Ciao a tutti,
il seguente script non mi funziona:
<?
require "include/template.inc";
require "include/side.inc";
require "include/dbms.inc";
require "include/auth.inc";
$oid = mysql_query("SELECT * FROM private WHERE id='$id'");
$data = mysql_fetch_array($oid);
if ($data) {
header("Content-type: image/jpeg");
echo $data[file];
} else echo "error";
?>
mozilla ritorna il seguente errore:
"The image open_foto.php?id=1 cannot be displayed, because it contains errors"
Ecco lo script per inserire foto:
//contenuto del file
$image = addslashes(fread(fopen($_FILES["file"]["tmp_name"], "rb"), $_FILES["file"]["size"]));
$id = md5(time());
$nome = $_FILES["file"]["name"];
$size = $_FILES["file"]["size"];
$data= date(y)."-".date(m)."-".date(d);
$oid = mysql_query("INSERT INTO private VALUES ('$id','$nome','$descrizione','$data','$image','$s ize','$type')");
Mi sapreste dire dove sbaglio?!?! Sto impazzendo!!!![]()
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