Re: BUG #2644: pgadmin III foreign key

=20

> -----Original Message-----
> From: pgsql-bugs-owner@postgresql.org=20
> [mailto:pgsql-bugs-owner@postgresql.org] On Behalf Of=20
> Massimiliano Citterio
> Sent: 22 September 2006 09:21
> To: pgsql-bugs@postgresql.org
> Subject: [BUGS] BUG #2644: pgadmin III foreign key
>=20
>=20
> The following bug has been logged online:
>=20
> Bug reference:      2644
> Logged by:          Massimiliano Citterio
> Email address:      massicitte@libero.it
> PostgreSQL version: 8.1.4
> Operating system:   Windows 2003
> Description:        pgadmin III  foreign key
> Details:=20
>=20
> The referencing column dropdown listbox not filled with fields from
> referenced table.
>=20
> The case appear after doing the following actions.
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> Create a schema named like $user (ex. "postgres").
> Delete Schema "Public".
> Refresh the Database View...
> Now the default Schema for the database is "postgres"
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> create tables, the try to create foreign keys using the GUI.
> There is no way.
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> If you create a new schema "public" it does not works.
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> If you create a new schema "public" and rename the "postgres"=20
> schema to
> something else the it works.
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> If you drop schema "postgres" create schema "public" and then recreate
> schema postgres and his tables, then foreign key GUI works.
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> May be the GUI search for schema "public" but it must have an=20
> OID lesser of
> all other schemas in the searchpath.
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> Infact the default search path is $user, public.
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> 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.