FTP question

[Richard Hulbert]

Ok Its getting late and i am getting tired. what stupid mistake am I making
here:

<?php
$filepath = $_REQUEST['filepath']; // this is the path to say music/preview
or /cd_covers
$ftp_server = 'ftp.XXXXX.co.uk';
$ftp_user_name = 'xxxxxxx';
$ftp_user_pass = 'xxxxxxxx';
$file = $_FILES['userfile']['name'];
$remote_file = $filepath.$file;
// set up a connection or die
$conn_id = ftp_connect($ftp_server) or die("Couldn't connect to
$ftp_server");
$login_result = ftp_login($conn_id, $ftp_user_name, $ftp_user_pass) or
die("login failed"); // logs the user or sets an error message

if(ftp_chdir($conn_id, $filepath)){
}else{
echo "Directory doesnot exist";
exit;
}
// upload a file\
if (ftp_put($conn_id, $remote_file, $file, FTP_BINARY)) {
echo "successfully uploaded $file\n";
} else {
echo "There was a problem while uploading $file\n";
}

ftp_close($conn_id);
?>
straight out of the manual what gives?
Cheers Richard

[Cameron]

Richard Hulbert wrote:

Ok Its getting late and i am getting tired. what stupid mistake am I
making here:

<code snip>

The actual error message or what ever would be helpful.

~Cameron

[Pedro Graca]

Richard Hulbert wrote:

Ok Its getting late and i am getting tired. what stupid mistake am I making
here:

<?php
$filepath = $_REQUEST['filepath']; // this is the path to say music/preview
or /cd_covers
$ftp_server = 'ftp.XXXXX.co.uk';
$ftp_user_name = 'xxxxxxx';
$ftp_user_pass = 'xxxxxxxx';
$file = $_FILES['userfile']['name'];
$remote_file = $filepath.$file;

(snip)

if I update a file named XXX, seems like $remote_file will have either
music/previewXXX or /cd_coversXXX

Are you sure that is what you want?
What exact error do you get? At which instruction?

This is like asking us to guess what is the license plate on your car
just be telling us it's a FOOBAR model!
--
--= my mail box only accepts =--
--= Content-Type: text/plain =--
--= Size below 10001 bytes =--

[Richard Hulbert]

On 2004-02-23 19:06:53 +0000, Pedro Graca <he****@hotpop.com> said:

Are you sure that is what you want?
What exact error do you get? At which instruction?

This is like asking us to guess what is the license plate on your car
just be telling us it's a FOOBAR model!
-- --= my mail box only accepts =--
--= Content-Type: text/plain =--
--= Size below 10001 bytes =--

Firstly Thanks for answering.

Ok Sorry let me step back a bit. firstly here is what i am trying to do:

I am building a CMS that needs to upload two supporting files to
directories on the public server. I want to do this from another site, at
the moment I am doing this from localhost on my dev machine.

I thought FTP option would be better in terms of preventing php actually
writing files to the server.

Ok so the Error - I don't get an error as such, i connect in, can navigate
directories,etc it is the last call that is giving me a problem.

if (ftp_put($conn_id, $remote_file, $file, FTP_BINARY)) {
echo "successfully uploaded $file\n";
} else {
echo "There was a problem while uploading $file\n";
}

now $remote file is a string right? somthing like /music/previews/abc.mp3
and $file is from $file = $_FILES['userfile']['name'];

basically ftp_put fails and I Echo There was a problem...

am i missunderstanding how to use the $_FILES object in this context?
thanks Richard

[Richard Hulbert]

SNIP

aha I have was indeed missing how to reference the file object. so to make
it all work really well.

$upload = (ftp_put($conn_id, $filepath . $_FILES['userfile']['name'],
$_FILES['userfile']['tmp_name'], FTP_BINARY));

the key being $_FILES['userfile']['tmp_name']

i am now going to write one hundred times 'php references everything from
the server !'

thanks Richard

[Dan Tripp]

> i am now going to write one hundred times 'php references everything

from the server !'

Nah, just write a loop with a counter and that'll do it. ;)

- Dan
http://blog.dantripp.com/