create table p_range(a int, b int) partition by range (a,b); create table p_range1 partition of p_range for values from (1,1) to (3,3); create table p_range2 partition of p_range for values from (4,4) to (6,6); explain select * from p_range where b =2;
QUERY PLAN
--------------------------------------------------------------------------
Append (cost=0.00..76.61 rows=22 width=8)
-> Seq Scan on p_range1 p_range_1 (cost=0.00..38.25 rows=11 width=8)
Filter: (b = 2)
-> Seq Scan on p_range2 p_range_2 (cost=0.00..38.25 rows=11 width=8)
Filter: (b = 2)
(5 rows)
The result of EXPLAIN shows that no partition prune happened.
And gen_prune_steps_from_opexps() has comments that can answer the result.
/*
* For range partitioning, if we have no clauses for the current key,
* we can't consider any later keys either, so we can stop here.
*/
if (part_scheme->strategy == PARTITION_STRATEGY_RANGE &&
clauselist == NIL)
break;
But I want to know why we don't prune when just have latter partition key in whereClause.
Thanks.
