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Infinite loading on conn.recv python socket

I'm currently trying to make a socket connection between a server and multiple clients in Python.
In short, I have a function that is used to test if the clients are still online or dead.
To do so, i'm sending a white space to the client, and if no exception occures, it's all good, and if there is an exception I'm deleting the client from the client list.
Everything works fine, except when the client is switching network (for exemple I switch from a 2GHz network, to the same network in 5GHz). Then, the conn.recv is creating an infinite loading because the client is not receiving the data that was sent in conn.send. As a precision, both of the network where I was connected works well separately. The only solution that I found for now is restarting the server when such an event occurs..
Here is the part of the code that doesn't work (server):
# Check for clients availability
def check_clients(self, UUID):
invalid_conn = []
print(f"(UUID: {UUID}) Checking for dead clients")
if UUID not in self.all_connections: return False
# Delete dead clients for specific user (UUID)
self.socket.settimeout(2) # Also tried to set a timeout, doesn't work
print('setting timeout')
for i, conn in enumerate(self.all_connections[UUID]):
try:
print('sending blank cmd')
conn.send(str.encode(' '))
cwd = str(conn.recv(20480)) # This is the line that causes infinite loop
print('got cwd:',cwd) # This is what we are excepting from the client
except:
print('appening invalid conn to list:',conn)
invalid_conn.append(conn)
continue
print('removing timeout')
self.socket.settimeout(None)
for i in invalid_conn:
client_id = self.all_connections[UUID].index(i)
print(f'(UUID: {UUID}) Could not reach client on conn list, removing {self.all_addresses[UUID][client_id]}')
del self.all_connections[UUID][client_id]
del self.all_addresses[UUID][client_id]
print(f"(UUID: {UUID}) Dead clients check finished")
return
And here is the part of the client that should receive the server information:
# Receive commands from server
def receive_commands(self):
while True:
print("Listening for commands")
output_str = None
data = self.sock.recv(1024)
print('data:',data)
if len(data) > 0:
try:
cmd = data[:].decode("cp850")
# Do such and such events with cmd, this is a shortened version for demo
output_str = str(f'received: {cmd} from server', "cp850")
except Exception as e:
output_str = f"Command failed to execute: {str(e)}"
if output_str == None:
raise ConnectionError ("No response from server")
else:
self.print_output(output_str)
print(f"Input command: {data[:].decode('cp850')}")
print(f"Output -> {output_str}\n")
self.sock.close()
return
Please ask me if you need any more info, thank you for your help.

Trigger event when ethernet adapter state changes Python

Trying to create a program that runs 24x7 in the background and undertakes actions triggered by change in the up/down i.e connected/disconnected state of the ethernet adapter port in PCs/Laptops. How to implement the same in Python?
For ex:
conn = True
def conCheck()
if conn == False:
<trigger onward events>
else:
<else statements>
Any help will be greatly appreciated. Trying to take up and learn python.
Thanks and regards

You can write a function that can request a web page for checking the internet connection.
If there is no internet, You might be getting a ConnectionError.
Otherwise, You'll get a response from the URL. i.e. You're connected with the internet.
Use the following code & try to interrupt the connection your code is executing. You'll see the difference when your internet gets interrupted.
import time
import requests
url = "http://www.kite.com"
timeout = 5
conn = True
def conCheck():
try:
request = requests.get(url, timeout=timeout)
print("Connected to the Internet")
return True
except (requests.ConnectionError, requests.Timeout) as exception:
print("You're not connected to the internet. ")
return False
while True:
conCheck()
time.sleep(1)

i am trying to develop a port scanner in python but my code is not giving any open port give me some suggestions about that?

This is my code it's just starting the scan but it is not completing ,where is the error in it. i need output as port number and port side by side.when i run in command prompt it gives like that,please give suggetions on that
from socket import *
import sys,time
from datetime import datetime
host =''
max_port=100
min_port=1
def scan_host(host,port,r_code=1):
try:
s=socket(AF_INET,SOCK_STREAM)
code=s.connect_ex((host,port))
if code==0:
r_code=code
s.close()
except Exception,e:
pass
return r_code
try:
host=raw_input("Enter Host address:")
except KeyboardInterrupt:
print("\n Application shtdown")
sys.exit(1)
hostip=gethostbyname(host)
print("\n Host:%s IP:%s" %(host,hostip))
print("Scanning Started At %s...\n" %(time.strftime("%H:%M:%S")))
start_time=datetime.now()
for port in range(min_port,max_port):
try:
response=scan_host(host,port)
if response ==0:
print("Port %d: Open" %(port))
except Exception,e:
pass
stop_time=datetime.now()
total_time_duration=stop_time -start_time
print("\n Scanning Finished At %s ..." % (time.strftime("%H:%M:%S")))
print("Scanning Duration:%s..." %(total_time_duration))
print("Have a nice day ...Sergeant Exploiter (Sploit)")

