yobo Posted March 26, 2007 Share Posted March 26, 2007 hey all, i am having problems with this script the first part works ok but if i add another mysql_fetch_array it gives me an error Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Program Files\xampp\htdocs\phpbb3hacks\profile.php on line 61 anyways here is my code <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>Untitled Document</title> </head> <body> <h1> Viewing Profile For <?php $userid=$_GET['userid']; echo "$userid"; ?> </h1> <br /> Deatils for <?php $userid=$_GET['userid']; echo "$userid"; ?><br /> <br /> <table width="50%" border="0" cellspacing="0" cellpadding="3" align="left" > <tr> <th align="left">Name</th> <th align="left">E-Mail</th> <th align="left">Website</th> </tr> <?php include("config.php"); // connect to the mysql server $link = mysql_connect($server, $db_user, $db_pass) or die ("Could not connect to mysql because ".mysql_error()); // select the database mysql_select_db($database) or die ("Could not select database because ".mysql_error()); $sql=mysql_query("SELECT * FROM users"); $result=mysql_fetch_array($sql); $userid = $result['userid']; $name = $result['name']; $email = $result['email']; $website = $result['website']; echo "<td>$name</td>\n"; echo "<td>$email</td>\n"; echo "<td>$website</td>\n"; ?> </table> <br /> <br /> <br /> <br /> Hacks Submmited By <?php $userid=$_GET['userid']; echo "$userid"; ?> <br /> <table width="50%" border="0" cellspacing="0" cellpadding="3" align="left" > <tr> <th align="left">Hack Name</th> <th align="left">Description</th> </tr> <?php $sql2=mysql_query("SELECT * FROM hacks WHERE 'username'={$_GET['userid']}"); $result=mysql_fetch_array($sql2); $userid = $result['userid']; $hackname = $result['hackname']; $description = $result['description']; echo "<td>$hackname</td>\n"; echo "<td>$description</td>\n"; ?> </table> </body> </html> Link to comment https://forums.phpfreaks.com/topic/44393-multiple-mysql_fetch_array-problems/ Share on other sites More sharing options...
per1os Posted March 26, 2007 Share Posted March 26, 2007 Use that or DIE(mysql_error()) after your mysql_query see why it is throwing an error. I am betting it is in the WHERE statement due to single quotes ( ' ) and not ( ` ). IE: WHERE 'username' SHOULD BE WHERE `username`='".$_GET['userid']."' Note single quotes around the inputed data. Link to comment https://forums.phpfreaks.com/topic/44393-multiple-mysql_fetch_array-problems/#findComment-215595 Share on other sites More sharing options...
yobo Posted March 26, 2007 Author Share Posted March 26, 2007 thanks mate it works Link to comment https://forums.phpfreaks.com/topic/44393-multiple-mysql_fetch_array-problems/#findComment-215604 Share on other sites More sharing options...
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