The following bug has been logged online:

Bug reference: 2080
Logged by: Janko Richter
Email address: jankorichter@yahoo.de
PostgreSQL version: 8.1
Operating system: Linux
Description: Partitioning does not work with PREPARE
Details:

Partitioning does not work with PREPARE. The following
statements show the effect:

CREATE TABLE master
(
p_id int2 NOT NULL,
PRIMARY KEY (p_id)
) WITHOUT OIDS;

CREATE TABLE partition_0
(
PRIMARY KEY (p_id),
CONSTRAINT mod CHECK ( p_id = 0::smallint )
) INHERITS (master)
WITHOUT OIDS;

CREATE TABLE partition_1
(
PRIMARY KEY (p_id),
CONSTRAINT mod CHECK ( p_id = 1::smallint )
) INHERITS (master)
WITHOUT OIDS;

INSERT INTO partition_0 VALUES (0);

INSERT INTO partition_1 VALUES (1);



EXPLAIN SELECT p_id FROM master WHERE p_id=1::smallint ;
QUERY PLAN

----------------------------------------------------------------------------
---------------------------
Result (cost=0.00..10.65 rows=2 width=2)
-> Append (cost=0.00..10.65 rows=2 width=2)
-> Index Scan using master_pkey on master (cost=0.00..4.82 rows=1
width=2)
Index Cond: (p_id = 1::smallint)
-> Index Scan using partition_1_pkey on partition_1 master
(cost=0.00..5.82 rows=1 width=2)
Index Cond: (p_id = 1::smallint)
(6 rows)



PREPARE foo(int2) AS SELECT p_id FROM master WHERE p_id=$1::smallint;
EXPLAIN EXECUTE foo(1);
QUERY PLAN
----------------------------------------------------------------------------
---------------------------
Result (cost=0.00..16.47 rows=3 width=2)
-> Append (cost=0.00..16.47 rows=3 width=2)
-> Index Scan using master_pkey on master (cost=0.00..4.82 rows=1
width=2)
Index Cond: (p_id = $1)
-> Index Scan using partition_0_pkey on partition_0 master
(cost=0.00..5.82 rows=1 width=2)
Index Cond: (p_id = $1)
-> Index Scan using partition_1_pkey on partition_1 master
(cost=0.00..5.82 rows=1 width=2)
Index Cond: (p_id = $1)
(8 rows)


Regards,
Janko Richter

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"Janko Richter" <jankorichter@yahoo.de> writes:
> Partitioning does not work with PREPARE.

Well, no. How would you expect the prepared plan to know which
partition to look in, for a key value it doesn't have yet?

regards, tom lane

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