Re: Invalid MySQL result resource

On Thu, 13 Sep 2001 14:15, Dan Harrington wrote: > > -----Original Message----- > > From: speedboy [mailto:speedboy@nomicrosoft.org] > > Sent: Wednesday, September 12, 2001 10:27 PM > > To: php-general@lists.php.net > > Subject: Re: [PHP] Invalid MySQL result resource > > > > > "Supplied argument is not a valid MySQL result resource in ..... on > > > line 52" > > > > > > Line 52 is the line where I fetch the row from the second query. > > > > Don't keep doing a mysql_connect on each loop. > > ....thats whats weird about this, > > I'm not. > > Here's what's going on. > ======================================================== > > $connection = mysql_connect($server, $user, $pass); > mysql_select_db($db); > > $query = "select * from shoppingcart where PHPSESSID = '$PHPSESSID'"; > $result = mysql_query($query,$connection); > $numrows = mysql_num_rows($result); > > $count = 0; > > while ($count < $numrows){ > $cartitem = mysql_fetch_array($result); > $productquery = "select * from products where productid = > '$cartitem[productid]'"; $productresult = mysql_query($productquery); Ah. Should that be $cartitem[$productid] ??? And perhaps better written as $productquery = "select * from products where productid = '" . $cartitem[productid] . "'"; to be sure that the array element is dereferenced properly. > $productinfo = mysql_fetch_array($productresult); //THIS line produces > the error $count++; > } > > mysql_close($connection); > > ======================================================== > > I don't get it. I must be blind or something because I just don't > see whats causing this. > > Thanks again, > Dan -- David Robley Techno-JoaT, Web Maintainer, Mail List Admin, etc CENTRE FOR INJURY STUDIES Flinders University, SOUTH AUSTRALIA Half the lies they tell me aren't true. --Yogi Berra