Salve ragazzi ho un problema, ho creato il seguente script:
Codice PHP:
include("class.php");
$db = new Database;
$db->connect();
$sql = "SELECT * FROM news";
$result = $db->query($sql) or die ("Errore query: " . mysql_error());
$num_rows = $db->numRows();
for ($k = 0; $num_rows > $k; $k++)
{
// Ricaviamo i valori
$id = mysql_result($result, $k, 'id_news');
$testo = mysql_result($result, $k, 'testo');
$data = mysql_result($result, $k, 'data');
if ($k % 2 == 0)
$color = "#F2F2F2";
else
$color = "FFFFFF";
echo "<p style=\"background-color: ".$color."\">";
echo "[b]<img src=\"img/freccia.png\" width=\"13\" height=\"9\" />".$data."[/b]
";
echo $testo;
echo "</p>";
}
Non riesco a capire perchè mi dà il seguente errore:
codice:
Warning: mysql_result(): supplied argument is not a valid MySQL result resource in D:\Inetpub\webs\fitnutritionit\index.php on line 97
Warning: mysql_result(): supplied argument is not a valid MySQL result resource in D:\Inetpub\webs\fitnutritionit\index.php on line 98
Warning: mysql_result(): supplied argument is not a valid MySQL result resource in D:\Inetpub\webs\fitnutritionit\index.php on line 99
Ecco le righe 97,98 e 99:
Codice PHP:
$id = mysql_result($result, $k, 'id_news');
$testo = mysql_result($result, $k, 'testo');
$data = mysql_result($result, $k, 'data');
Grazie anticipatamente,
Gaten