A bug in PostgreSQL 8.0 rc1 has been submitted to the Developement team.

PostgreSQL version8.0

Operating system: Microsoft Windows 2003

¤ Short description: Pg_get_serial_sequence can not find a table that existing.

¤ Details:

I created one table via SQL command like this:

Create table tTestTable {

FId bigint not null default (-1 ),

CONSTRAINT tTestTable_pkey primary key (FId)

);

And then I created another table via pgAdminIII 1.2 that installed with PostgreSQL 8.0 rc1. SQL like:

Create table tTestTableAnother {

FId bigint not null default nextval ('Public. "tTestTableAnother_FId_seq" ': text ),

CONSTRAINT tTestTableAnother_pkey primary key (FId)

);

I type the following sqls:

> Select pg_get_serial_sequence ('ttesttable', 'fid ')

Result is null, as I expected.

> Select pg_get_serial_sequence ('ttesttable', 'fid ')

Result is null, as I expected.

> Select pg_get_serial_sequence ('ttesttableanother ', 'fid ')

ERROR: relation "ttesttableanother" does not exist

> Select pg_get_serial_sequence ('ttesttableanother ', 'fid ')

ERROR: relation "ttesttableanother" does not exist

Open the pg_class, found 'ttesttableanother 'existing.

Caution, not 'ttesttableanother '.

And more, no table like 'ttesttable' existing while 'ttesttable' appearance.

I guess PostgreSQL 8.0 does not offer a unique interface for creating a table. Or pgAdminIII 1.2.0 calling the old function to create a table? But no matter how, PostgreSQL shoshould offer a unique way. over.

PostgreSQL's reply list:

1. The report (reference: 1356) will be forwarded to the development team for further investigation.

¤ Links:

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