Dunque...
Ho un file form.php
Codice PHP:
<form name="dataEntry" id="dataEntry" method="POST" enctype="multipart/form-data" action="<? echo "ins.php"; ?>">
<tr>
<td class="label"><label id="linfo" for="info">Order</label></td>
<td class="label"><input type="file" name="info" id="info"></td>
<td class="status"></td>
</tr>
<input type=submit value=Add><input type=reset value=Clear>
</form>
ed il relativo ins.php
Codice PHP:
if ($_FILES["info"]["error"] = 0){
$temp_file = $_FILES['info']['tmp_name'];
$name_file = $_FILES['info']['name'];
$type_file = $_FILES['info']['type'];
$size_file = $_FILES['info']['size'];
$fp = fopen($temp_file, 'r');
$content_info = fread($fp, filesize($temp_file));
$content_info = mysql_escape_string($content_info);
fclose($fp);
$name_file = mysql_escape_string($name_file);
}
echo "$content_info";
echo "$name_file";
mysql_query("INSERT INTO supplies (order, nameorder, typeorder, sizeorder) VALUES ('$content_info','$name_file','$type_file','$size_file')")or die("Insert Error: ".mysql_error());
mi restituisce questo errore:
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 'order, nameorder, typeorder, sizeorder, companies_idcompanies) VALUES ('','','',' at line 1
Non riesco a capire perchè!!! Ma la cosa incredibile è che in un altro file ho lo stesso identico codice e non mi da nessun errore...
In pratica è come se $_FILES fosse vuoto...
gli echo che ho messo echo "$content_info"; echo "$name_file"; non restituiscono nulla!!!
Chi mi sa dire perchè? Cosa devo fare?
Grazie!!!