On Jan 14, 2006, at 0:45 , Michael Glaesemann wrote:


On Jan 14, 2006, at 0:22 , Adam Rosi-Kessel wrote:

id1 will always have a value but id2 can be NULL. So should I do a left JOIN on id2 but a plain JOIN on id1? Is there a disadvantage to using a left JOIN
where it is not necessary?

In that case, yes, I'd JOIN on id1 and LEFT JOIN on id2. I'm not sure if there's a penalty or not in query planning, though there might be.


You can always use EXPLAIN ANALYZE to compare query plans. It can be very useful to see how your query is executed by the planner.

Of course, this sample is probably much smaller than your actual dataset, but here it does appear that JOIN is a (tiny) bit more efficient than LEFT [OUTER] JOIN. I'm a novice at reading these, but the things I notice are:

* The topmost hash join in the first query is faster than the topmost merge left join in the second. * The second query needs to do an extra sort and sequential scan that is unnecessary in the first.

In *this* case, I think the difference may be just noise. You'll get slightly different times each run, so you'll have to see what runs better on average if they're close. However, it does look like the left join forces at least an extra 2 steps, which should take more time, on average, so I think the LEFT JOIN will be slower than the JOIN.

All those more knowledgeable, feel free to jump in and correct me :)

Michael Glaesemann
grzm myrealbox com

test=# explain analyze
select some_date
    , x.some_name as name1
    , y.some_name as name2
from table1
join table2 as x on id1 = x.id
join table2 as y on id2 = y.id
where (some_date is not null and (id1 = 1 or id2 = 1))
order by some_date;
                                                          QUERY PLAN
------------------------------------------------------------------------ ------------------------------------------------------- Sort (cost=3.20..3.21 rows=2 width=18) (actual time=0.270..0.272 rows=3 loops=1)
   Sort Key: table1.some_date
-> Hash Join (cost=2.12..3.19 rows=2 width=18) (actual time=0.219..0.227 rows=3 loops=1)
         Hash Cond: ("outer".id = "inner".id1)
-> Seq Scan on table2 x (cost=0.00..1.03 rows=3 width=11) (actual time=0.016..0.019 rows=3 loops=1) -> Hash (cost=2.12..2.12 rows=2 width=15) (actual time=0.141..0.141 rows=3 loops=1) -> Hash Join (cost=1.05..2.12 rows=2 width=15) (actual time=0.109..0.118 rows=3 loops=1)
                     Hash Cond: ("outer".id = "inner".id2)
-> Seq Scan on table2 y (cost=0.00..1.03 rows=3 width=11) (actual time=0.004..0.008 rows=3 loops=1) -> Hash (cost=1.04..1.04 rows=2 width=12) (actual time=0.053..0.053 rows=3 loops=1) -> Seq Scan on table1 (cost=0.00..1.04 rows=2 width=12) (actual time=0.017..0.027 rows=3 loops=1) Filter: ((some_date IS NOT NULL) AND ((id1 = 1) OR (id2 = 1)))
Total runtime: 0.594 ms
(13 rows)

test=# explain analyze
select some_date
    , x.some_name as name1
    , y.some_name as name2
from table1
join table2 as x on id1 = x.id
left join table2 as y on id2 = y.id
where (some_date is not null and (id1 = 1 or id2 = 1))
order by some_date;
                                                          QUERY PLAN
------------------------------------------------------------------------ ------------------------------------------------------- Sort (cost=3.24..3.25 rows=2 width=18) (actual time=0.281..0.282 rows=3 loops=1)
   Sort Key: table1.some_date
-> Merge Left Join (cost=3.19..3.23 rows=2 width=18) (actual time=0.242..0.251 rows=3 loops=1)
         Merge Cond: ("outer".id2 = "inner".id)
-> Sort (cost=2.13..2.14 rows=2 width=15) (actual time=0.165..0.165 rows=3 loops=1)
               Sort Key: table1.id2
-> Hash Join (cost=1.05..2.12 rows=2 width=15) (actual time=0.128..0.137 rows=3 loops=1)
                     Hash Cond: ("outer".id = "inner".id1)
-> Seq Scan on table2 x (cost=0.00..1.03 rows=3 width=11) (actual time=0.016..0.020 rows=3 loops=1) -> Hash (cost=1.04..1.04 rows=2 width=12) (actual time=0.052..0.052 rows=3 loops=1) -> Seq Scan on table1 (cost=0.00..1.04 rows=2 width=12) (actual time=0.017..0.027 rows=3 loops=1) Filter: ((some_date IS NOT NULL) AND ((id1 = 1) OR (id2 = 1))) -> Sort (cost=1.05..1.06 rows=3 width=11) (actual time=0.067..0.068 rows=3 loops=1)
               Sort Key: y.id
-> Seq Scan on table2 y (cost=0.00..1.03 rows=3 width=11) (actual time=0.005..0.010 rows=3 loops=1)
Total runtime: 0.600 ms
(16 rows)


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