Paul DuBois wrote:
> 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: mysql@stripped
>> 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.
