Re: Optimizing this count query

I have to do all the joins because in the where cause I can also have other
conditions that are related to the other tables.
For example:
....WHERE o.id_status>3 AND o.id_ag=72 AND v.id_worker=5 AND z.id=10.

Now if these search functions are IN then the query runs faster.

One thing I could do at this point is not to make the join if that table is
not needed in the where clause.

This is the explain analize for the first query.
Aggregate  (cost=37182.56..37182.56 rows=1 width=4) (actual
time=3032.126..3032.126 rows=1 loops=1)
  ->  Hash Join  (cost=27279.22..37079.68 rows=41154 width=4) (actual
time=662.600..2999.845 rows=42835 loops=1)
        Hash Cond: ("outer".id_endkunde = "inner".id)
        ->  Hash Join  (cost=25770.48..34068.10 rows=41153 width=8) (actual
time=561.112..2444.574 rows=42835 loops=1)
              Hash Cond: ("outer".id_worker = "inner".id)
              ->  Hash Join  (cost=25759.54..33326.98 rows=41151 width=12)
(actual time=560.514..2361.776 rows=42835 loops=1)
                    Hash Cond: ("outer".id_ag = "inner".id)
                    ->  Hash Join  (cost=25587.07..32331.51 rows=41150
width=16) (actual time=551.505..2240.217 rows=42835 loops=1)
                          Hash Cond: ("outer".id_status = "inner".id)
                          ->  Hash Join  (cost=25586.00..31713.18 rows=41150
width=20) (actual time=551.418..2150.224 rows=42835 loops=1)
                                Hash Cond: ("outer".id_zufriden =
"inner".id)
                                ->  Hash Join  (cost=25584.85..31094.78
rows=41150 width=24) (actual time=551.341..2057.142 rows=42835 loops=1)
                                      Hash Cond: ("outer".id_plannung =
"inner".id)
                                      ->  Hash Join
(cost=24135.60..27869.53 rows=41149 width=24) (actual time=415.189..1162.429
rows=42835 loops=1)
                                            Hash Cond: ("outer".id =
"inner".id_order)
                                            ->  Seq Scan on orders o
(cost=0.00..2058.54 rows=42527 width=20) (actual time=0.046..93.692
rows=42835 loops=1)
                                                  Filter: (id_status > 3)
                                            ->  Hash
(cost=23860.48..23860.48 rows=42848 width=8) (actual time=414.923..414.923
rows=0 loops=1)
                                                  ->  Seq Scan on report r
(cost=0.00..23860.48 rows=42848 width=8) (actual time=282.905..371.401
rows=42848 loops=1)
                                      ->  Hash  (cost=1050.80..1050.80
rows=62180 width=8) (actual time=133.505..133.505 rows=0 loops=1)
                                            ->  Seq Scan on plannung v
(cost=0.00..1050.80 rows=62180 width=8) (actual time=0.034..73.048
rows=62180 loops=1)
                                ->  Hash  (cost=1.12..1.12 rows=12 width=4)
(actual time=0.048..0.048 rows=0 loops=1)
                                      ->  Seq Scan on zufriden z
(cost=0.00..1.12 rows=12 width=4) (actual time=0.027..0.040 rows=12 loops=1)
                          ->  Hash  (cost=1.06..1.06 rows=6 width=4) (actual
time=0.045..0.045 rows=0 loops=1)
                                ->  Seq Scan on status s  (cost=0.00..1.06
rows=6 width=4) (actual time=0.032..0.037 rows=6 loops=1)
                    ->  Hash  (cost=161.57..161.57 rows=4357 width=4)
(actual time=8.973..8.973 rows=0 loops=1)
                          ->  Seq Scan on contact c  (cost=0.00..161.57
rows=4357 width=4) (actual time=0.032..5.902 rows=4357 loops=1)
              ->  Hash  (cost=10.84..10.84 rows=42 width=4) (actual
time=0.557..0.557 rows=0 loops=1)
                    ->  Hash Join  (cost=1.51..10.84 rows=42 width=4)
(actual time=0.182..0.523 rows=41 loops=1)
                          Hash Cond: ("outer".id = "inner".id_person)
                          ->  Seq Scan on person p  (cost=0.00..7.66
rows=166 width=4) (actual time=0.027..0.216 rows=166 loops=1)
                          ->  Hash  (cost=1.41..1.41 rows=41 width=8)
(actual time=0.125..0.125 rows=0 loops=1)
                                ->  Seq Scan on mpsworker w
(cost=0.00..1.41 rows=41 width=8) (actual time=0.038..0.086 rows=41 loops=1)
        ->  Hash  (cost=1245.99..1245.99 rows=44299 width=4) (actual
time=101.257..101.257 rows=0 loops=1)
              ->  Seq Scan on endkunde e  (cost=0.00..1245.99 rows=44299
width=4) (actual time=0.050..59.641 rows=44301 loops=1)
Total runtime: 3033.230 ms

Thanks for help.
Andy.

----- Original Message -----
From: "Tom Lane" <tgl@sss.pgh.pa.us>
To: "Andrei Bintintan" <klodoma@ar-sd.net>
Cc: <pgsql-performance@postgresql.org>
Sent: Monday, January 17, 2005 7:55 PM
Subject: Re: [PERFORM] Optimizing this count query


> "Andrei Bintintan" <klodoma@ar-sd.net> writes:
>> SELECT count(o.id) FROM orders o
>>       INNER JOIN report r ON o.id=r.id_order
>>       INNER JOIN status s ON o.id_status=s.id
>>       INNER JOIN contact c ON o.id_ag=c.id
>>       INNER JOIN endkunde e ON o.id_endkunde=e.id
>>       INNER JOIN zufriden z ON r.id_zufriden=z.id
>>       INNER JOIN plannung v ON v.id=o.id_plannung
>>       INNER JOIN mpsworker w ON v.id_worker=w.id
>>       INNER JOIN person p ON p.id = w.id_person
>>       WHERE o.id_status>3
>
>> I'm asking how can I improve the execution time of this query, because =
>> these tables are always increasing. And this count sometimes takes more =
>> than 10 secs and I need to run this count very offen.
>
> Unless you've increased the default value of join_collapse_limit, this
> construction will be forcing the join order; see
> http://www.postgresql.org/docs/7.4/static/explicit-joins.html
>
> I'm not sure if you can improve the join order at all --- since you only
> showed EXPLAIN and not EXPLAIN ANALYZE, it's hard to be sure whether any
> of the steps are producing large intermediate results.  But it's
> something to look into.
>
> You should also ask yourself if you need to be joining so many tables at
> all.  The planner seems to think that only the o/r join is really going
> to affect the result row count.  I can't tell if it's right or not, but
> if this is a star schema and the other seven tables are just detail
> tables, you don't need them in order to obtain a count.
>
> regards, tom lane
>
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