How can I get the IP address of eth0 in Python?

Question

When an error occurs in a Python script on Unix , an email is sent.

I have been asked to add {Testing Environment} to the subject line of the email if the IP address is 192.168.100.37 which is the testing server. This way we can have one version of a script and a way to tell if the email is coming from messed up data on the testing server.

However, when I google I keep finding this code:

import socket
socket.gethostbyname(socket.gethostname())

However, that's giving me the IP address of 127.0.1.1. When I use ifconfig I get this

eth0      Link encap:Ethernet  HWaddr 00:1c:c4:2c:c8:3e
          inet addr:192.168.100.37  Bcast:192.168.100.255  Mask:255.255.255.0
          UP BROADCAST RUNNING MULTICAST  MTU:1500  Metric:1
          RX packets:75760697 errors:0 dropped:411180 overruns:0 frame:0
          TX packets:23166399 errors:0 dropped:0 overruns:0 carrier:0
          collisions:0 txqueuelen:1000
          RX bytes:59525958247 (59.5 GB)  TX bytes:10142130096 (10.1 GB)
          Interrupt:19 Memory:f0500000-f0520000

lo        Link encap:Local Loopback
          inet addr:127.0.0.1  Mask:255.0.0.0
          UP LOOPBACK RUNNING  MTU:16436  Metric:1
          RX packets:25573544 errors:0 dropped:0 overruns:0 frame:0
          TX packets:25573544 errors:0 dropped:0 overruns:0 carrier:0
          collisions:0 txqueuelen:0
          RX bytes:44531490070 (44.5 GB)  TX bytes:44531490070 (44.5 GB)

Firstly, I don't know where it got 127.0.1.1 from, but either way that's not what I want. When I google I keep coming to the same syntax, Bash scripts or netifaces and I'm trying to use standard libraries.

So how can I get the IP address of eth0 in Python?

Answer 1

Since most of the answers use ifconfig to extract the IPv4 from the eth0 interface, which is deprecated in favor of ip addr, the following code could be used instead:

import os

ipv4 = os.popen('ip addr show eth0 | grep "\<inet\>" | awk \'{ print $2 }\' | awk -F "/" \'{ print $1 }\'').read().strip()
ipv6 = os.popen('ip addr show eth0 | grep "\<inet6\>" | awk \'{ print $2 }\' | awk -F "/" \'{ print $1 }\'').read().strip()

UPDATE:

Alternatively, you can shift part of the parsing task to the python interpreter by using split() instead of grep and awk, as @serg points out in the comment:

import os

ipv4 = os.popen('ip addr show eth0').read().split("inet ")[1].split("/")[0]
ipv6 = os.popen('ip addr show eth0').read().split("inet6 ")[1].split("/")[0]

But in this case you have to check the bounds of the array returned by each split() call.

UPDATE 2:

Another version using regex:

import os
import re

ipv4 = re.search(re.compile(r'(?<=inet )(.*)(?=\/)', re.M), os.popen('ip addr show eth0').read()).groups()[0]
ipv6 = re.search(re.compile(r'(?<=inet6 )(.*)(?=\/)', re.M), os.popen('ip addr show eth0').read()).groups()[0]

Answer 2

try below code, it works for me in Mac10.10.2:

import subprocess

if __name__ == "__main__":
    result = subprocess.check_output('ifconfig en0 |grep -w inet', shell=True) # you may need to use eth0 instead of en0 here!!!
    print 'output = %s' % result.strip()
    # result = None
    ip = ''
    if result:
        strs = result.split('\n')
        for line in strs:
            # remove \t, space...
            line = line.strip()
            if line.startswith('inet '):
                a = line.find(' ')
                ipStart = a+1
                ipEnd = line.find(' ', ipStart)
                if a != -1 and ipEnd != -1:
                    ip = line[ipStart:ipEnd]
                    break
    print 'ip = %s' % ip

Answer 3

It worked for me

 import subprocess
 my_ip = subprocess.Popen(['ifconfig eth0 | awk "/inet /" | cut -d":" -f 2 | cut -d" " -f1'], stdout=subprocess.PIPE, shell=True)
 (IP,errors) = my_ip.communicate()
 my_ip.stdout.close()
 print IP

Answer 4

Two methods:

Method #1 (use external package)

You need to ask for the IP address that is bound to your eth0 interface. This is available from the netifaces package

import netifaces as ni
ni.ifaddresses('eth0')
ip = ni.ifaddresses('eth0')[ni.AF_INET][0]['addr']
print ip  # should print "192.168.100.37"

You can also get a list of all available interfaces via

ni.interfaces()

Method #2 (no external package)

Here's a way to get the IP address without using a python package:

import socket
import fcntl
import struct

def get_ip_address(ifname):
    s = socket.socket(socket.AF_INET, socket.SOCK_DGRAM)
    return socket.inet_ntoa(fcntl.ioctl(
        s.fileno(),
        0x8915,  # SIOCGIFADDR
        struct.pack('256s', ifname[:15])
    )[20:24])

get_ip_address('eth0')  # '192.168.0.110'

Note: detecting the IP address to determine what environment you are using is quite a hack. Almost all frameworks provide a very simple way to set/modify an environment variable to indicate the current environment. Try and take a look at your documentation for this. It should be as simple as doing

if app.config['ENV'] == 'production':
  #send production email
else:
  #send development email

Answer 5

Building on the answer from @jeremyjjbrown, another version that cleans up after itself as mentioned in the comments to his answer. This version also allows providing a different server address for use on private internal networks, etc..

import socket

def get_my_ip_address(remote_server="google.com"):
    """
    Return the/a network-facing IP number for this system.
    """
    with socket.socket(socket.AF_INET, socket.SOCK_DGRAM) as s: 
        s.connect((remote_server, 80))
        return s.getsockname()[0]

Answer 6

Alternatively, if you want to get the IP address of whichever interface is used to connect to the network without having to know its name, you can use this:

import socket
def get_ip_address():
    s = socket.socket(socket.AF_INET, socket.SOCK_DGRAM)
    s.connect(("8.8.8.8", 80))
    return s.getsockname()[0]

I know it's a little different than your question, but others may arrive here and find this one more useful. You do not have to have a route to 8.8.8.8 to use this. All it is doing is opening a socket, but not sending any data.