RE: [PHP-WIN] Re: MySQL Errors

This might also be of some help. Change your result and fetch lines to read: $result = mysql_query($sql,$db) or die (mysql_error()); $myrow=mysql_fetch_array($result) or die(mysql_error()); David -----Original Message----- From: Robert Schipper [mailto:froschi82@web.de] Sent: Thursday, October 21, 2004 7:49 AM To: php-windows@lists.php.net Subject: [PHP-WIN] Re: MySQL Errors hello mikea, looks like there is something missing in your sourcecode, which you should have provided. do you give any arguments to mysql_fetch_array() ? Do you check if it is not false? your code should look like this: $sql = "SELECT * FROM example"; $result = mysql_query($sql,$db); if ($result) { $myrow=mysql_fetch_array($result); } else { echo "ERROR: SQL-query returned false!"; } Robert Mikea wrote: > I am getting the following errors when I execute a php script. > > Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result > resource > Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result > resource > Warning: mysql_num_rows(): supplied argument is not a valid MySQL result > resource > > What do I need to make this work? I am new to PHP. I read something about > mysql.class.php somewhere. Do I need that somewhere? Do I put an include > in the main script? Where does mysql.class.php go directory-wise? > > I am sure I am not giving enough information and may not be explaining it > good enough. My apologies. Again, I am new at this. > > I appreciate any help. Thank you in advance. > > Mike -- PHP Windows Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php