python - Regular Expression: (or not) Looking to print only data after the header -

some homework appreciated.

using socket, need parse data website (http://www.py4inf.com/code/romeo.txt).

i'm using regular expression '^\s*$' locate first blank line after header , above data.

any tips on how extract data (and not print header)?

import socket import re  mysock = socket.socket(socket.af_inet, socket.sock_stream)  try:     userurl = raw_input('enter url: ')      d = userurl.split('/')     d.remove("")      host = d[1]      mysock.connect((host, 80))     mysock.send('get %s http/1.0\n\n'%(userurl))       while true:         data = mysock.recv(3000)         if len(data) < 1: break         print (''.join([x x in re.findall(**'^\s*$'**,data,re.dotall)]))              except exception e:     print (str(e)) 

i'm assuming since it's homework problem have to use socket , can't use more user-friendly requests.

i first loop until have complete response in string, , iterate on this:

... response = "" while true:     data = mysock.recv(3000)     if len(data) < 1: break     response += data  iterator = iter(response.split("\n"))  line in iterator:     if not line.strip():  # empty line         break  body = "\n".join(iterator)  # put rest of data in string