Cannot list FTP directory using ftplib – but FTP client works

I’m trying to connect to an FTP but I am unable to run any commands.

ftp_server = ip
ftp_username = username
ftp_password = password

ftp = ftplib.FTP(ftp_server)
ftp.login(ftp_username, ftp_password)
'230 Logged on'

ftp.nlst()

The ftp.nlst throws this error:

Error:
[WinError 10060] A connection attempt failed because the connected party did not properly respond after a period of time, or established connection failed because connected host has failed to respond

I’ve tested the connection using FileZilla (running on the same machine) and it works fine.

This is FileZilla log:

Status: Connection established, waiting for welcome message...
Status: Insecure server, it does not support FTP over TLS.
Status: Logged in Status: Retrieving directory listing...
Status: Server sent passive reply with unroutable address. Using server address instead.
Status: Directory listing of "/" successful

Answers:

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Method 1

Status: Server sent passive reply with unroutable address

The above means that the FTP server is misconfigured. It sends its internal network IP to outside network (to the client – FileZilla or Python ftplib), where it is invalid. FileZilla can detect that and automatically fall back to the original IP address of the server.

Python ftplib does not do this kind of detection.

You need to fix your FTP server to return the correct IP address.

If it is not feasible to fix the server (it’s not yours and the admin is not cooperative), you can make ftplib ignore the returned (invalid) IP address and use the original address instead by overriding FTP.makepasv:

class SmartFTP(FTP):
    def makepasv(self):
        invalidhost, port = super(SmartFTP, self).makepasv()
        return self.host, port

ftp = SmartFTP(ftp_server)

# the rest of the code is the same

Another solution may be to use IPv6. See Python 3.8.5 FTPS connection.

For a different problem with similar consequences, see vsftpd returns 0,0,0,0 in response to PASV.

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0