Test if a network is online by using urllib2

Question

My concerns:

1. Is the code pythonic
2. Is the code practical

I tend on expanding the usefulness of this script, if you don't think it's stupid idea. One thing I will implement is command-line arguments by way of argparse.

#!/usr/bin/env python
'''
Test if network is online by sending and http request to a remote
web-server. I wrote this because my network connection sucks.
'''
import urllib2
import sys
def is_online(**kwargs):
 '''
 Determine weather or not your network is online
 by sending a HTTP GET request. If the network is
 online the function will return true. If the network
 is not online the function will return false
 '''
 # Set the default request time out.
 # This will be passed as a keyword argument
 # to urllib2.urlopen.
 kwargs.setdefault('timeout', 0.30)
 kwargs.setdefault('url', 'http://www.google.com')
 timeout = kwargs.get('timeout')
 url = kwargs.get('url')
 try:
 print 'Sending request to {}'.format(url)
 response = urllib2.urlopen(url, timeout=timeout)
 except KeyboardInterrupt:
 sys.exit(0)
 except Exception:
 # Generally using a catch-all is a bad practice but
 # I think it's ok in this case
 response = False
 print 'Request to {} failed'.format(url)
 if response:
 return True
 else:
 return False
def main():
 '''
 Continually check if the network is online.
 If the network is online stop checking.
 '''
 while True:
 if is_online():
 print 'Net is up'
 break
 else:
 print 'Net down.'
if __name__ == '__main__':
 main()

Answer 1

I would use Python's socket library instead of urllib. Then, you could test like this:

import socket
REMOTE_SERVER = "www.google.com"
def is_connected():
 try:
 host = socket.gethostbyname(REMOTE_SERVER)
 s = socket.create_connection((host, 80), 2)
 return True
 except:
 pass
 return False
print is_connected()

_{Borrowed from here}

---

I think the most obvious thing to point out is this:

if response:
 return True
else:
 return False

You can actually just return the variable, so just return response would replace all those lines.

---

There's not really any reason to use .format for one variable.

'Sending request to {}'.format(url)

Plain string concatenation is fine, for one variable.

---

Anyway, what's the point of printing out that you're requesting google.com?

'Request to {} failed'.format(url)

The level of abstraction that you put into this, is that it'll return whether you're connected or not.

Printing out anything other than: 'Net is up' or 'Net down' (Which could better be: 'Net is down, retrying connection') seems superfluous.

---

kwargs.setdefault('timeout', 0.30)

I would avoid the magic numbers (undefined numbers: 0.30), define it as a constant (in all caps) to improve readability.

---

Is this code Pythonic?

I'd recommend reading PEP8, Python's official style guide. It has many guidelines to follow on just about everything; it should really be a Python programmer's Bible. There's also a nice little Online PEP8 checker to help/show you where your code doesn't meet the standards of PEP8.

The term 'Pythonic' is not so easily defined, so I'm not really going to give you a 'Yes or No' answer, but, I'd just really recommend reading over PEP8.

Is this code practical?

Yes, it's practical.

However, one thing more to point out; While you keep retrying and retrying:

• Put a some kind of wait in between network connection tests.
• Every time you call is_online, when the Net is down, you redo the url setup, consider keeping that seperate so you can only do the bare minimum for the continual tests:

kwargs.setdefault('timeout', 0.30)
kwargs.setdefault('url', 'http://www.google.com')
timeout = kwargs.get('timeout')
url = kwargs.get('url')

• janos: Nicely done, I especially like the part about the mismatched level of abstraction. Though, I'd say the OP's dependency on urllib2 is not practical, when the job can be done without it
• Sanjiv: Is there any other way to check internet running status. Because if you pining google.com again again, then they will block our IP. So Is there any other way.?

Answer 2

This try/except block serves no purpose.

try:
 print 'Sending request to {}'.format(url)
 response = urllib2.urlopen(url, timeout=timeout)
except KeyboardInterrupt:
 sys.exit(0)

When the user presses CTRL+c the program will exit. There is no need to catch the error and then manually exit with sys.exit.

• Daniel: Won't it suppress the traceback usually associated with a KeyboardInterrupt exception?