How to isolate a part of HTML page in Python 3 - html

Question

I made a simple script to retrieve sourcecode of a page, but I'd like to "isolate" the part of ips so that I can save to proxy.txt file. Any suggestions?
import urllib.request
sourcecode = urllib.request.urlopen("https://www.inforge.net/xi/threads/dichvusocks-us-15h10-pm-update-24-24-good-socks.455588/")
sourcecode = str(sourcecode.read())
out_file = open("proxy.txt","w")
out_file.write(sourcecode)
out_file.close()

Answer 1

I've added a couple of lines to your code, the only problem is that the UI version (check the page source) is being added as an IP address.
import urllib.request
import re
sourcecode = urllib.request.urlopen("https://www.inforge.net/xi/threads/dichvusocks-us-15h10-pm-update-24-24-good-socks.455588/")
sourcecode = str(sourcecode.read())
out_file = open("proxy.txt","w")
out_file.write(sourcecode)
out_file.close()
with open('proxy.txt') as fp:
for line in fp:
ip = re.findall('(?:[\d]{1,3})\.(?:[\d]{1,3})\.(?:[\d]{1,3})\.(?:[\d]{1,3})', line)
for addr in ip:
print(addr)
UPDATE:
This is what you are looking for, BeatifulSoup can extract only the data we need from the page using CSS classes, however it needs to be installed with pip. You don't need to save the page to a file.
from bs4 import BeautifulSoup
import urllib.request
import re
url = urllib.request.urlopen('https://www.inforge.net/xi/threads/dichvusocks-us-15h10-pm-update-24-24-good-socks.455588/').read()
soup = BeautifulSoup(url, "html.parser")
# Searching the CSS class name
msg_content = soup.find_all("div", class_="messageContent")
ips = re.findall('(?:[\d]{1,3})\.(?:[\d]{1,3})\.(?:[\d]{1,3})\.(?:[\d]{1,3})', str(msg_content))
for addr in ips:
print(addr)

Answer 2

Why won't you use re?
I need the source code to say exactly how.