CODEWITHSUNDEEP
pythonpython-3.xlinuxsocketsnetwork-programming
KMG
KMG

Reputation: 1501

Why my python client hangs when recieving data from internet?

I'm trying to make a simple port scanner using python but it seems that it hangs duringrecv that the default timeout raise an exception and terminate the program.

import socket, argparse

parser = argparse.ArgumentParser(description="A Port scanner",
                                 usage="command -w <website> -p <port>",
                                 prog="Port Scanner")

parser.add_argument("-w", nargs=1, type=str, required=True, dest="site",
                    help="The website to scan")
parser.add_argument("-p", nargs=1, type=int, dest="port",
                    help="The port to scan", default=80)

namespace = parser.parse_args()
def connect(host: str, port:int) -> bool:
    ip = socket.gethostbyname(host)
    socket.setdefaulttimeout(10)
    conn_sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
    try:
        conn_sock.connect((ip, port))
    except:
        print("[+] port {} is closed".format(port))
        return False
    else:
        print("[+] Port {} is open".format(port))
        conn_sock.sendall(b"HTTP GET /")
        response = conn_sock.recv(100)
        print("[+] Got 100 bytes as {}".format(response))
        return True
connect(namespace.site[0], namespace.port[0])

running as prog -w www.google.com -p 80(also tried with different websites) it says that port 80 is opened and connected as supposed also sending data doesn't seem to cause any problem. Trackback is

    response = conn_sock.recv(100)
socket.timeout: timed out

Upvotes: 0

Views: 59

Answers (1)

mhawke
mhawke

Reputation: 87074

The server is not responding because you have not sent it a complete or valid request. Try:

conn_sock.sendall(b"HTTP GET /\r\n\r\n")

A blank line is required between the (optional) headers and the (optional) body. You don't set any headers so you simply need two CRLFs at the end of the request. This will get you communicating with the server.

But you should change your request to:

GET / HTTP/1.1\r\n\r\n

if you want to get a valid HTTP/1.1 200 OK response.

Upvotes: 1

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