Before using the following port scanner, you may want to check a few things first:
Is the firewall on your computer blocking the port scanner?
Is the device your computer connected to blocking certain ports?
Is the computer you are trying to scan blocking ports with its firewall?
Do you know the correct name of the host that you are trying to scan?
Can you create a server on one computer and connect to it with a client on the other?
If none of the above points are cause for your problem, the program shown below may work for you:
#! /usr/bin/env python3
import argparse
import collections
import itertools
import multiprocessing
import operator
import socket
PURPOSE = 'Scan for open ports on a computer.'
PORTS = range(1 << 16)
POOL_SIZE = 1 << 8
TIMEOUT = 0.01
def main():
"""Get computer to scan, connect with process pool, and show open ports."""
parser = argparse.ArgumentParser(description=PURPOSE)
parser.add_argument('host', type=str, help='computer you want to scan')
host = parser.parse_args().host
with multiprocessing.Pool(POOL_SIZE, socket.setdefaulttimeout, [TIMEOUT]) \
as pool:
results = pool.imap_unordered(test, ((host, port) for port in PORTS))
servers = filter(operator.itemgetter(0), results)
numbers = map(operator.itemgetter(1), servers)
ordered = sorted(numbers)
print(f'Ports open on {host}:', *format_ports(ordered), sep='\n ')
field_names = 'family', 'socket_type', 'protocol', 'canon_name', 'address'
AddressInfo = collections.namedtuple('AddressInfo', field_names)
del field_names
def test(address):
"""Try connecting to the server and return whether or not it succeeded."""
host, port = address
for info in itertools.starmap(AddressInfo, socket.getaddrinfo(host, port)):
try:
probe = socket.socket(info.family, info.socket_type, info.protocol)
except OSError:
pass
else:
try:
probe.connect(info.address)
except OSError:
pass
else:
probe.shutdown(socket.SHUT_RDWR)
return True, port
finally:
probe.close()
return False, port
def format_ports(ports):
"""Convert port numbers into strings and show all associated services."""
if ports:
for port in ports:
try:
service = socket.getservbyport(port)
except OSError:
service = '?'
yield f'{port:<5} = {service}'
else:
yield 'None'
if __name__ == '__main__':
main()

Checking internet connection before polling? (Telegram bot) [duplicate]

I want to see if I can access an online API, but for that, I need to have Internet access.
How can I see if there's a connection available and active using Python?

Perhaps you could use something like this:
import urllib2
def internet_on():
try:
urllib2.urlopen('http://216.58.192.142', timeout=1)
return True
except urllib2.URLError as err:
return False
Currently, 216.58.192.142 is one of the IP addresses for google.com. Change http://216.58.192.142 to whatever site can be expected to respond quickly.
This fixed IP will not map to google.com forever. So this code is
not robust -- it will need constant maintenance to keep it working.
The reason why the code above uses a fixed IP address instead of fully qualified domain name (FQDN) is because a FQDN would require a DNS lookup. When the machine does not have a working internet connection, the DNS lookup itself may block the call to urllib_request.urlopen for more than a second. Thanks to #rzetterberg for pointing this out.
If the fixed IP address above is not working, you can find a current IP address for google.com (on unix) by running
% dig google.com +trace
...
google.com. 300 IN A 216.58.192.142

