Re: Foreign keys and pg_user table

Well, it's not a table! It's a view:

dbs=# \d pg_user
View "pg_catalog.pg_user"
Column | Type | Modifiers
-------------+---------+-----------
usename | name |
usesysid | integer |
usecreatedb | boolean |
usesuper | boolean |
usecatupd | boolean |
passwd | text |
valuntil | abstime |
useconfig | text[] |
View definition:
SELECT pg_shadow.usename, pg_shadow.usesysid, pg_shadow.usecreatedb,
pg_shadow.usesuper, pg_shadow.usecatupd, '********'::text AS passwd,
pg_shadow.valuntil, pg_shadow.useconfig
FROM pg_shadow;

So, you really want to use the pg_shadow table.

C G wrote:

>
> but I get told that "...pg_user is not a table."
>
> Is there another way of doing what I want?