while help

[ctcp]

<?
mysql_connect("$db_host","$db_username","$db_password");
mysql_select_db("$db_name");

$query=mysql_real_escape_string($_GET['']);

$query_for_result=mysql_query("SELECT * FROM $db_tb_name WHERE 

$db_tb_atr_name like '%".$query."%' LIMIT 0, 1 ");

//echo "<h2>Search Results</h2><ol>";
while($data_fetch=mysql_fetch_array($query_for_result))
{
   	echo "<li>";
echo "Order ID | #03355" . $data_fetch['id'];
echo "</li><hr/>";
    echo "<li>";
    echo "URL | "  . substr($data_fetch[$db_tb_atr_name], 0,160);
echo "</li><hr/>";
echo "<li>";
echo "Order added | " . substr($data_fetch['date_added'], 0,10);
echo "</li><hr/>";
echo "<li>";
echo "Credits | " . $data_fetch['total_credits'];
echo "</li><hr/>";
	echo "<li>";
echo "Available  | " . $data_fetch['available_credits'];
echo "</li><hr/>";
    echo "<li>";
echo "Total Likes | " . $data_fetch['total_likes'];
    echo "</li><hr/>";

echo "</ol>";
$c=$data_fetch['total_credits'];
$l=$data_fetch['total_likes'];
$stats = $l / $c * 100 ;
//echo "$stats % Complete"; 
//echo $c;
//echo $l;
include('includes/progress.php');
$progress6 = new Progress();
//echo "<li>";
//echo "Campaign progress  $stats %";
$progress6->percent =  $stats . "% Campaign progress";
$progress6->width = 800;
$progress6->height = 22;
$progress6->style = "image";
$progress6->makeBar();
}


$result_1=mysql_query("SELECT $db_tb_name FROM $db_name");     
while($row=mysql_fetch_array($result_1))   //line 76
{
echo "Order ID | #" . $row['id'];	
}
mysql_close();
}

?>

 

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/content/89/7774789/html/xxx/like/clientstats.php on line 76

 

what is wrong here ?

[ManiacDan]

mysql_query returns false when the query fails.  mysql_fetch_array expects a mysql resource, and will throw this error when it receives a false.

 

Your query is wrong.  Look at the line above the error.  That query is missing a field list.

[ctcp]

thank you mate