How to get the physical interface IP address from an interface
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
- Is there a way to distinguish between them?
- Can this be done with ordinary Python, instead of using PyQt classes?
- How can I get the IPv6 address as well?
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
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'
For more
I use this solution. It's a little bit tricky actually, and it's only works on linux family.
import commands
intf = 'eth0'
intf_ip = commands.getoutput("ip address show dev " + intf).split()
intf_ip = intf_ip[intf_ip.index('inet') + 1].split('/')[0]
print intf_ip
Those code utilize ip
command on linux family operating system. It split the output from ip
command and only take the IPv4 address of those interface. You can change the value intf
to eth1
or p2p1
.