Jacbey Posted May 22, 2011 Share Posted May 22, 2011 hey guys; Another problem with sql. { echo "Welcome " . $row['username']; echo "<br />"; echo "<br />"; echo "<br />"; $msgquery = "SELECT * FROM spotty_messages WHERE (id_receiver = '" . $userid . "') AND message_read = '0'"; $messageres = mysql_query($msgquery); $messrow = mysql_fetch_array($messageres); $messagenum = mysql_num_rows($messageres); } $i = 0; while ($i < $messagenum) { $f1 = "From:" . mysql_result($messrow,$i,"sender"); echo " <tr> <td>" . $f1 ."</font></td> " ; $i++; } This is returning the error : Warning: mysql_result() expects parameter 1 to be resource, null given in /customers/klueless.net/klueless.net/httpd.www/daisysite/messages.php on line 106 Please help, thanks! Link to comment https://forums.phpfreaks.com/topic/237110-mysql-error/ Share on other sites More sharing options...
wildteen88 Posted May 22, 2011 Share Posted May 22, 2011 You're using the wrong variable here $f1 = "From:" . mysql_result($messrow,$i,"sender"); You want to use the $messageres variable (this is the result resource) Link to comment https://forums.phpfreaks.com/topic/237110-mysql-error/#findComment-1218663 Share on other sites More sharing options...
Jacbey Posted May 22, 2011 Author Share Posted May 22, 2011 You're using the wrong variable here $f1 = "From:" . mysql_result($messrow,$i,"sender"); You want to use the $messageres variable (this is the result resource) Whoo Thanks dude; I owe you one! Link to comment https://forums.phpfreaks.com/topic/237110-mysql-error/#findComment-1218664 Share on other sites More sharing options...
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