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

ninjagecko回答的变体。这应该在任何允许UDP广播的LAN上工作,并且不需要访问LAN或internet上的地址。

import socket
def getNetworkIp():
    s = socket.socket(socket.AF_INET, socket.SOCK_DGRAM)
    s.setsockopt(socket.SOL_SOCKET, socket.SO_BROADCAST, 1)
    s.connect(('<broadcast>', 0))
    return s.getsockname()[0]

print (getNetworkIp())

其他回答

如果计算机有到Internet的路由,即使/etc/hosts没有正确设置,这也将始终工作以获得首选的本地ip地址。

import socket

s = socket.socket(socket.AF_INET, socket.SOCK_DGRAM)
s.connect(('8.8.8.8', 1))  # connect() for UDP doesn't send packets
local_ip_address = s.getsockname()[0]

Socket API方法

参见https://stackoverflow.com/a/28950776/711085

缺点:

Not cross-platform. Requires more fallback code, tied to existence of particular addresses on the internet This will also not work if you're behind a NAT Probably creates a UDP connection, not independent of (usually ISP's) DNS availability (see other answers for ideas like using 8.8.8.8: Google's (coincidentally also DNS) server) Make sure you make the destination address UNREACHABLE, like a numeric IP address that is spec-guaranteed to be unused. Do NOT use some domain like fakesubdomain.google.com or somefakewebsite.com; you'll still be spamming that party (now or in the future), and spamming your own network boxes as well in the process.


反射器方法

(请注意,这并没有回答OP的本地IP地址问题,例如192.168…;它会给你你的公共IP地址,根据用例,这可能更可取。)

你可以查询一些网站,如whatismyip.com(但有一个API),例如:

from urllib.request import urlopen
import re
def getPublicIp():
    data = str(urlopen('http://checkip.dyndns.com/').read())
    # data = '<html><head><title>Current IP Check</title></head><body>Current IP Address: 65.96.168.198</body></html>\r\n'

    return re.compile(r'Address: (\d+\.\d+\.\d+\.\d+)').search(data).group(1)

或者如果使用python2:

from urllib import urlopen
import re
def getPublicIp():
    data = str(urlopen('http://checkip.dyndns.com/').read())
    # data = '<html><head><title>Current IP Check</title></head><body>Current IP Address: 65.96.168.198</body></html>\r\n'

    return re.compile(r'Address: (\d+\.\d+\.\d+\.\d+)').search(data).group(1)

优点:

这种方法的一个优点是它是跨平台的 它从丑陋的nat(例如你的家用路由器)后面工作。

缺点(和变通方法):

要求网站正常运行,格式不变(几乎肯定不会),DNS服务器正常工作。在失败的情况下,还可以通过查询其他第三方IP地址反射器来缓解这个问题。 如果您不查询多个反射器(以防止一个受损害的反射器告诉您您的地址不是某个东西),或者如果您不使用HTTPS(以防止假装是服务器的中间人攻击),则可能的攻击向量


edit: Though initially I thought these methods were really bad (unless you use many fallbacks, the code may be irrelevant many years from now), it does pose the question "what is the internet?". A computer may have many interfaces pointing to many different networks. For a more thorough description of the topic, google for gateways and routes. A computer may be able to access an internal network via an internal gateway, or access the world-wide web via a gateway on for example a router (usually the case). The local IP address that the OP asks about is only well-defined with respect to a single link layer, so you have to specify that ("is it the network card, or the ethernet cable, which we're talking about?"). There may be multiple non-unique answers to this question as posed. However the global IP address on the world-wide web is probably well-defined (in the absence of massive network fragmentation): probably the return path via the gateway which can access the TLDs.

import netifaces as ni 

ni.ifaddresses('eth0')
ip = ni.ifaddresses('eth0')[ni.AF_INET][0]['addr']
print(ip)

这将返回你的IP地址在Ubuntu系统和MacOS。输出将是系统IP地址,如我的IP: 192.168.1.10。

在拥有iproute2实用程序的现代*NIX系统上,您可以通过subprocess.run()调用它,因为您可以使用-j开关在JSON中输出,然后使用JSON .loads()模块和方法将其转换为python数据结构。下面的代码显示第一个非环回IP地址。

import subprocess
import json

ip = json.loads(subprocess.run('ip -j a'.split(),capture_output=True).stdout.decode())[1]['addr_info'][0]['local'] 

print(ip)

或者,如果你有多个IP,并且想要找到连接到特定目的地的IP,你可以使用IP -j route get 8.8.8.8,如下所示:

import subprocess 
import json 

ip = json.loads(subprocess.run('ip -j route get 8.8.8.8'.split(),capture_output=True).stdout.decode())[0]['prefsrc']

print(ip)

如果你在寻找所有的IP地址,你可以遍历IP -j a返回的字典列表

import subprocess
import json

list_of_dicts = json.loads(subprocess.run('ip -j a'.split(),capture_output=True).stdout.decode())

for interface in list_of_dicts:
    try:print(f"Interface: {interface['ifname']:10} IP: {interface['addr_info'][0]['local']}")
    except:pass

这与之前发布的答案非常相似,但我找不到任何与这种调用用法有关的答案。这是我用于ipv4的。对于ipv6,更改'。' in to ':' in

import socket
print next(i[4][0] for i in socket.getaddrinfo(
    socket.gethostname(), 80) if '127.' not in i[4][0] and '.' in i[4][0]);"