> -----Original Message-----
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of
> Massimiliano Citterio
> Sent: 22 September 2006 09:21
> To: pgsql-bugs@postgresql.org
> Subject: [BUGS] BUG #2644: pgadmin III foreign key
>
>
> The following bug has been logged online:
>
> Bug reference: 2644
> Logged by: Massimiliano Citterio
> Email address: [EMAIL PROTECTED]
> PostgreSQL version: 8.1.4
> Operating system: Windows 2003
> Description: pgadmin III foreign key
> Details:
>
> The referencing column dropdown listbox not filled with fields from
> referenced table.
>
> The case appear after doing the following actions.
>
> Create a schema named like $user (ex. "postgres").
> Delete Schema "Public".
> Refresh the Database View...
> Now the default Schema for the database is "postgres"
>
> create tables, the try to create foreign keys using the GUI.
> There is no way.
>
> If you create a new schema "public" it does not works.
>
> If you create a new schema "public" and rename the "postgres"
> schema to
> something else the it works.
>
> If you drop schema "postgres" create schema "public" and then recreate
> schema postgres and his tables, then foreign key GUI works.
>
> May be the GUI search for schema "public" but it must have an
> OID lesser of
> all other schemas in the searchpath.
>
> Infact the default search path is $user, public.
>
> By the way it does not work if a schema named public does not exist.
I cannot reproduce this in the 1.6 development code - can you test beta
1 please?
Regards, Dave.
---------------------------(end of broadcast)---------------------------
TIP 9: In versions below 8.0, the planner will ignore your desire to
choose an index scan if your joining column's datatypes do not
match
