Thread: remove indexes on a column?

remove indexes on a column?

From
"Vance Maverick"
Date:
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.  (I know what indexes are *supposed* to be there, but depending on the
migrationhistory of the specific instance, the names may vary.)
 

I tried writing this logic using the system catalogs (pg_index, etc.), and it works up to a point.  But when some of
theindexes involve expressions, e.g.
 
 
   CREATE INDEX foo_lower_value ON foo(lower(value));
 
it's not so easy to do the lookup.  In this case, the column index is coded deep in an expression string ("in
nodeToString()representation"), and I don't see how to parse that.
 

Alternatively, I could take the brute-force approach:
- create a new column with the same type
- copy the values from the old column to the new
- drop the old column, presumably killing all the indices
- rename the new column to the old name
But that involves a lot of data copying, table restructuring, etc.

Is there a good way to do this?  Thanks,

    Vance

Re: remove indexes on a column?

From
Tom Lane
Date:
"Vance Maverick" <vmaverick@pgp.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

Re: remove indexes on a column?

From
Vance Maverick
Date:
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@pgp.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

Re: remove indexes on a column?

From
Tom Lane
Date:
Vance Maverick <vmaverick@pgp.com> writes:
> 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';

Too tired/lazy to check right now, but you should also look into
what the pg_depend representation is for constraints: I have a feeling
that a unique or primary key constraint yields a pg_depend structure
with an indirect linkage through a pg_constraint entry.

Also, the above query doesn't seem very schema-safe: what if there
are multiple tables named mytable?  Personally I'd try something
like tab.oid = 'mytable'::regclass instead of the relname test.

            regards, tom lane