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How do I make a HTTP call to a particular URL in Python and then figure out that whether the server is UP and RUNNING or not.

As I recently started working with Python so not sure of any such libraries.

If I am opening the below URL - 

http://dbx45.dc1.host.com:8082/console

then some page gets opened to me which means my server is UP and RUNNING. But if it is not UP and RUNNING, then it shows me like this on CHROME - 

Oops! Google Chrome could not connect to http://dbx45.dc1.host.com:8082/console

And on FIREFOX - 

Unable to connect
Firefox can't establish a connection to the server at dbx45.dc1.host.com:8082

So how do I figure out this thing from the Python that whether my server is UP and RUNNING or not?

UPDATE:-

This is what I have tried so far - 

status = urllib2.urlopen("http://dbx45.dc1.host.com:8082/console").read()
print status

How do I simplify this to understand whether the server is UP and RUNNING or not?

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  • Have you tried using urllib: docs.python.org/2/library/urllib2.html? Commented Jan 1, 2014 at 9:21
  • I recently started working with Python so don't know about these libraries.. Commented Jan 1, 2014 at 9:22
  • Have added an example. Commented Jan 1, 2014 at 9:33

4 Answers 4

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5

an easy library to use is requests.

import requests
page = requests.get('http://dbx45.dc1.host.com:8082/console')

then you can check the status like this

page.status_code

200 means it's ok and something in the 400's or 500's means something went wrong.

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6 Comments

In my editor PyCharm it says No module named requests Any idea why?
here are the directions to download this python module docs.python-requests.org/en/latest/user/install But it works for many python modules in general
I just ran the code. It gives me like this for a successfully server which is up and running as <Response [200]> but if I just change the port to 8083 instead of 8082, it throws an exception instead of returning any number you just told me.. Should we catch those and then indentify whether those are up or not?
What is the exception?
This is what I am getting requests.exceptions.ConnectionError: HTTPConnectionPool(host='dbx45.dc1.host.com', port=8083): Max retries exceeded with url: /console (Caused by <class 'socket.error'>: [Errno 10061] No connection could be made because the target machine actively refused it)
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2

The error the browsers show is that they cannot connect. So I think what you are after is whether or not the server is actually listening on the port. For that you don't really need the whole URL, just the host and port and a simple socket connection test.

import sys
import socket

def servertest(argv):
    host = argv[1]
    port = int(argv[2])

    args = socket.getaddrinfo(host, port, socket.AF_INET, socket.SOCK_STREAM)
    for family, socktype, proto, canonname, sockaddr in args:
        s = socket.socket(family, socktype, proto)
        try:
            s.connect(sockaddr)
        except socket.error:
            return False
        else:
            s.close()
            return True


if __name__ == "__main__":
    if servertest(sys.argv):
        print("Server is UP")
    else:
        print("Server is DOWN")
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2

You can use urllib which raises an error in case the URL cannot be opened or gives an error code.

exception urllib2.URLError http://docs.python.org/2/library/urllib2.html#urllib2.URLError

It also has a reason parameter with which you can narrow down the issue.

Example:

>>> import urllib2
>>> urllib2.urlopen("http://dbx45.dc1.host.com:8082/console")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 126, in urlopen
    return _opener.open(url, data, timeout)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 394, in open
    response = self._open(req, data)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 412, in _open
    '_open', req)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 372, in _call_chain
    result = func(*args)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 1199, in http_open
    return self.do_open(httplib.HTTPConnection, req)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 1174, in do_open
    raise URLError(err)
urllib2.URLError: <urlopen error [Errno 60] Operation timed out>

To see if it is up:

try:
    urllib2.urlopen("http://dbx45.dc1.host.com:8082/console")
except urllib2.URLError:
    print "Some error"
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2 Comments

You want to surround your code in try except clause and if there is an exception, your server is up. Or it is not.
The code is slightly different sing Python 3. Instead of urllib2.urlopen use urllib.request.urlopen (same goes to urllib2.error: urllib.error).
0 
import os

def is_server_up(ip_addr):
    return os.system('ping -c 1 ' + ip_addr + ' > /dev/null') == 0
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