If we can connect to some Internet server, then we indeed have connectivity. However, for the fastest and most reliable approach, all solutions should comply with the following requirements, at the very least:
Avoid DNS resolution (we will need an IP that is well-known and guaranteed to be available for most of the time)
Avoid application layer connections (connecting to an HTTP/FTP/IMAP service)
Avoid calls to external utilities from Python or other language of choice (we need to come up with a language-agnostic solution that doesn't rely on third-party solutions)
To comply with these, one approach could be to, check if one of the Google's public DNS servers is reachable. The IPv4 addresses for these servers are 8.8.8.8 and 8.8.4.4. We can try connecting to any of them.
A quick Nmap of the host 8.8.8.8 gave below result:
$ sudo nmap 8.8.8.8
Starting Nmap 6.40 ( http://nmap.org ) at 2015-10-14 10:17 IST
Nmap scan report for google-public-dns-a.google.com (8.8.8.8)
Host is up (0.0048s latency).
Not shown: 999 filtered ports
PORT STATE SERVICE
53/tcp open domain
Nmap done: 1 IP address (1 host up) scanned in 23.81 seconds
As we can see, 53/tcp is open and non-filtered. If you are a non-root user, remember to use sudo or the -Pn argument for Nmap to send crafted probe packets and determine if a host is up.
Before we try with Python, let's test connectivity using an external tool, Netcat:
$ nc 8.8.8.8 53 -zv
Connection to 8.8.8.8 53 port [tcp/domain] succeeded!
Netcat confirms that we can reach 8.8.8.8 over 53/tcp. Now we can set up a socket connection to 8.8.8.8:53/tcp in Python to check connection:
import socket
def internet(host="8.8.8.8", port=53, timeout=3):
"""
Host: 8.8.8.8 (google-public-dns-a.google.com)
OpenPort: 53/tcp
Service: domain (DNS/TCP)
"""
try:
socket.setdefaulttimeout(timeout)
socket.socket(socket.AF_INET, socket.SOCK_STREAM).connect((host, port))
return True
except socket.error as ex:
print(ex)
return False
internet()
Another approach could be to send a manually crafted DNS probe to one of these servers and wait for a response. But, I assume, it might prove slower in comparison due to packet drops, DNS resolution failure, etc. Please comment if you think otherwise.
UPDATE #4: This listing of public nameservers is a good reference for IPs to test against.
UPDATE #3: Tested again after the exception handling change:
defos.py
True
00:00:00:00.410
iamaziz.py
True
00:00:00:00.240
ivelin.py
True
00:00:00:00.109
jaredb.py
True
00:00:00:00.520
kevinc.py
True
00:00:00:00.317
unutbu.py
True
00:00:00:00.436
7h3rAm.py
True
00:00:00:00.030
UPDATE #2: I did quick tests to identify the fastest and most generic implementation of all valid answers to this question. Here's the summary:
$ ls *.py | sort -n | xargs -I % sh -c 'echo %; ./timeit.sh %; echo'
defos.py
True
00:00:00:00.487
iamaziz.py
True
00:00:00:00.335
ivelin.py
True
00:00:00:00.105
jaredb.py
True
00:00:00:00.533
kevinc.py
True
00:00:00:00.295
unutbu.py
True
00:00:00:00.546
7h3rAm.py
True
00:00:00:00.032
And once more:
$ ls *.py | sort -n | xargs -I % sh -c 'echo %; ./timeit.sh %; echo'
defos.py
True
00:00:00:00.450
iamaziz.py
True
00:00:00:00.358
ivelin.py
True
00:00:00:00.099
jaredb.py
True
00:00:00:00.585
kevinc.py
True
00:00:00:00.492
unutbu.py
True
00:00:00:00.485
7h3rAm.py
True
00:00:00:00.035
True in the above output signifies that all these implementations from respective authors correctly identify connectivity to the Internet. Time is shown with milliseconds resolution.
UPDATE #1: Thanks to #theamk's comment, timeout is now an argument and initialized to 3s by default.

It will be faster to just make a HEAD request so no HTML will be fetched.
try:
import httplib # python < 3.0
except:
import http.client as httplib
def have_internet() -> bool:
conn = httplib.HTTPSConnection("8.8.8.8", timeout=5)
try:
conn.request("HEAD", "/")
return True
except Exception:
return False
finally:
conn.close()

As an alternative to ubutnu's/Kevin C answers, I use the requests package like this:
import requests
def connected_to_internet(url='http://www.google.com/', timeout=5):
try:
_ = requests.head(url, timeout=timeout)
return True
except requests.ConnectionError:
print("No internet connection available.")
return False
Bonus: this can be extended to this function that pings a website.
def web_site_online(url='http://www.google.com/', timeout=5):
try:
req = requests.head(url, timeout=timeout)
# HTTP errors are not raised by default, this statement does that
req.raise_for_status()
return True
except requests.HTTPError as e:
print("Checking internet connection failed, status code {0}.".format(
e.response.status_code))
except requests.ConnectionError:
print("No internet connection available.")
return False

Just to update what unutbu said for new code in Python 3.2
def check_connectivity(reference):
try:
urllib.request.urlopen(reference, timeout=1)
return True
except urllib.request.URLError:
return False
And, just to note, the input here (reference) is the url that you want to check: I suggest choosing something that connects fast where you live -- i.e. I live in South Korea, so I would probably set reference to http://www.naver.com.

