python - Handling bad URLs with requests -

sorry in advance beginner question. i'm learning how access web data in python, , i'm having trouble understanding exception handling in requests package.

so far, when accessing web data using urllib package, wrap urlopen call in try/except structure catch bad urls, this:

import urllib, sys  url = 'https://httpbintypo.org/' # note typo in url  try: uh=urllib.urlopen(url) except:     print 'failed open url.'     sys.exit()  text = uh.read() print text 

this kind of crude way it, can mask kinds of problems other bad urls.

from documentation, had sort of gathered avoid try/except structure when using requests package, this:

import requests, sys  url = 'https://httpbintypo.org/' # note typo in url  r = requests.get(url) if r.raise_for_status() not none:     print 'failed open url.'     sys.exit()  text = r.text print text 

however, doesn't work (throws error , traceback). what's "right" (i.e., simple, elegant, pythonic) way this?

try catch connection error:

from requests.exceptions import connectionerror  try:     requests.get('https://httpbintypo.org/') except connectionerror:     print 'failed open url.'