Python persistent socket connection

Question

Hello,

I'm new to python :) I would like to create persistent socket. I tried to do this using file descriptors. What I tried is:

1. Open a socket socket connection s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)

2. Get it's file descriptor number fd = s.fileno()

3. Open the file descriptor as I/O os.open(fd)

But I get OSError: [Errno 9] Bad file descriptor

I said I'm new to python and maybe I'm wrong with the implementation. But I tried a simpler example os.close(s.fileno()) and I get the same error OSError: [Errno 9] Bad file descriptor

I found an example written in ruby and I tested it, it works. http://stackoverflow.com/questions/38592/how-do-i-make-persistent-network-sockets-on-unix-in-ruby

Can any one write this into python for me, what I want to achieve is: socket_start.py google.com (thie will print the fd number) sleep 10 socket_write.py fd_number 'something' sleep 10 socket_read.py fd_number 1024

I hope you understand what I want to do. Thanks in advice!

---

After your responses I tried next code:

1

#!/usr/bin/python

import socket

s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.connect(('google.com', 80))
s.send('GET /\n')
print s.fileno()

2

#!/usr/bin/python

import os
import sys

fd = int(sys.argv[1])
os.read(fd, 1024)

And the error is OSError: [Errno 9] Bad file descriptor

I'm sure what the problem is (I'm a php developer). I think same as php, python deletes garbage after closing a script. How to solve this in python ?

Answer 1

You use os.fdopen() to open file descriptors.

I'm actually surprised that you got that far because os.open requires a filename and flags stating which mode to open the file in. for example fd = os.open('foo.txt', os.O_RONLY). As my example indicates, it returns a file descriptor rather than accepting one.

Answer 2

If you want to re-create the socket by a parameter received on the commandline (i.e., the socket), use socket.fromfd