it show me this:PHP Code:<?php
$con = mysql_connect("localhost", "username", "password");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("sjpayme_search", $con);
$result = mysql_query("SELECT * FROM search WHERE type='web' DESC");
while($row = mysql_fetch_array($result))
{
echo $row[query];
}
mysql_close($con);
?>
what is the problem?Code:Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /public_html/main/xyz.php on line 18


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