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
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
其他回答
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'
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。
好吧,这是Windows特定的,需要安装python WMI模块,但这似乎比不断尝试调用外部服务器要简单得多。这只是另一种选择,因为已经有很多好的选择,但它可能非常适合您的项目。
Import WMI
def getlocalip():
local = wmi.WMI()
for interface in local.Win32_NetworkAdapterConfiguration(IPEnabled=1):
for ip_address in interface.IPAddress:
if ip_address != '0.0.0.0':
localip = ip_address
return localip
>>>getlocalip()
u'xxx.xxx.xxx.xxx'
>>>
顺便说一下,WMI非常强大……如果你正在做任何窗口机器的远程管理,你一定要看看它能做什么。
使用新引入的asyncio包的Python 3.4版本。
async def get_local_ip():
loop = asyncio.get_event_loop()
transport, protocol = await loop.create_datagram_endpoint(
asyncio.DatagramProtocol,
remote_addr=('8.8.8.8', 80))
result = transport.get_extra_info('sockname')[0]
transport.close()
return result
这是基于UnkwnTech的精彩回答。
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())
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