You can just try to download data, and if connection fail you will know that somethings with connection isn't fine.
Basically you can't check if computer is connected to internet. There can be many reasons for failure, like wrong DNS configuration, firewalls, NAT. So even if you make some tests, you can't have guaranteed that you will have connection with your API until you try.

import urllib
def connected(host='http://google.com'):
try:
urllib.urlopen(host)
return True
except:
return False
# test
print( 'connected' if connected() else 'no internet!' )
For python 3, use urllib.request.urlopen(host)

Try the operation you were attempting to do anyway. If it fails python should throw you an exception to let you know.
To try some trivial operation first to detect a connection will be introducing a race condition. What if the internet connection is valid when you test but goes down before you need to do actual work?

Here's my version
import requests
try:
if requests.get('https://google.com').ok:
print("You're Online")
except:
print("You're Offline")

This might not work if the localhost has been changed from 127.0.0.1
Try
import socket
ipaddress=socket.gethostbyname(socket.gethostname())
if ipaddress=="127.0.0.1":
print("You are not connected to the internet!")
else:
print("You are connected to the internet with the IP address of "+ ipaddress )
Unless edited , your computers IP will be 127.0.0.1 when not connected to the internet.
This code basically gets the IP address and then asks if it is the localhost IP address .
Hope that helps

A modern portable solution with requests:
import requests
def internet():
"""Detect an internet connection."""
connection = None
try:
r = requests.get("https://google.com")
r.raise_for_status()
print("Internet connection detected.")
connection = True
except:
print("Internet connection not detected.")
connection = False
finally:
return connection
Or, a version that raises an exception:
import requests
from requests.exceptions import ConnectionError
def internet():
"""Detect an internet connection."""
try:
r = requests.get("https://google.com")
r.raise_for_status()
print("Internet connection detected.")
except ConnectionError as e:
print("Internet connection not detected.")
raise e

Best way to do this is to make it check against an IP address that python always gives if it can't find the website. In this case this is my code:
import socket
print("website connection checker")
while True:
website = input("please input website: ")
print("")
print(socket.gethostbyname(website))
if socket.gethostbyname(website) == "92.242.140.2":
print("Website could be experiencing an issue/Doesn't exist")
else:
socket.gethostbyname(website)
print("Website is operational!")
print("")

my favorite one, when running scripts on a cluster or not
import subprocess
def online(timeout):
try:
return subprocess.run(
['wget', '-q', '--spider', 'google.com'],
timeout=timeout
).returncode == 0
except subprocess.TimeoutExpired:
return False
this runs wget quietly, not downloading anything but checking that the given remote file exists on the web

Taking unutbu's answer as a starting point, and having been burned in the past by a "static" IP address changing, I've made a simple class that checks once using a DNS lookup (i.e., using the URL "https://www.google.com"), and then stores the IP address of the responding server for use on subsequent checks. That way, the IP address is always up to date (assuming the class is re-initialized at least once every few years or so). I also give credit to gawry for this answer, which showed me how to get the server's IP address (after any redirection, etc.). Please disregard the apparent hackiness of this solution, I'm going for a minimal working example here. :)
Here is what I have:
import socket
try:
from urllib2 import urlopen, URLError
from urlparse import urlparse
except ImportError: # Python 3
from urllib.parse import urlparse
from urllib.request import urlopen, URLError
class InternetChecker(object):
conn_url = 'https://www.google.com/'
def __init__(self):
pass
def test_internet(self):
try:
data = urlopen(self.conn_url, timeout=5)
except URLError:
return False
try:
host = data.fp._sock.fp._sock.getpeername()
except AttributeError: # Python 3
host = data.fp.raw._sock.getpeername()
# Ensure conn_url is an IPv4 address otherwise future queries will fail
self.conn_url = 'http://' + (host[0] if len(host) == 2 else
socket.gethostbyname(urlparse(data.geturl()).hostname))
return True
# Usage example
checker = InternetChecker()
checker.test_internet()

Taking Six' answer I think we could simplify somehow, an important issue as newcomers are lost in highly technical matters.
Here what I finally will use to wait for my connection (3G, slow) to be established once a day for my PV monitoring.
Works under Pyth3 with Raspbian 3.4.2
from urllib.request import urlopen
from time import sleep
urltotest=http://www.lsdx.eu # my own web page
nboftrials=0
answer='NO'
while answer=='NO' and nboftrials<10:
try:
urlopen(urltotest)
answer='YES'
except:
essai='NO'
nboftrials+=1
sleep(30)
maximum running: 5 minutes if reached I will try in one hour's time but its another bit of script!

