CODEWITHSUNDEEP
postgresqlpermissionsprimary-keyinformation-schema
Alan Moore
Alan Moore

Reputation: 319

Find primary key of table in Postgresql from information_schema with only SELECT

I am using the following query to discover (1) the primary key columns and (2) if the columns have a default value from the information_schema in Postgresql 9.1.

SELECT kcu.column_name, (c.column_default is not null) AS has_default 
FROM information_schema.key_column_usage kcu 
JOIN information_schema.table_constraints tc ON tc.constraint_name = kcu.constraint_name 
JOIN information_schema.columns c on c.column_name = kcu.column_name and c.table_name = kcu.table_name  
WHERE tc.constraint_type = 'PRIMARY KEY' AND kcu.table_name like :tablename

It works fine when run as the database owner, but when I run it as a "read-only" user (which I need to do in my application), it returns no data. Some research revealed that the problem is the information.table_constraints view; from the documentation:

The view table_constraints contains all constraints belonging to tables that the current user owns or has some non-SELECT privilege on.

So in order to retrieve table_constraints, my login role needs more than SELECT on the table? Is there no way to get the information from information_schema without giving write permissions to the login role?

Upvotes: 2

Views: 15785

Answers (3)

Alex Yu
Alex Yu

Reputation: 3537

There is another way to provide access for data in information_schema.

A. Envelop SQL in a function with SECURITY DEFINER modifier

CREATE FUNCTION fn_inf(name)
RETURNS TABLE (column_name information_schema.sql_identifier, has_default bool)
LANGUAGE SQL
SECURITY DEFINER
AS $$
SELECT kcu.column_name, (c.column_default is not null) AS has_default 
    FROM information_schema.key_column_usage kcu 
        JOIN information_schema.table_constraints tc ON tc.constraint_name = kcu.constraint_name 
        JOIN information_schema.columns c on c.column_name = kcu.column_name and c.table_name = kcu.table_name  
    WHERE tc.constraint_type = 'PRIMARY KEY' AND kcu.table_name like $1;
$$;

B. GRANT EXECUTE to user read_only

GRANT EXECUTE ON FUNCTION fn_inf to read_only;

C. Use as user read_only

SELECT * FROM fn_inf('spatial_ref_sys');
column_name has_default
srid false

Upvotes: 0

krokodilko
krokodilko

Reputation: 36127

Use pg_* views instead of information_schema views.
pg_* views display all information regardles of granted privileges.

Try this query:

select
    t.relname as table_name,
    i.relname as index_name,
    a.attname as column_name,
    d.adsrc   as default_value
from
    pg_class t
    join pg_attribute a on a.attrelid = t.oid
    join pg_index ix    on t.oid = ix.indrelid AND a.attnum = ANY(ix.indkey)
    join pg_class i     on i.oid = ix.indexrelid
    left join pg_attrdef d on d.adrelid = t.oid and d.adnum = a.attnum  
where
    t.relkind = 'r'
    and t.relname in ( 'aa', 'bb', 'cc' )
order by
    t.relname,
    i.relname,
    a.attnum;

An example of the query results:

create table aa(
  x int primary KEY
);

create table bb(
  x int default 1,
  constraint pk primary key ( x )
);

create table cc(
  x int default 20,
  y varchar(10) default 'something',
  constraint cc_pk primary key ( x, y )
);

 table_name | index_name | column_name |         default_value
------------+------------+-------------+--------------------------------
 aa         | aa_pkey    | x           |
 bb         | pk         | x           | 1
 cc         | cc_pk      | x           | 20
 cc         | cc_pk      | y           | 'something'::character varying

Upvotes: 5

user3094383
user3094383

Reputation: 594

This is correct, the official postgresql query is below

http://wiki.postgresql.org/wiki/Retrieve_primary_key_columns

if schema is needed the query is as follows

SELECT               
  pg_attribute.attname, 
  format_type(pg_attribute.atttypid, pg_attribute.atttypmod) 
FROM pg_index, pg_class, pg_attribute, pg_namespace 
WHERE 
  pg_class.oid = 'MY TABLE'::regclass AND 
  indrelid = pg_class.oid AND 
  nspname = 'MY CLASS' AND 
  pg_class.relnamespace = pg_namespace.oid AND 
  pg_attribute.attrelid = pg_class.oid AND 
  pg_attribute.attnum = any(pg_index.indkey)
  AND indisprimary

The difference can be up to 6000~7000 times. The pg_ one runs often in 0.56ms where the schema based one can run up 6500ms. This is a huge difference especially if you have a high load on the server.

Upvotes: 0

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