RE: [PHP] mysql problems

I agree with Hugh. I think it is always a good idea to check for return values of any function call. So, I'd replace this line: $result = mysql_query("select * from table"); with $result = mysql_query("select * from table"); or die("Unable to connect to SQL server: ". mysql_error()); -Teresa -----Original Message----- From: hugh danaher [mailto:hdanaher@earthlink.net] Sent: Tuesday, March 12, 2002 8:27 PM To: Maarten Weyn; php Subject: Re: [PHP] mysql problems Maarten, Perhaps "table" isn't the name of the table you want. If mysql can't find the table (in line 13?), your $result variable is empty and this causes (line 17?) to fail also. Hope this helps, Hugh ----- Original Message ----- From: "Maarten Weyn" <maarten.weyn@wesdec.com> To: <php-general@lists.php.net> Sent: Tuesday, March 12, 2002 5:19 PM Subject: [PHP] mysql problems > Hi on this code: > > $link = mysql_connect("localhost", "login", "passwd"); > mysql_select_db("table"); > $result = mysql_query("select * from table"); > while ($row = mysql_fetch_object($result)) { > echo $row->ID; > echo $row->Drank; > } > mysql_free_result($result); > > mysql_close($link); > > It resulst in > Warning: Supplied argument is not a valid MySQL result resource in > c:\program files\apache group\apache\htdocs\index.php on line 13 > Warning: Supplied argument is not a valid MySQL result resource in > c:\program files\apache group\apache\htdocs\index.php on line 17 > > line 13 = while ($row = mysql_fetch_object($result)) { > line 17 = mysql_free_result($result); > > I'm running an Apache/1.3.23 on a win 2000 with PHP Version 4.1.2 and mysql > 3.23.39. > > > Does it not recoginze this mysql_... statements? > > > Maarten > > > > -- > 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