Как ще стане ъплоуд скрипта с 5 фаила на веднъж като след секи успешно качен файл искам да добавя в базата данни ето и скрипта
Код:
// Update Additional Picture //
///////////////////////////////////////
$ss = "-";
$BASEURL = "http://" . $_SERVER["HTTP_HOST"];
if (isset($_POST['update_additional_picture'])){
$ext = strstr($_FILES['additional_picture']['name'],".");
$temp_number = rand(2, 999999999);
$temp_number_small = rand(2, 999999);
$name = $_FILES['additional_picture']['name'];
$file = "pictures/".$temp_number."$ss".$temp_number_small.$ext;
if (($ext!= ".jpg") && ($ext!= ".gif") && ($ext!= ".bmp") && ($ext!= ".jpeg") && ($ext!= ".png") && ($ext!= ".GIF") && ($ext!= ".JPG")){
$error = "<BR><span class=\"error\">Не позволен формат на изображението</span>";
} else if (copy($_FILES['additional_picture']['tmp_name'], "pictures/".$temp_number."$ss".$temp_number_small.$ext)){
//mysql_query("INSERT INTO gallery (id, path) VALUES ( 'NULL', '".$file."')");
$picture_error = "".$lang['picture_additional_uploaded']."<BR>
<textarea style=\"width:400px; height:40px;\" id=\"copy\" readonly=\"readonly\">$BASEURL/pictures/$temp_number-$temp_number_small$ext</textarea>
<input type=\"button\" onfocus=\"document.getElementById('copy').select();\" value=\"Маркирай\" /> ";
//echo "<br> <a href=$BASEURL/pictures/$temp_number-$temp_number_small$ext >$BASEURL/pictures/$temp_number-$temp_number_small$ext</a>";
}
}?>
<table width="280" border="0" class="title_text">
<?php
echo "<form method=\"post\" action=\"upload.php\" enctype=\"multipart/form-data\">";
?>
<tr>
<td height="21"><?=$lang['picture_additional_new'];?></td>
</tr>
<tr>
<td height="21"><input name="additional_picture" type="file" class="main_text_normal" size="38"></td>
</tr>
<tr>
<td height="21"><input type="submit" name="update_additional_picture" value="Качи" class="main_text_normal"><?=$error;?></td>
</tr>
<tr>
<td height="21" class="error"><?=$picture_error;?></td>
</tr>
</form>
</table>