-selects.
So in your case, just use:
SELECT *
FROM (SELECT DISTINCT ON (b_id) * FROM a) sub
LEFT JOIN b ON b.id = sub.b_id
ORDER BY sub.b_id, sub.id;
But why bother with a sub-select anyway? You can write it as:
SELECT DISTINCT ON (a.b_id) *
FROM a
LEFT JOIN b ON b.id = a.b_id
ORDER BY a.b_id, a.id;
--
Thom Brown
Twitter: @darkixion
IRC (freenode): dark_ixion
Registered Linux user: #516935
On 11/19/2010 03:21 AM, Thom Brown wrote:
You should always use ORDER BY on the outer-most part of the query
since that's what will be finally returning your data. Don't bother
with ordering sub-selects.
I definiatelly have to use the "order by" inside for two reasons.
When "distinct on (x)" is used then x must be in the first column in the
order by part.
The second column in the order by decides which records will I include
in the join so it is very important to use it for ordering.
So in your case, just use:
SELECT *
FROM (SELECT DISTINCT ON (b_id) * FROM a) sub
LEFT JOIN b ON b.id = sub.b_id
ORDER BY sub.b_id, sub.id;
select distinct on (id) * from b order by name;
ERROR: SELECT DISTINCT ON expressions must match initial ORDER BY
expressions
But why bother with a sub-select anyway? You can write it as:
SELECT DISTINCT ON (a.b_id) *
FROM a
LEFT JOIN b ON b.id = a.b_id
ORDER BY a.b_id, a.id;
I considered this, however the subquery is generated by an ORM. I wanted
to separate it.
Also the whole join affects many rows. I thought it's cheaper to
preselect them inside the subquery then do the join. I am not sure.
Explain analyze is my good friend but in this case I prefer to ask.
Mage
I considered this, however the subquery is generated by an ORM. I
wanted to separate it.Also the whole join affects many rows. I thought it's cheaper to
preselect them inside the subquery then do the join. I am not sure.
Explain analyze is my good friend but in this case I prefer to ask.
# EXPLAIN ANALYZE select * from (select distinct on (b_id) * from a
order by b_id, id) sub left join b on b.id = sub.b_id;
QUERY PLAN
-----------------------------------------------------------------------------------------------------------------
Hash Left Join (cost=187.45..243.70 rows=1230 width=44) (actual
time=0.000..0.000 rows=3 loops=1)
[...]
(11 rows)
# EXPLAIN ANALYZE SELECT DISTINCT ON (a.b_id) * FROM a LEFT JOIN b ON
b.id = a.b_id ORDER BY a.b_id, a.id;
QUERY PLAN
------------------------------------------------------------------------------------------------------------------------
Unique (cost=1339.24..1405.05 rows=200 width=44) (actual
time=0.000..0.000 rows=3 loops=1)
[...]
(15 rows)
mage=# EXPLAIN ANALYZE select * from (select distinct on (b_id) * from a
order by b_id, id) sub left join b on b.id = sub.b_id order by b.id;
QUERY PLAN
-----------------------------------------------------------------------------------------------------------------------
Sort (cost=306.83..309.90 rows=1230 width=44) (actual
time=0.000..0.000 rows=3 loops=1)
The subquery seems to be better choice even with double ordering. But is
the second order required?
Mage