Código PHP:
$enlace = mysql_connect(xxx);
mysql_select_db(xxx, $enlace);
$_pagi_sql = mysql_query("SELECT * FROM aprender");
include('paginator.php');
echo "<ul>";
while($array = mysql_fetch_array($_pagi_result)){
echo "<li><a href=veraprender.php?id=".$array['id']." target=_top>".$array['titulo']."</a>";
}
echo "</ul>";
echo $_pagi_navegacion;
Error en la consulta de conteo de registros: Resource id #9. Mysql dijo: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Resource id #9' at line 1
¿Por qué me sale?