Taking Ivelin's answer and add some extra check as my router delivers its ip address 192.168.0.1 and returns a head if it has no internet connection when querying google.com.
import socket
def haveInternet():
try:
# first check if we get the correct IP-Address or just the router's IP-Address
info = socket.getaddrinfo("www.google.com", None)[0]
ipAddr = info[4][0]
if ipAddr == "192.168.0.1" :
return False
except:
return False
conn = httplib.HTTPConnection("www.google.com", timeout=5)
try:
conn.request("HEAD", "/")
conn.close()
return True
except:
conn.close()
return False

This works for me in Python3.6
import urllib
from urllib.request import urlopen
def is_internet():
"""
Query internet using python
:return:
"""
try:
urlopen('https://www.google.com', timeout=1)
return True
except urllib.error.URLError as Error:
print(Error)
return False
if is_internet():
print("Internet is active")
else:
print("Internet disconnected")

I added a few to Joel's code.
import socket,time
mem1 = 0
while True:
try:
host = socket.gethostbyname("www.google.com") #Change to personal choice of site
s = socket.create_connection((host, 80), 2)
s.close()
mem2 = 1
if (mem2 == mem1):
pass #Add commands to be executed on every check
else:
mem1 = mem2
print ("Internet is working") #Will be executed on state change
except Exception as e:
mem2 = 0
if (mem2 == mem1):
pass
else:
mem1 = mem2
print ("Internet is down")
time.sleep(10) #timeInterval for checking

For my projects I use script modified to ping the google public DNS server 8.8.8.8. Using a timeout of 1 second and core python libraries with no external dependencies:
import struct
import socket
import select
def send_one_ping(to='8.8.8.8'):
ping_socket = socket.socket(socket.AF_INET, socket.SOCK_RAW, socket.getprotobyname('icmp'))
checksum = 49410
header = struct.pack('!BBHHH', 8, 0, checksum, 0x123, 1)
data = b'BCDEFGHIJKLMNOPQRSTUVWXYZ[\\]^_`abcdefghijklmnopqrstuvwx'
header = struct.pack(
'!BBHHH', 8, 0, checksum, 0x123, 1
)
packet = header + data
ping_socket.sendto(packet, (to, 1))
inputready, _, _ = select.select([ping_socket], [], [], 1.0)
if inputready == []:
raise Exception('No internet') ## or return False
_, address = ping_socket.recvfrom(2048)
print(address) ## or return True
send_one_ping()
The select timeout value is 1, but can be a floating point number of choice to fail more readily than the 1 second in this example.

Make sure your pip is up to date by running
pip install --upgrade pip
Install the requests package using
pip install requests
import requests
import webbrowser
url = "http://www.youtube.com"
timeout = 6
try:
request = requests.get(url, timeout=timeout)
print("Connected to the Internet")
print("browser is loading url")
webbrowser.open(url)
except (requests.ConnectionError, requests.Timeout) as exception:
print("poor or no internet connection.")

import requests and try this simple python code.
def check_internet():
url = 'http://www.google.com/'
timeout = 5
try:
_ = requests.get(url, timeout=timeout)
return True
except requests.ConnectionError:
return False

I just want to refer to Ivelin's solution, because I can't comment there.
In python 2.7 with an old SSL certificate (in my case, not possible to update, which is another story), there is a possibility of a Certificate Error. In that case, replacing '8.8.8.8' with 'dns.google' or '8888.google' can help.
Hope this will someone helps too.
try:
import httplib # python < 3.0
except:
import http.client as httplib
def have_internet():
conn = httplib.HTTPSConnection("8888.google", timeout=5)
try:
conn.request("HEAD", "/")
return True
except Exception:
return False
finally:
conn.close()

How can I see if there's an available and active network connection in Python?

I want to see if I can access an online API, but for that, I need to have Internet access.
How can I see if there's a connection available and active using Python?

Perhaps you could use something like this:
import urllib2
def internet_on():
try:
urllib2.urlopen('http://216.58.192.142', timeout=1)
return True
except urllib2.URLError as err:
return False
Currently, 216.58.192.142 is one of the IP addresses for google.com. Change http://216.58.192.142 to whatever site can be expected to respond quickly.
This fixed IP will not map to google.com forever. So this code is
not robust -- it will need constant maintenance to keep it working.
The reason why the code above uses a fixed IP address instead of fully qualified domain name (FQDN) is because a FQDN would require a DNS lookup. When the machine does not have a working internet connection, the DNS lookup itself may block the call to urllib_request.urlopen for more than a second. Thanks to #rzetterberg for pointing this out.
If the fixed IP address above is not working, you can find a current IP address for google.com (on unix) by running
% dig google.com +trace
...
google.com. 300 IN A 216.58.192.142

