From: | Michael Fuhr <mike(at)fuhr(dot)org> |
---|---|
To: | Miguel Angel Tribaldos Hervas <mitriher(at)teleco(dot)upv(dot)es> |
Cc: | pgsql-general(at)postgresql(dot)org |
Subject: | Re: inherited type |
Date: | 2005-04-01 15:42:16 |
Message-ID: | 20050401154216.GA59714@winnie.fuhr.org |
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Thread: | |
Lists: | pgsql-general |
On Fri, Apr 01, 2005 at 12:42:19PM +0200, Miguel Angel Tribaldos Hervas wrote:
>
> I am working with inheritance in postgresql 7.4. If I create a table named A,
> and another named B that inherits from A, if I perform a query such a
> "SELECT * FROM A" (without ONLY clause),
> how can I get the type (identifier from pg_type) of the returned records??
You can look at the tableoid system column to get the oid from
pg_class:
CREATE TABLE parent (pid integer);
CREATE TABLE child (cid integer) INHERITS (parent);
INSERT INTO parent (pid) VALUES (1);
INSERT INTO child (pid, cid) VALUES (2, 3);
SELECT tableoid, tableoid::regclass, * FROM parent;
tableoid | tableoid | pid
----------+----------+-----
39455 | parent | 1
39457 | child | 2
(2 rows)
If necessary, you could join tableoid against pg_type.typrelid
or pg_class.oid:
SELECT t.typname, p.*
FROM parent AS p
JOIN pg_type AS t ON t.typrelid = p.tableoid;
typname | pid
---------+-----
parent | 1
child | 2
(2 rows)
SELECT c.relname, p.*
FROM parent AS p
JOIN pg_class AS c ON c.oid = p.tableoid;
relname | pid
---------+-----
parent | 1
child | 2
(2 rows)
--
Michael Fuhr
http://www.fuhr.org/~mfuhr/
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