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
对于*nix系统上的IP地址列表,
import subprocess
co = subprocess.Popen(['ifconfig'], stdout = subprocess.PIPE)
ifconfig = co.stdout.read()
ip_regex = re.compile('((25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-4]|2[0-5][0-9]|[01]?[0-9][0-9]?))')
[match[0] for match in ip_regex.findall(ifconfig, re.MULTILINE)]
虽然现在回答这个问题有点晚了,但我认为其他人可能会发现它有用:-)
PS:它会返回广播地址和网络掩码。
其他回答
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
这将在大多数linux盒子上工作:
import socket, subprocess, re
def get_ipv4_address():
"""
Returns IP address(es) of current machine.
:return:
"""
p = subprocess.Popen(["ifconfig"], stdout=subprocess.PIPE)
ifc_resp = p.communicate()
patt = re.compile(r'inet\s*\w*\S*:\s*(\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3})')
resp = patt.findall(ifc_resp[0])
print resp
get_ipv4_address()
我决定使用ipfy: https://www.ipify.org的服务和/或API。
#!/usr/bin/env python3
from urllib.request import urlopen
def public_ip():
data = urlopen('https://api.ipify.org').read()
return str(data, encoding='utf-8')
print(public_ip())
还可以以JSON和JSONP格式获得响应。
Github上有一个ipify Python库。
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()
@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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