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
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.
其他回答
在Debian上(经过测试)，我怀疑大多数Linux ..
import commands
RetMyIP = commands.getoutput("hostname -I")
在MS Windows上(已测试)
import socket
socket.gethostbyname(socket.gethostname())
这是UnkwnTech的答案的变体——它提供了一个get_local_addr()函数，该函数返回主机的主LAN ip地址。我发布它是因为这增加了一些东西:ipv6支持，错误处理，忽略localhost/linklocal地址，并使用TESTNET地址(rfc5737)来连接。
# imports
import errno
import socket
import logging
# localhost prefixes
_local_networks = ("127.", "0:0:0:0:0:0:0:1")
# ignore these prefixes -- localhost, unspecified, and link-local
_ignored_networks = _local_networks + ("0.", "0:0:0:0:0:0:0:0", "169.254.", "fe80:")
def detect_family(addr):
if "." in addr:
assert ":" not in addr
return socket.AF_INET
elif ":" in addr:
return socket.AF_INET6
else:
raise ValueError("invalid ipv4/6 address: %r" % addr)
def expand_addr(addr):
"""convert address into canonical expanded form --
no leading zeroes in groups, and for ipv6: lowercase hex, no collapsed groups.
"""
family = detect_family(addr)
addr = socket.inet_ntop(family, socket.inet_pton(family, addr))
if "::" in addr:
count = 8-addr.count(":")
addr = addr.replace("::", (":0" * count) + ":")
if addr.startswith(":"):
addr = "0" + addr
return addr
def _get_local_addr(family, remote):
try:
s = socket.socket(family, socket.SOCK_DGRAM)
try:
s.connect((remote, 9))
return s.getsockname()[0]
finally:
s.close()
except socket.error:
# log.info("trapped error connecting to %r via %r", remote, family, exc_info=True)
return None
def get_local_addr(remote=None, ipv6=True):
"""get LAN address of host
:param remote:
return LAN address that host would use to access that specific remote address.
by default, returns address it would use to access the public internet.
:param ipv6:
by default, attempts to find an ipv6 address first.
if set to False, only checks ipv4.
:returns:
primary LAN address for host, or ``None`` if couldn't be determined.
"""
if remote:
family = detect_family(remote)
local = _get_local_addr(family, remote)
if not local:
return None
if family == socket.AF_INET6:
# expand zero groups so the startswith() test works.
local = expand_addr(local)
if local.startswith(_local_networks):
# border case where remote addr belongs to host
return local
else:
# NOTE: the two addresses used here are TESTNET addresses,
# which should never exist in the real world.
if ipv6:
local = _get_local_addr(socket.AF_INET6, "2001:db8::1234")
# expand zero groups so the startswith() test works.
if local:
local = expand_addr(local)
else:
local = None
if not local:
local = _get_local_addr(socket.AF_INET, "192.0.2.123")
if not local:
return None
if local.startswith(_ignored_networks):
return None
return local
from netifaces import interfaces, ifaddresses, AF_INET
iplist = [ifaddresses(face)[AF_INET][0]["addr"] for face in interfaces() if AF_INET in ifaddresses(face)]
print(iplist)
['10.8.0.2', '192.168.1.10', '127.0.0.1']
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)]
@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);
}
})
}
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