Do not use double kotes here. $link_glob = \"mysql_connect(\'$host_glob\', \'$un_glob\', \'$pw_glob\')\";
mysql_connect is a function. Regards, Laercio Fortes Citando Jason Soza <[EMAIL PROTECTED]>: > If this is truly the code you\'re using, you\'re missing the closing > curly-brace after the else statement. I.e. this: > } else { > echo \"<font size=+1>Your Information has successfully been > entered into the database!</font><br>\"; > > Should be this: > } else { > echo \"<font size=+1>Your Information has successfully been > entered into the database!</font><br>\"; > } > > HTH, > Jason Soza > > ----- Original Message ----- > From: Jule Slootbeek <[EMAIL PROTECTED]> > Date: Monday, June 3, 2002 2:09 pm > Subject: stupid error, please kick me (and send me a solution) > > > Hey guys, > > i\'m getting this error with the following sql script using php: > > --error-- > > Warning: Supplied argument is not a valid MySQL-Link resource in > > /var/www/phpquiz/register_user.php on line 12 > > --error-- > > > > --script-- > > $link_glob = \"mysql_connect(\'$host_glob\', \'$un_glob\', \'$pw_glob\')\"; > > $query = \"INSERT INTO user values(\'0\', \'$fname\', \'$lname\', > > \'$email\', > > \'$username\', PASSWORD(\'$password\')\"; > > $result = mysql_db_query(\'$db_glob\', \'$query\', > > $link_glob\'); if (!$result) { > > echo \"<font size=+1>Your Information > > could not be entered into the database, > > Please contact the > > href=mailto:$webmaster>webmaster.</font><br><br>\" . mysql_errno() > > . > > \": \" . mysql_error() . \"<br><br>\"; > > } else { > > echo \"<font size=+1>Your Information > has > > successfully been entered into the > > database!</font><br>\"; > > > > -- > > Jule Slootbeek > > [EMAIL PROTECTED] > > > > http://blindtheory.cjb.net > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php