At 8:20 -0500 7/26/04, Victor Pendleton wrote:
Have you tried using the last insert id function instead? SET @backup_id = last_insert_id()
That'll give him the same result. I suspect the problem might be that user variables are not replicated well in MySQL 4.0.x.
Yes, same error.
Philippe, what version of MySQL are you using? If 4.0.x, you might try skipping the SET statement and just refer to LAST_INSERT_ID() or @@LAST_INSERT_ID() directly in your second INSERT statement.
the version is 4.0.15.
the pb is that I have two INSERT to do with the same id... any workaround for that ?
-----Original Message----- From: Philippe Poelvoorde To: [EMAIL PROTECTED] Sent: 7/26/04 7:03 AM Subject: Replication script pb
Hi,
We have an environnment with a master and a slave. We run a script every
hour (on the master only) that does something like this to backup some parameters : insert into backup(NULL,NULL) VALUES(NULL,NOW()) SET @backup_id = @@LAST_INSERT_ID INSERT INTO backup_param ( SELECT @backup_id, col1, col2 FROM param) It works perfectly on the master but the slave stop due to duplicate entries. the @backup_id do not pass the replication... any solution to have that script working ?
-- Philippe Poelvoorde COS Trading Ltd.
-- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe: http://lists.mysql.com/[EMAIL PROTECTED]
