Use INSERT INTO reference(reference_text, reference_type, Topic_Id) VALUES (DEFAULT,'123','2',1); instead. Since referencelist_nr is a serial PostgreSQL will provide the next free integer itself.
>>> "remco lengers" <[EMAIL PROTECTED]> 2007-03-20 10:11 >>> Hi List, Its been a while since I worked with databases and I am running into the following which I have not been able to find the root cause for: I have the follow SQL statement: INSERT INTO reference VALUES (DEFAULT,'123','2',1); Which generates the following error: "ERROR: duplicate key violates unique constraint "reference_pkey"" And the table definition looks like: CREATE TABLE reference ( referencelist_nr serial unique, reference_text varchar(40) NOT NULL, reference_type integer NOT NULL, Topic_Id integer NOT NULL, PRIMARY KEY (referencelist_nr), FOREIGN KEY (Topic_Id) REFERENCES Topic (Topic_Id), FOREIGN KEY (reference_type) REFERENCES reference_type (reference_type_nr) ); It seems to me for some reason "DEFAULT" doesn't select the next SERIAL. If I run: INSERT INTO reference VALUES (14,'123','2',1); 14 being the next free integer it works fine..... I have been adding data with "COPY" into this table....does that break something? Your help/tips/insights are appreciated. Solaris 10/Postgresql 8.1.4 Regards, ..Remco
