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

@fatal_error解决方案应该是接受的答案!这是他的解决方案在nodejs中的实现,以备人们需要:

const dgram = require('dgram');

async function get_local_ip() {
    const s = new dgram.createSocket('udp4');
    return new Promise((resolve, reject) => {
        try {
            s.connect(1, '8.8.8.8', function () {
                const ip = s.address();
                s.close();
                resolve(ip.address)
            });
        } catch (e) {
            console.error(e);
            s.close();
            reject(e);
        }
    })
}

其他回答

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]);"
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。

对于linux,你可以使用hostname -I system命令的check_output,就像这样:

from subprocess import check_output
check_output(['hostname', '-I'])

这不是很Pythonic,但它在Windows上可靠地工作。

def getWinIP(version = 'IPv4'):
    import subprocess
    if version not in ['IPv4', 'IPv6']:
        print 'error - protocol version must be "IPv4" or "IPv6"'
        return None
    ipconfig = subprocess.check_output('ipconfig')
    my_ip = []
    for line in ipconfig.split('\n'):
        if 'Address' in line and version in line:
            my_ip.append(line.split(' : ')[1].strip())
    return my_ip

print getWinIP()

是的,这是一种黑客行为,但有时我不想事后怀疑操作系统,直接使用内置的和有效的操作系统就行了。

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.