On Friday, May 10, 2013, Robert Haas wrote:
On Sat, Mar 23, 2013 at 8:12 PM, Misa Simic <misa.simic@gmail.com> wrote:
> but problem is - we don't know the thing id - we know calc_id:
>
> EXPLAIN ANALYZE
> SELECT * FROM t2_left_t3_volatile v INNER JOIN t1 USING (thing_id)
> WHERE calc_id = 20
With this query you've got to scan all three tables. The calc_id qual
can only be pushed down into the scan on t1, so you need the whole
t2/t3 join product.
> EXPLAIN ANALYZE
> SELECT v.no_index FROM t2_left_t3 v INNER JOIN t1 USING (thing_id)
> WHERE calc_id = 20
With this query you only need to scan 2 tables. The join between t2
and t3 is eliminated by the join removal code in favor of scanning
only t2, as shown in the plan you included:
> "Nested Loop (cost=437.49..13047.74 rows=12111 width=4) (actual
> time=6.360..71.818 rows=12038 loops=1)"
> " -> Seq Scan on t1 (cost=0.00..2.25 rows=1 width=4) (actual
> time=0.016..0.024 rows=1 loops=1)"
> " Filter: (calc_id = 20)"
> " Rows Removed by Filter: 99"
> " -> Bitmap Heap Scan on t2 (cost=437.49..12924.38 rows=12111 width=12)
> (actual time=6.330..69.063 rows=12038 loops=1)"
> " Recheck Cond: (thing_id = t1.thing_id)"
> " -> Bitmap Index Scan on t5_c2_idx (cost=0.00..434.46 rows=12111
> width=0) (actual time=4.372..4.372 rows=12038 loops=1)"
> " Index Cond: (thing_id = t1.thing_id)"
> "Total runtime: 72.461 ms"
The difference is that this query has only one column in its target list, not *.
--
Robert Haas
EnterpriseDB: http://www.enterprisedb.com
The Enterprise PostgreSQL Company
Thanks Robert,
That is a bit "old" problem to us...
Solution for that kind of problems is: "rephrase" the question. So we have added one more layer to transform input query to "better" query for postgres - very wierd...
However, there are no differences... Planer use the same bad plan for:
SELECT v.no_index FROM t2_left_t3_volatile v INNER JOIN t1 USING (thing_id)
WHERE calc_id = 20
In that is one column as well...
We basicaly above query transform to:
SELECT v.no_index FROM t2_left_t3_volatile v
WHERE v.thing_id = (
SELECT thing_id FROM t1
WHERE calc_id = 20
)
What give us good result... Very wierd....
Thanks,
Misa