How to get the physical interface IP address from an interface

Question

What I have done so far, using PyQt classes:

all_Addresses = QNetworkInterface.allAddresses()    #list-of-QHostAddress  for addr in all_Addresses:     print(addr.toString()) 

Output:

172.16.0.186 - Virtual Interface IP address 192.168.10.2 - Physical interface IP address. I want this one. 127.0.0.1 

Using socket:

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

Output:

172.16.0.186 - When openVPN is on 192.168.10.2 - When its off 

1. Is there a way to distinguish between them?
2. Can this be done with ordinary Python, instead of using PyQt classes?
3. How can I get the IPv6 address as well?

Answer 1

You should use netifaces. It is designed to be cross-platform and contains specialised code for Windows together with a variety of generic versions that work on different UNIX/UNIX-like platforms.

As of netifaces version 0.10.0, Python3 is supported.

Usage Summary

>>> from netifaces import AF_INET, AF_INET6, AF_LINK, AF_PACKET, AF_BRIDGE >>> import netifaces as ni >>> ni.interfaces() ['lo', 'eth0', 'eth1', 'vboxnet0', 'dummy1'] >>> >>> ni.ifaddresses('eth0')[AF_LINK]   # NOTE: AF_LINK is an alias for AF_PACKET [{'broadcast': 'ff:ff:ff:ff:ff:ff', 'addr': '00:02:55:7b:b2:f6'}] >>> ni.ifaddresses('eth0')[AF_INET] [{'broadcast': '172.16.161.7', 'netmask': '255.255.255.248', 'addr': '172.16.161.6'}] >>> >>> # eth0 ipv4 interface address >>> ni.ifaddresses('eth0')[AF_INET][0]['addr'] '172.16.161.6' >>>> 

Details

Windows Support:

No compiler required for most MS Windows installs. If you get warnings about installing MS Visual C++ for Windows, be very careful because you need to match the version of compiler used for your python with that used for the module.

Detailed example of netifaces data structures:

>>> import netifaces as ni >>> ni.interfaces() ['lo', 'eth0', 'eth1', 'vboxnet0', 'dummy1'] >>> ni.ifaddresses('eth0') {     17: [         {             'broadcast': 'ff:ff:ff:ff:ff:ff',             'addr': '00:02:55:7b:b2:f6'         }     ],     2: [         {             'broadcast': '172.16.161.7',             'netmask': '255.255.255.248',             'addr': '172.16.161.6'         }     ],     10: [         {             'netmask': 'ffff:ffff:ffff:ffff::',             'addr': 'fe80::202:55ff:fe7b:b2f6%eth0'         }     ] } >>>  >>> print(ni.ifaddresses.__doc__) Obtain information about the specified network interface.  Returns a dict whose keys are equal to the address family constants, e.g. netifaces.AF_INET, and whose values are a list of addresses in that family that are attached to the network interface. >>> >>> # for the IPv4 address of eth0 >>> ni.ifaddresses('eth0')[2][0]['addr'] '172.16.161.6' 

The numbers used to index protocols are from /usr/include/linux/socket.h (in Linux)... EDIT: my 3.2 kernel has them here: /usr/src/linux-headers-3.2.0-4-common/include/linux/socket.h

#define AF_INET         2       /* Internet IP Protocol         */ #define AF_INET6        10      /* IP version 6                 */ #define AF_PACKET       17      /* Packet family                */ 

The good news is that you don't have to remember all those header constants, they are included with netifaces:

>>> from netifaces import AF_INET, AF_INET6, AF_LINK, AF_PACKET, AF_BRIDGE >>> import netifaces as ni 

Answer 2

Uses the Linux SIOCGIFADDR ioctl to find the IP address associated with a network interface, given the name of that interface, e.g. "eth0". The address is returned as a string containing a dotted quad.

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('lo') '127.0.0.1'  >>> get_ip_address('eth0') '38.113.228.130' 

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