Thomas, I changed it so that the two connections use different username/password combination, and everything went smoothly.
Thank you very much for your rapid response. Tim "Thomas Seifert" <[EMAIL PROTECTED]> wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... On Sat, 1 Feb 2003 13:28:00 -0800 [EMAIL PROTECTED] (Tim Lan) wrote: > The following code is supposed to migrate data from an old database to a new > one, but produces the error: > > Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result > resource in C:\...\convertdb.php on line 17 > > The code: > > <?php > > /* connect to database */ > if(!($srcDB = mysql_pconnect('localhost', 'root')) || > !mysql_select_db('temp', $srcDB)) { > echo 'Unable to connect to source database'; > exit; > } > > if(!($desDB = mysql_pconnect('localhost', 'root')) || > !mysql_select_db('cgu', $desDB)) { > echo 'Unable to connect to destination database'; > exit; > } > ah yes and another thing. you can't work on two databases with the same connection parameters (host, username,password) to have different connection-ids. PHP only keeps one connection and not two with the same parameters. They are the same connection for php which is expected behaviour there (although I don't like it too). Regards, -- Thomas Seifert mailto:[EMAIL PROTECTED] http://www.MyPhorum.de -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
