PostGIS – Using Spatial Indexes on Foreign Tables

The (lack of) documentation suggests that this behaviour is an implementation detail, and is therefore undefined and subject to change at any time.

This is in stark contrast to CREATE FULLTEXT INDEX, where you have to specify the name of an index to attach to -- AFAIK, there is no undocumented FOREIGN KEY syntax to do the equivalent (though theoretically, there could be in the future).

As mentioned, it does make sense that SQL Server chooses the smallest physical index with which to associate the foreign key. If you change the script to create the unique constraint as CLUSTERED, the script "works" on 2008 R2. But that behaviour is still undefined and should not be relied upon.

As with most legacy applications, you'll just have to get down to the nitty-gritty and clean things up.

Given the following setup:

regress=> CREATE SCHEMA A;
CREATE SCHEMA
regress=> CREATE SCHEMA B;
CREATE SCHEMA
regress=> SET search_path = B, public;
SET
regress=> CREATE TABLE bar(email text);
CREATE TABLE
CREATE UNIQUE INDEX index_bar_on_email ON bar USING btree (email);
CREATE INDEX

I cannot reproduce the problem you report in PostgreSQL 9.2:

regress=> SET search_path = A, B;
SET
regress=> CREATE TABLE bar(email text);
CREATE TABLE
regress=> CREATE UNIQUE INDEX index_bar_on_email ON bar USING btree (email);
CREATE INDEX

However, rather than using the search_path, it's safer to use explicit schema-qualification. For example, I'd re-write the above as:

regress=> RESET search_path;
RESET
regress=> SHOW search_path ;
  search_path   
----------------
 "$user",public
(1 row)

CREATE TABLE B.bar(email text);
CREATE UNIQUE INDEX b.index_bar_on_email ON b.bar USING btree (email);
CREATE TABLE A.bar(email text);
CREATE UNIQUE INDEX index_bar_on_email ON A.bar USING btree (email);

The indexes are automatically created in the schema of their associated table; see:

regress=> \di B.
                  List of relations
 Schema |        Name        | Type  | Owner | Table 
--------+--------------------+-------+-------+-------
 b      | index_bar_on_email | index | craig | bar
(1 row)

regress=> \di A.
                  List of relations
 Schema |        Name        | Type  | Owner | Table 
--------+--------------------+-------+-------+-------
 a      | index_bar_on_email | index | craig | bar
(1 row)

Update based on question change:

Yes, what you've shown does look like a Rails adapter issue. It's checking to see whether the index exists in any schema. It should be checking to see whether the first table of the given name in the search_path has the named index.

I would write the query differently. I'd leave off the join on pg_class entirely, instead using a cast to regclass to handle search_path resolution for me. I'd use the resulting oid to search for the index. Compare original, then updated, below. Note that the updated query does require search_path to be set first.

regress=> SELECT DISTINCT i.relname, d.indisunique, d.indkey, t.oid, am.amname
 FROM pg_class t, pg_class i, pg_index d, pg_am am
 WHERE i.relkind = 'i'
 AND d.indexrelid = i.oid
 AND d.indisprimary = 'f'
 AND t.oid = d.indrelid
 AND t.relname = 'bar'
 AND i.relnamespace IN (SELECT oid FROM pg_namespace WHERE nspname IN ('b','a') )
 AND i.relam = am.oid
 ORDER BY i.relname
;
      relname       | indisunique | indkey |  oid  | amname 
--------------------+-------------+--------+-------+--------
 index_bar_on_email | t           | 1      | 28585 | btree
 index_bar_on_email | t           | 1      | 28592 | btree
(2 rows)

regress=> SELECT DISTINCT i.relname, d.indisunique, d.indkey, 'bar'::regclass::oid, am.amname
 FROM pg_class i, pg_index d, pg_am am
 WHERE i.relkind = 'i'
 AND d.indexrelid = i.oid
 AND d.indisprimary = 'f'
 AND 'bar'::regclass = d.indrelid
 AND i.relam = am.oid
 ORDER BY i.relname                                                              
;
      relname       | indisunique | indkey | regclass | amname 
--------------------+-------------+--------+----------+--------
 index_bar_on_email | t           | 1      | 28585    | btree
(1 row)