If we can connect to some Internet server, then we indeed have connectivity. However, for the fastest and most reliable approach, all solutions should comply with the following requirements, at the very least:
Avoid DNS resolution (we will need an IP that is well-known and guaranteed to be available for most of the time)
Avoid application layer connections (connecting to an HTTP/FTP/IMAP service)
Avoid calls to external utilities from Python or other language of choice (we need to come up with a language-agnostic solution that doesn't rely on third-party solutions)
To comply with these, one approach could be to, check if one of the Google's public DNS servers is reachable. The IPv4 addresses for these servers are 8.8.8.8 and 8.8.4.4. We can try connecting to any of them.
A quick Nmap of the host 8.8.8.8 gave below result:
$ sudo nmap 8.8.8.8
Starting Nmap 6.40 ( http://nmap.org ) at 2015-10-14 10:17 IST
Nmap scan report for google-public-dns-a.google.com (8.8.8.8)
Host is up (0.0048s latency).
Not shown: 999 filtered ports
PORT STATE SERVICE
53/tcp open domain
Nmap done: 1 IP address (1 host up) scanned in 23.81 seconds
As we can see, 53/tcp is open and non-filtered. If you are a non-root user, remember to use sudo or the -Pn argument for Nmap to send crafted probe packets and determine if a host is up.
Before we try with Python, let's test connectivity using an external tool, Netcat:
$ nc 8.8.8.8 53 -zv
Connection to 8.8.8.8 53 port [tcp/domain] succeeded!
Netcat confirms that we can reach 8.8.8.8 over 53/tcp. Now we can set up a socket connection to 8.8.8.8:53/tcp in Python to check connection:
import socket
def internet(host="8.8.8.8", port=53, timeout=3):
"""
Host: 8.8.8.8 (google-public-dns-a.google.com)
OpenPort: 53/tcp
Service: domain (DNS/TCP)
"""
try:
socket.setdefaulttimeout(timeout)
socket.socket(socket.AF_INET, socket.SOCK_STREAM).connect((host, port))
return True
except socket.error as ex:
print(ex)
return False
internet()
Another approach could be to send a manually crafted DNS probe to one of these servers and wait for a response. But, I assume, it might prove slower in comparison due to packet drops, DNS resolution failure, etc. Please comment if you think otherwise.
UPDATE #4: This listing of public nameservers is a good reference for IPs to test against.
UPDATE #3: Tested again after the exception handling change:
defos.py
True
00:00:00:00.410
iamaziz.py
True
00:00:00:00.240
ivelin.py
True
00:00:00:00.109
jaredb.py
True
00:00:00:00.520
kevinc.py
True
00:00:00:00.317
unutbu.py
True
00:00:00:00.436
7h3rAm.py
True
00:00:00:00.030
UPDATE #2: I did quick tests to identify the fastest and most generic implementation of all valid answers to this question. Here's the summary:
$ ls *.py | sort -n | xargs -I % sh -c 'echo %; ./timeit.sh %; echo'
defos.py
True
00:00:00:00.487
iamaziz.py
True
00:00:00:00.335
ivelin.py
True
00:00:00:00.105
jaredb.py
True
00:00:00:00.533
kevinc.py
True
00:00:00:00.295
unutbu.py
True
00:00:00:00.546
7h3rAm.py
True
00:00:00:00.032
And once more:
$ ls *.py | sort -n | xargs -I % sh -c 'echo %; ./timeit.sh %; echo'
defos.py
True
00:00:00:00.450
iamaziz.py
True
00:00:00:00.358
ivelin.py
True
00:00:00:00.099
jaredb.py
True
00:00:00:00.585
kevinc.py
True
00:00:00:00.492
unutbu.py
True
00:00:00:00.485
7h3rAm.py
True
00:00:00:00.035
True in the above output signifies that all these implementations from respective authors correctly identify connectivity to the Internet. Time is shown with milliseconds resolution.
UPDATE #1: Thanks to #theamk's comment, timeout is now an argument and initialized to 3s by default.

