Python (qpython) socket programming

Question

import socket
import sys

# creating socket object
s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)

host = 'www.google.com'
port = "80"


x = "GET  /  HTTP/1.1\r\nHost :  "+ host + "\n\n"

s.connect((host, port)
s.send(request.encode()) # the line has error invalid syntax 

response  = s.recv(4096)

print (response)

Answer 1

Error in this line:

s.send(request.encode()) # the line has error invalid syntax

Reason: you have defined request with variable name x so this should work for you,

s.send(x)

Answer 2

Adding to the existing answer about the problems with your code:

x = "GET  /  HTTP/1.1\r\nHost :  "+ host + "\n\n"

You are trying to construct a HTTP request here but get it wrong:

1. there should be no space after Host, i.e. it should be Host: hostname and not Host: hostname as you do.
2. The end of line should be \r\n and not \n. You get it right after the first line but not after the Host header and not for the end-of-header marker.