CODEWITHSUNDEEP
pythonsocketsipport
Codieroot
Codieroot

Reputation: 747

Scan a port of an ip address and print the response status using python

I am trying to hit a port of an IP address and print the status code (i.e 200 OK , 502 Bad Gateway etc). Suppose, let's say port 80 is open at ip address xx.xx.xx.xxx . So, xx.xx.xx.xxx:80 should give us 200 OK. In my code, I am using port 811450 , which is definitely not opened and no application is running there. But it is not showing any error.

import httplib  
import urllib
import socket


class mysocket:


    def __init__(self, sock=None):
        if sock is None:
            self.sock = socket.socket(
                socket.AF_INET, socket.SOCK_STREAM)
        else:
            self.sock = sock

    host = "xx.xx.xx.xxx"
    port = "811450"

    def connect(self, host, port):
        self.sock.connect((host, port))

    def mysend(self, msg):
        totalsent = 0
        while totalsent < MSGLEN:
            sent = self.sock.send(msg[totalsent:])
            if sent == 0:
                raise RuntimeError("socket connection broken")
            totalsent = totalsent + sent

    def myreceive(self):
        chunks = []
        bytes_recd = 0
        while bytes_recd < MSGLEN:
            chunk = self.sock.recv(min(MSGLEN - bytes_recd, 2048))
            if chunk == '':
                raise RuntimeError("socket connection broken")
            chunks.append(chunk)
            bytes_recd = bytes_recd + len(chunk)
        return ''.join(chunks)

What is the recommended way to achieve it.

Upvotes: 0

Views: 1517

Answers (1)

Erik Cederstrand
Erik Cederstrand

Reputation: 10220

First of all, 811450 isn't a valid port. The largest valid TCP port range is 65535.

Second, you're mixing up the TCP protocol and the HTTP protocol. If you want an HTTP response from the socket, you need to send a valid HTTP request. You can easily do this with the requests package:

import requests

r = requests.get('http://xx.xx.xx.xxx:811450')
print(r.status_code)

Upvotes: 1

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