Simple TCP socket in Python that accepts URL(s) as input and return its http status code

url_status.py

#!/usr/bin/env python3

import socket
import requests


def url_status(listening_ip, listening_port, *args, **kwargs):
    """
    This a simple TCP socket that accept url(s) as input and return url(s), and
    its status code.
    Args:
        listening_ip (str): Binding IP
        listening_port str: Binding port
        *args: Variable length argument list
        **kwargs: Arbitrary keyword arguments:
            status_log (str): Path of status log
            error_log (str): Path of error log
    """

    status_log = kwargs['status_log'] if 'status_log' in kwargs else 'status.log'
    error_log = kwargs['error_log'] if 'error_log' in kwargs else 'error.log'

    sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
    sock.bind((listening_ip, listening_port))
    sock.listen(5)
    while True:
        connection, address = sock.accept()
        buf = connection.recv(1024)
        connection.send(buf)
        urls = buf.decode().strip().split("\n")
        for url in urls:
            try:
                req = requests.get(url)
            except Exception as e:
                with open(error_log, 'a') as error_log_file:
                    error_log_file.write(f'{url}, {e}\n')
                connection.send(
                    f' => {url} is not reachable\n'.encode())
                continue
            else:
                with open(status_log, 'a') as status_log_file:
                    status_log_file.write(f'{url},{req.status_code}\n')
                connection.send(
                    f' => {url} status code is {req.status_code}\n'.encode())

        connection.close()


url_status('127.0.0.1', 8080)