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

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)]

其他回答

一台计算机可以有多个网络接口(包括您提到的本地环回127.0.0.1)。就操作系统而言,它也是一个“真实IP地址”。

如果你想跟踪所有的接口,看看下面的Python包,参见:http://alastairs-place.net/netifaces/

我认为,如果您从主机文件中删除环回条目,就可以避免gethostbyname返回127.0.0.1。(有待核实)。

我必须解决“判断一个IP地址是否是本地的”这个问题,我的第一个想法是建立一个本地IP列表,然后与之匹配。这让我想到了这个问题。然而,我后来意识到有一种更直接的方法:尝试绑定该IP,看看它是否有效。

_local_ip_cache = []
_nonlocal_ip_cache = []
def ip_islocal(ip):
    if ip in _local_ip_cache:
        return True
    if ip in _nonlocal_ip_cache:
        return False
    s = socket.socket()
    try:
        try:
            s.bind((ip, 0))
        except socket.error, e:
            if e.args[0] == errno.EADDRNOTAVAIL:
                _nonlocal_ip_cache.append(ip)
                return False
            else:
                raise
    finally:
        s.close()
    _local_ip_cache.append(ip)
    return True

我知道这并不能直接回答问题,但是这对于任何试图解决相关问题的人以及遵循相同思路的人都是有帮助的。我认为这是一个跨平台的解决方案。

要获取ip地址,可以直接在python中使用shell命令:

import socket, subprocess

def get_ip_and_hostname():
    hostname =  socket.gethostname()

    shell_cmd = "ifconfig | awk '/inet addr/{print substr($2,6)}'"
    proc = subprocess.Popen([shell_cmd], stdout=subprocess.PIPE, shell=True)
    (out, err) = proc.communicate()

    ip_list = out.split('\n')
    ip = ip_list[0]

    for _ip in ip_list:
        try:
            if _ip != "127.0.0.1" and _ip.split(".")[3] != "1":
                ip = _ip
        except:
            pass
    return ip, hostname

ip_addr, hostname = get_ip_and_hostname()

此方法返回本地盒子上的“主”IP(具有默认路由的IP)。

不需要可路由的网络访问或任何连接。 即使所有接口都从网络断开,也能正常工作。 不需要甚至不尝试去其他地方。 工作与NAT,公共,私有,外部和内部IP 没有外部依赖的纯Python 2(或3)。 支持Linux、Windows和OSX。

Python 3或2:

    import socket
    def get_ip():
        s = socket.socket(socket.AF_INET, socket.SOCK_DGRAM)
        s.settimeout(0)
        try:
            # doesn't even have to be reachable
            s.connect(('10.254.254.254', 1))
            IP = s.getsockname()[0]
        except Exception:
            IP = '127.0.0.1'
        finally:
            s.close()
        return IP
    print(get_ip())

这将返回一个主IP(具有默认路由的IP)。如果您需要将所有IP附加到所有接口(包括localhost等),请参见类似这样的回答。

如果你在一个NAT防火墙后面,比如你家里的wifi路由器,那么这将不会显示你的公共NAT IP,而是显示你在本地网络上的私有IP,它有一个默认路由到你的本地wifi路由器。如果你需要外部IP:

在那个外部设备(wifi路由器)上运行这个功能,或者 连接到外部服务(如https://www.ipify.org/),可以反射从外部世界看到的IP

... 但这些想法与最初的问题完全不同。:)

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.