PostgreSQL 8.0 rc1 的一個BUG，已經提交給Developement team

¤PostgreSQL version: 8.0

¤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 should 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:

PostgreSQL Official Site

PostgreSQL 中文站