Possibilities: 1)There was an error when mysql_query(). Possibly the SQL statement is not correct. Trace with var_dump()/mysql_errno(). 2)You give as parameter to some of mysql_* functions a variable not returned from mysql_query().
HTH Regards, Andrey Hristov ----- Original Message ----- From: "Dan McCullough" <[EMAIL PROTECTED]> To: "PHP General List" <[EMAIL PROTECTED]> Sent: Tuesday, January 08, 2002 6:47 PM Subject: [PHP] Supplied argument is not a valid MySQL result resource > I know this is a common error, but what does it mean. This snippet of code is >something I use all > the time. > > help please. > > dan > > __________________________________________________ > Do You Yahoo!? > Send FREE video emails in Yahoo! Mail! > http://promo.yahoo.com/videomail/ > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, e-mail: [EMAIL PROTECTED] > For additional commands, e-mail: [EMAIL PROTECTED] > To contact the list administrators, e-mail: [EMAIL PROTECTED] > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
