Re: remove indexes on a column?

Perfect! Looks like I can get the names of the existing indexes by
doing

SELECT dep.relname
FROM pg_attribute col, pg_class tab, pg_depend pd, pg_class dep
WHERE tab.relname = 'mytable'
AND col.attname = 'mycolumn'
AND col.attrelid = tab.oid
AND pd.refobjid = tab.oid
AND pd.refobjsubid = col.attnum
AND pd.objid = dep.oid
AND dep.relkind = 'i';

Thanks.

Vance

On Wed, 2008-09-10 at 00:23 -0400, Tom Lane wrote:
> "Vance Maverick" <vmaverick(at)pgp(dot)com> writes:
> > I'd like to write a SQL script, possibly with some PL/pgSQL, that can
> > find all indexes on a column -- so I can remove them, and set up
> > exactly the indexes I want.
>
> Yeah, this seems a bit tricky if you have expression indexes involving
> the column. I concur that trying to parse the expressions is a bad
> idea --- even if your code works today, it'll probably break in future
> PG releases, because the nodetree representation is not very stable.
>
> What I'd look for is pg_depend entries showing indexes that depend on
> the column. Here's a hint:
>
> regression=# create table foo (f1 int);
> CREATE TABLE
> regression=# create index fooi on foo (abs(f1));
> CREATE INDEX
> regression=# select * from pg_depend where refobjid = 'foo'::regclass;
> classid | objid | objsubid | refclassid | refobjid | refobjsubid | deptype
> ---------+--------+----------+------------+----------+-------------+---------
> 1247 | 534605 | 0 | 1259 | 534603 | 0 | i
> 1259 | 534606 | 0 | 1259 | 534603 | 1 | a
> (2 rows)
>
> regression=# select 534606::regclass;
> regclass
> ----------
> fooi
> (1 row)
>
>
> regards, tom lane