It will be faster to just make a HEAD request so no HTML will be fetched.
try:
import httplib # python < 3.0
except:
import http.client as httplib
def have_internet() -> bool:
conn = httplib.HTTPSConnection("8.8.8.8", timeout=5)
try:
conn.request("HEAD", "/")
return True
except Exception:
return False
finally:
conn.close()

As an alternative to ubutnu's/Kevin C answers, I use the requests package like this:
import requests
def connected_to_internet(url='http://www.google.com/', timeout=5):
try:
_ = requests.head(url, timeout=timeout)
return True
except requests.ConnectionError:
print("No internet connection available.")
return False
Bonus: this can be extended to this function that pings a website.
def web_site_online(url='http://www.google.com/', timeout=5):
try:
req = requests.head(url, timeout=timeout)
# HTTP errors are not raised by default, this statement does that
req.raise_for_status()
return True
except requests.HTTPError as e:
print("Checking internet connection failed, status code {0}.".format(
e.response.status_code))
except requests.ConnectionError:
print("No internet connection available.")
return False

Just to update what unutbu said for new code in Python 3.2
def check_connectivity(reference):
try:
urllib.request.urlopen(reference, timeout=1)
return True
except urllib.request.URLError:
return False
And, just to note, the input here (reference) is the url that you want to check: I suggest choosing something that connects fast where you live -- i.e. I live in South Korea, so I would probably set reference to http://www.naver.com.

You can just try to download data, and if connection fail you will know that somethings with connection isn't fine.
Basically you can't check if computer is connected to internet. There can be many reasons for failure, like wrong DNS configuration, firewalls, NAT. So even if you make some tests, you can't have guaranteed that you will have connection with your API until you try.

import urllib
def connected(host='http://google.com'):
try:
urllib.urlopen(host)
return True
except:
return False
# test
print( 'connected' if connected() else 'no internet!' )
For python 3, use urllib.request.urlopen(host)

Try the operation you were attempting to do anyway. If it fails python should throw you an exception to let you know.
To try some trivial operation first to detect a connection will be introducing a race condition. What if the internet connection is valid when you test but goes down before you need to do actual work?

Here's my version
import requests
try:
if requests.get('https://google.com').ok:
print("You're Online")
except:
print("You're Offline")

This might not work if the localhost has been changed from 127.0.0.1
Try
import socket
ipaddress=socket.gethostbyname(socket.gethostname())
if ipaddress=="127.0.0.1":
print("You are not connected to the internet!")
else:
print("You are connected to the internet with the IP address of "+ ipaddress )
Unless edited , your computers IP will be 127.0.0.1 when not connected to the internet.
This code basically gets the IP address and then asks if it is the localhost IP address .
Hope that helps

A modern portable solution with requests:
import requests
def internet():
"""Detect an internet connection."""
connection = None
try:
r = requests.get("https://google.com")
r.raise_for_status()
print("Internet connection detected.")
connection = True
except:
print("Internet connection not detected.")
connection = False
finally:
return connection
Or, a version that raises an exception:
import requests
from requests.exceptions import ConnectionError
def internet():
"""Detect an internet connection."""
try:
r = requests.get("https://google.com")
r.raise_for_status()
print("Internet connection detected.")
except ConnectionError as e:
print("Internet connection not detected.")
raise e

Best way to do this is to make it check against an IP address that python always gives if it can't find the website. In this case this is my code:
import socket
print("website connection checker")
while True:
website = input("please input website: ")
print("")
print(socket.gethostbyname(website))
if socket.gethostbyname(website) == "92.242.140.2":
print("Website could be experiencing an issue/Doesn't exist")
else:
socket.gethostbyname(website)
print("Website is operational!")
print("")

my favorite one, when running scripts on a cluster or not
import subprocess
def online(timeout):
try:
return subprocess.run(
['wget', '-q', '--spider', 'google.com'],
timeout=timeout
).returncode == 0
except subprocess.TimeoutExpired:
return False
this runs wget quietly, not downloading anything but checking that the given remote file exists on the web

Taking unutbu's answer as a starting point, and having been burned in the past by a "static" IP address changing, I've made a simple class that checks once using a DNS lookup (i.e., using the URL "https://www.google.com"), and then stores the IP address of the responding server for use on subsequent checks. That way, the IP address is always up to date (assuming the class is re-initialized at least once every few years or so). I also give credit to gawry for this answer, which showed me how to get the server's IP address (after any redirection, etc.). Please disregard the apparent hackiness of this solution, I'm going for a minimal working example here. :)
Here is what I have:
import socket
try:
from urllib2 import urlopen, URLError
from urlparse import urlparse
except ImportError: # Python 3
from urllib.parse import urlparse
from urllib.request import urlopen, URLError
class InternetChecker(object):
conn_url = 'https://www.google.com/'
def __init__(self):
pass
def test_internet(self):
try:
data = urlopen(self.conn_url, timeout=5)
except URLError:
return False
try:
host = data.fp._sock.fp._sock.getpeername()
except AttributeError: # Python 3
host = data.fp.raw._sock.getpeername()
# Ensure conn_url is an IPv4 address otherwise future queries will fail
self.conn_url = 'http://' + (host[0] if len(host) == 2 else
socket.gethostbyname(urlparse(data.geturl()).hostname))
return True
# Usage example
checker = InternetChecker()
checker.test_internet()

