Hi not very experienced at using PHP but i am trying to a bring up a query result which the user has typed in the query themselves into a textarea.

this is my code

<?php
include("dbandy.php");

$conn = mysql_connect($dbhost, $dbuser, $dbpass) or die('Error connecting to mysql');

$dbname = 'db_andrewb';
mysql_select_db($dbname)or die('error finding DB');

$QSTRING = $_SERVER['QUERY_STRING'];

$QSTRING = str_replace('queries=', '',$QSTRING);


echo urldecode($QSTRING);


$query = $QSTRING;
$result = mysql_query($query) or die(mysql_error());


$number = mysql_numrows($result);

if ($number == 0) {
print "Sorry, there were no records matching those criteria";
} else {
for($i=0; $i<$number; $i++){
$degreeid = mysql_result($result,$i,"degreeid");
$degree_name = mysql_result($result,$i,"degree_name");
$modid = mysql_result($result,$i,"modid");
$mod_name = mysql_result($result,$i,"mod_name");
$studentid = mysql_result($result,$i,"studentid");
$stud_fname = mysql_result($result,$i,"stud_fname");
$stud_sname = mysql_result($result,$i, "stud_sname");
$tutorid = mysql_result($result,$i,"tutorid");
$tutor_fname = mysql_result($result,$i,"tutor_fname");
$tutor_sname = mysql_result($result,$i,"tutor_sname");

print "$degreeid, $degree_name, $modid, $mod_name, $studentid, $stud_fname, $stud_sname, $tutorid, $tutor_fname, $tutor_sname<br>";
}
}

mysql_close();

?>

and when trying to put in a query into the text area like SELECT * FROM student this error message comes up:

Warning: mysql_numrows(): supplied argument is not a valid MySQL result resource

Can anyone help me?