Hi.
This code shows a list of news from a db.
snippet "news"
$db_nome = "test";
$nome_tabella = "news";
$email = "
<a href=\"mailto:
[email protected]\">info</a>";
$connessione = @mysql_connect("localhost", "root", "")
or die("Impossibile stabilire una connessione.");
$db = @mysql_select_db($db_nome, $connessione)
or die("Impossibile selezionare il database.");
$sql = "SELECT * FROM $nome_tabella ORDER BY data DESC";
$risultato = @mysql_query($sql,$connessione)
or die("Impossibile eseguire l’interrogazione.");
while ($riga = mysql_fetch_array($risultato)) {
$id = $riga[’id’];
$titolo = $riga[’titolo’];
$data = $riga[’data’];
$note = $riga[’note’];
$visualizza_blocco .= "
<P><strong>$titolo
</strong>
$note<em>$email</em>
$data</P> ";
}
return $visualizza_blocco;
For each news I want to show a random image and so I want to run the random_img snippet
snippet "random_img"
$dir=opendir("./imm");
$directory=("imm/");
$pattern=".(GIF|JPG|gif|jpg|jpeg|JPEG|png|PNG)$";
$s=readdir($dir);
$count=0;
$image = array();
while($s)
{
if(ereg($pattern, $s)) {
$image[$count]=$s;
$count++;
}
$s=readdir($dir);
}
closedir($dir);
//Spit it out
srand(time());
$limit=count($image);
$limit--;
$randNum=mt_rand(0,$limit);
$filename = "$image[$randNum]";
return "<img src=’$directory$filename’ alt=’$filename’ />";
How to run random_img snippet into news snippet?
Tnk in advance