php upload multiple files via ftp

Question

I am currently using the following code to upload images to an ftp server:

<input name="userfile" type="file" size="50">

...

$filep=$_FILES['userfile']['tmp_name'];
$name=$_FILES['userfile']['name']; 

$upload = ftp_put($conn_id, $paths.'/'.$name, $filep, FTP_BINARY);

This works fine except that it only uploads one image.

Any help with multiple (3-4) image uploads pls.

Thanks in advance

Take all images name in array and make all code in loop

softsdev
– softsdev

2012-10-31 12:59:04 +00:00
Commented Oct 31, 2012 at 12:59 — softsdev
– softsdev, Commented Oct 31, 2012 at 12:59
I was hoping for some code hint

echez
– echez

2012-10-31 13:03:48 +00:00
Commented Oct 31, 2012 at 13:03 — echez
– echez, Commented Oct 31, 2012 at 13:03
Did any of these answers help you?

James
– James

2012-10-31 13:26:47 +00:00
Commented Oct 31, 2012 at 13:26 — James
– James, Commented Oct 31, 2012 at 13:26

Mihai Matei · Accepted Answer · 2012-10-31 13:03:02Z

4

<input name="userfile[]" type="file" size="50" />
<input name="userfile[]" type="file" size="50" />
<input name="userfile[]" type="file" size="50" />

for($i = 0; $i < count($_FILES['userfile]); $i++) {
   $filep=$_FILES['userfile']['tmp_name'][$i];
   $name=$_FILES['userfile']['name'][$i]; 

   $upload = ftp_put($conn_id, $paths.'/'.$name, $filep, FTP_BINARY);
}

answered Oct 31, 2012 at 13:03

Mihai Matei

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Comments

James · Accepted Answer · 2012-10-31 13:04:41Z

Here's one I had used. All the image fields have the same name property value of files

<input type="file" name="files[]" class="files" multiple /> can be used for html5, or add multiple of this line for non-html5. I used jquery for my uploader, but you can just have multiple upload fields visible instead.

Obviously you can change my code to use your ftp code, but here's another way too.

        $files = array();

        $allowed_filetypes = array('.jpg','.jpeg','.gif','.bmp','.png','.tif');
        $max_filesize = 1524288;
        $upload_path = 'images/image_uploads/';

        for ($i = 0; $i < count($_FILES['files']['name']); $i++){
            if($_FILES['files']['name'][$i] != "") {
                $filename = $_FILES['files']['name'][$i];
                $ext = substr($filename, strpos($filename,'.'), strlen($filename)-1); 
                $ext = strtolower($ext);

                if(!in_array($ext,$allowed_filetypes))
                    die("The file you attempted to upload ($filename) is not allowed.");

                if(filesize($_FILES['files']['tmp_name'][$i]) > $max_filesize)
                    die("The file you attempted to upload ($filename) is too large.");

                if(!is_writable($upload_path))
                    die("You cannot upload to the specified directory, please CHMOD it to 777.");

                $ran = rand();
                $filename = $ran.$ext;
                if(move_uploaded_file($_FILES['files']['tmp_name'][$i],$upload_path.$filename)) {
                    $result = mysql_query("Insert Into image_uploads_images (upload_id, image, original_name) Values ('$id', '$filename', '".$_FILES['files']['name'][$i]."');");

                    if($result){
                        array_push($files, "http://www.site.com/images/image_uploads/$filename => ".$_FILES['files']['name'][$i]);
                    }else{
                        echo "<p style=\"color:#cc3333;\">Unable to upload ".$_FILES['files']['name'][$i]."</p>";
                    }
                }else{
                    echo "<p style=\"color:#cc3333;\">Unable to upload ".$_FILES['files']['name'][$i]."</p>";
                }
            }
        }

DJafari · Accepted Answer · 2012-10-31 13:13:01Z

1

you should upload files async one by one :)

HTML :

<input name="userfile[]" type="file" size="50" />
<input name="userfile[]" type="file" size="50" />
<input name="userfile[]" type="file" size="50" />

PHP :

for( $i = 0 ; $i < count( $_FILES['userfile'] ) ; $i++ )
{
   $filep = $_FILES['userfile']['tmp_name'][$i];
   $name = $_FILES['userfile']['name'][$i];

   $upload = ftp_put( $conn_id , $paths . '/' . $name , $filep , FTP_BINARY );
}

edited Oct 31, 2012 at 13:13

answered Oct 31, 2012 at 13:02

DJafari

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