Algo sigue estando mal y de cualquier manera GRACIAS por todo lo que ya me han ayudado......
Le anexo mi script y tal vez podamos encontrar la solucion.... :)
Código PHP:
<?php
$link = mysql_connect("localhost", "root", "xxx");
mysql_select_db("articles", $link);
$sql = "SELECT DATE_FORMAT(date,'%b, %Y') AS fecha_formateada, section, locate2, title, description, url FROM articles WHERE (section='hotels.php') AND (locate2='PV') ORDER BY date";
//esta es otra opcion que intente YO....
//$sql = "SELECT section, locate2, title, description, url DATE_FORMAT(date,'%b, %Y') AS fecha_formateada FROM articles WHERE (section='hotels.php') AND (locate2='PV')";
$result = mysql_query($sql, $link);
if ($row = mysql_fetch_array($result)){
echo "<img src='../images/hotels.gif'>\n";
echo "<table width='100%' border='0' cellspacing='8'>\n";
do {
echo "<tr>\n";
echo "<td valign='top'><div><a href='".$row["url"]."' class='links'>".$row["title"]."</a> <span class='date'>".$row["fecha_formateada"]."</span></div>\n";
echo "<div class='articlestext'>".$row["description"]."</div>\n";
echo "</td>\n";
echo "</tr>\n";
} while ($row = mysql_fetch_array($result));
echo "</table>\n";
}
?>
Este es el error:: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /usr/local/etc/httpd/htdocs/articles/fecha.php on line 7