<?php include ("db.php");?>
<HTML>
<HEAD>
<TITLE> </TITLE>
<META NAME="Keywords" CONTENT="">
<META NAME="Description" CONTENT="">
</HEAD>

<BODY>

<?
$nume_tag=$row["nume_tag"];

$sql = "select *, MATCH (titlu, description) AGAINST ('nume_tag') AS id_unic
FROM produs WHERE MATCH (titlu, description) AGAINST ('nume_tag')
ORDER BY id_unic DESC LIMIT 10";


$result = mysql_query ($sql);
while($row = mysql_fetch_array($result)){
$id_unic=$row["id_unic"];
$titlu = $row["titlu"];
$description = $row["description"];

echo"

$titlu : $description

";}?>

</BODY>
</HTML>
Imi da eroarea: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in line 82 (while($row = mysql_fetch_array($result)){ )

Cred ca e gresit ceva in cod, o virgula ceva. Am tot schimbat,dar tot asa imi apare .

Acum folosesc selectul de mai jos in acelasi cod,care nu da eroare(dar nu prea afiseaza rezultate,cum va spuneam)
$sql = "select * from produs WHERE description LIKE '%$nume_tag%' ORDER BY id_unic DESC LIMIT 20";