Re: BUG #16395: error when selecting generated column in a foreigntable

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От Peter Eisentraut
Тема Re: BUG #16395: error when selecting generated column in a foreigntable
Дата
Msg-id a4938e39-8369-f321-a691-1912e7a62b13@2ndquadrant.com
обсуждение исходный текст
Ответ на BUG #16395: error when selecting generated column in a foreign table  (PG Bug reporting form <noreply@postgresql.org>)
Список pgsql-bugs
On 2020-04-27 16:24, PG Bug reporting form wrote:
> -- bd: postgres
> CREATE TABLE public.test
> (
>      id integer NOT NULL,
>      field1 integer NOT NULL,
>      field2 character varying COLLATE pg_catalog."default",
>      CONSTRAINT test_pkey PRIMARY KEY (id)
> );
> 
> -- bd: test
> CREATE FOREIGN TABLE public.test(
>      id integer NULL,
>      field1 integer NULL,
>      field2 character varying NULL COLLATE pg_catalog."default",
>      field_calc integer generated always as (field1 * 2) stored
> )
>      SERVER fs_postgres;
> 
> test=# SELECT id, field1, field2, field_calc FROM public.test;
> ERROR:  column "field_calc" does not exist
> CONTEXT:  remote SQL command: SELECT id, field1, field2, field_calc FROM
> public.test

Well, your public.test base table does not in fact have a "field_calc" 
column.  You need to add that.

Note that these are *stored* generated columns, so the value that is 
generated by the foreign table on insert is meant to be stored somewhere 
in the backing store of the foreign table.

-- 
Peter Eisentraut              http://www.2ndQuadrant.com/
PostgreSQL Development, 24x7 Support, Remote DBA, Training & Services



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