Taking Six' answer I think we could simplify somehow, an important issue as newcomers are lost in highly technical matters.
Here what I finally will use to wait for my connection (3G, slow) to be established once a day for my PV monitoring.
Works under Pyth3 with Raspbian 3.4.2
from urllib.request import urlopen
from time import sleep
urltotest=http://www.lsdx.eu # my own web page
nboftrials=0
answer='NO'
while answer=='NO' and nboftrials<10:
try:
urlopen(urltotest)
answer='YES'
except:
essai='NO'
nboftrials+=1
sleep(30)
maximum running: 5 minutes if reached I will try in one hour's time but its another bit of script!

Taking Ivelin's answer and add some extra check as my router delivers its ip address 192.168.0.1 and returns a head if it has no internet connection when querying google.com.
import socket
def haveInternet():
try:
# first check if we get the correct IP-Address or just the router's IP-Address
info = socket.getaddrinfo("www.google.com", None)[0]
ipAddr = info[4][0]
if ipAddr == "192.168.0.1" :
return False
except:
return False
conn = httplib.HTTPConnection("www.google.com", timeout=5)
try:
conn.request("HEAD", "/")
conn.close()
return True
except:
conn.close()
return False

This works for me in Python3.6
import urllib
from urllib.request import urlopen
def is_internet():
"""
Query internet using python
:return:
"""
try:
urlopen('https://www.google.com', timeout=1)
return True
except urllib.error.URLError as Error:
print(Error)
return False
if is_internet():
print("Internet is active")
else:
print("Internet disconnected")

I added a few to Joel's code.
import socket,time
mem1 = 0
while True:
try:
host = socket.gethostbyname("www.google.com") #Change to personal choice of site
s = socket.create_connection((host, 80), 2)
s.close()
mem2 = 1
if (mem2 == mem1):
pass #Add commands to be executed on every check
else:
mem1 = mem2
print ("Internet is working") #Will be executed on state change
except Exception as e:
mem2 = 0
if (mem2 == mem1):
pass
else:
mem1 = mem2
print ("Internet is down")
time.sleep(10) #timeInterval for checking

For my projects I use script modified to ping the google public DNS server 8.8.8.8. Using a timeout of 1 second and core python libraries with no external dependencies:
import struct
import socket
import select
def send_one_ping(to='8.8.8.8'):
ping_socket = socket.socket(socket.AF_INET, socket.SOCK_RAW, socket.getprotobyname('icmp'))
checksum = 49410
header = struct.pack('!BBHHH', 8, 0, checksum, 0x123, 1)
data = b'BCDEFGHIJKLMNOPQRSTUVWXYZ[\\]^_`abcdefghijklmnopqrstuvwx'
header = struct.pack(
'!BBHHH', 8, 0, checksum, 0x123, 1
)
packet = header + data
ping_socket.sendto(packet, (to, 1))
inputready, _, _ = select.select([ping_socket], [], [], 1.0)
if inputready == []:
raise Exception('No internet') ## or return False
_, address = ping_socket.recvfrom(2048)
print(address) ## or return True
send_one_ping()
The select timeout value is 1, but can be a floating point number of choice to fail more readily than the 1 second in this example.

Make sure your pip is up to date by running
pip install --upgrade pip
Install the requests package using
pip install requests
import requests
import webbrowser
url = "http://www.youtube.com"
timeout = 6
try:
request = requests.get(url, timeout=timeout)
print("Connected to the Internet")
print("browser is loading url")
webbrowser.open(url)
except (requests.ConnectionError, requests.Timeout) as exception:
print("poor or no internet connection.")

import requests and try this simple python code.
def check_internet():
url = 'http://www.google.com/'
timeout = 5
try:
_ = requests.get(url, timeout=timeout)
return True
except requests.ConnectionError:
return False

I just want to refer to Ivelin's solution, because I can't comment there.
In python 2.7 with an old SSL certificate (in my case, not possible to update, which is another story), there is a possibility of a Certificate Error. In that case, replacing '8.8.8.8' with 'dns.google' or '8888.google' can help.
Hope this will someone helps too.
try:
import httplib # python < 3.0
except:
import http.client as httplib
def have_internet():
conn = httplib.HTTPSConnection("8888.google", timeout=5)
try:
conn.request("HEAD", "/")
return True
except Exception:
return False
finally:
conn.close()