如何找到本地IP地址(即192.168.x。x或10.0.x.x)在Python平台独立,只使用标准库?


当前回答

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

这并不总是有效(在/etc/hosts主机名为127.0.0.1的机器上返回127.0.0.1),gimel显示的是一个缓和的方法,使用socket.getfqdn()代替。当然,您的机器需要一个可解析的主机名。

其他回答

另一个变体的前面的答案,可以保存到一个名为my-ip-to的可执行脚本:

#!/usr/bin/env python

import sys, socket

if len(sys.argv) > 1:
    for remote_host in sys.argv[1:]:
        # determine local host ip by outgoing test to another host
        # use port 9 (discard protocol - RFC 863) over UDP4
        with socket.socket(socket.AF_INET, socket.SOCK_DGRAM) as s:
            s.connect((remote_host, 9))
            my_ip = s.getsockname()[0]
            print(my_ip, flush=True)
else:
    import platform

    my_name = platform.node()
    my_ip = socket.gethostbyname(my_name)
    print(my_ip)

它需要任意数量的远程主机,并打印出本地ip,逐个到达它们:

$ my-ip-to z.cn g.cn localhost
192.168.11.102
192.168.11.102
127.0.0.1
$

并在没有给出arg时打印best-bet。

$ my-ip-to
192.168.11.102

稍微改进了使用IP命令的命令版本,并返回IPv4和IPv6地址:

import commands,re,socket

#A generator that returns stripped lines of output from "ip address show"
iplines=(line.strip() for line in commands.getoutput("ip address show").split('\n'))

#Turn that into a list of IPv4 and IPv6 address/mask strings
addresses1=reduce(lambda a,v:a+v,(re.findall(r"inet ([\d.]+/\d+)",line)+re.findall(r"inet6 ([\:\da-f]+/\d+)",line) for line in iplines))
#addresses1 now looks like ['127.0.0.1/8', '::1/128', '10.160.114.60/23', 'fe80::1031:3fff:fe00:6dce/64']

#Get a list of IPv4 addresses as (IPstring,subnetsize) tuples
ipv4s=[(ip,int(subnet)) for ip,subnet in (addr.split('/') for addr in addresses1 if '.' in addr)]
#ipv4s now looks like [('127.0.0.1', 8), ('10.160.114.60', 23)]

#Get IPv6 addresses
ipv6s=[(ip,int(subnet)) for ip,subnet in (addr.split('/') for addr in addresses1 if ':' in addr)]

通过命令行utils产生“干净”输出的一个简单方法:

import commands
ips = commands.getoutput("/sbin/ifconfig | grep -i \"inet\" | grep -iv \"inet6\" | " +
                         "awk {'print $2'} | sed -ne 's/addr\:/ /p'")
print ips

它将显示系统上的所有IPv4地址。

在Linux上:

>>> import socket, struct, fcntl
>>> sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
>>> sockfd = sock.fileno()
>>> SIOCGIFADDR = 0x8915
>>>
>>> def get_ip(iface = 'eth0'):
...     ifreq = struct.pack('16sH14s', iface, socket.AF_INET, '\x00'*14)
...     try:
...         res = fcntl.ioctl(sockfd, SIOCGIFADDR, ifreq)
...     except:
...         return None
...     ip = struct.unpack('16sH2x4s8x', res)[2]
...     return socket.inet_ntoa(ip)
... 
>>> get_ip('eth0')
'10.80.40.234'
>>> 

一个我不相信已经发布的版本。 我在Ubuntu 12.04上使用python 2.7进行测试。

找到这个解决方案:http://code.activestate.com/recipes/439094-get-the-ip-address-associated-with-a-network-inter/

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])

结果示例:

>>> get_ip_address('eth0')
'38.113.228.130'