给定一个IP地址(192.168.0.1说),我怎么检查它是否在Python的网络(比如192.168.0.0/24)吗?
有没有在Python一般工具的IP地址操纵?这样的东西主机查找,IP地址经过为int,与网络掩码的网络地址为int等?希望在标准Python库2.5。
This article显示您可以用socket
和struct
模块做没有太多额外的努力。我加了一点文章如下:
import socket,struct
def makeMask(n):
"return a mask of n bits as a long integer"
return (2L<<n-1) - 1
def dottedQuadToNum(ip):
"convert decimal dotted quad string to long integer"
return struct.unpack('L',socket.inet_aton(ip))[0]
def networkMask(ip,bits):
"Convert a network address to a long integer"
return dottedQuadToNum(ip) & makeMask(bits)
def addressInNetwork(ip,net):
"Is an address in a network"
return ip & net == net
address = dottedQuadToNum("192.168.1.1")
networka = networkMask("10.0.0.0",24)
networkb = networkMask("192.168.0.0",24)
print (address,networka,networkb)
print addressInNetwork(address,networka)
print addressInNetwork(address,networkb)
这种输出:
False
True
如果你只是想一个函数,它串它是这样的:
import socket,struct
def addressInNetwork(ip,net):
"Is an address in a network"
ipaddr = struct.unpack('L',socket.inet_aton(ip))[0]
netaddr,bits = net.split('/')
netmask = struct.unpack('L',socket.inet_aton(netaddr))[0] & ((2L<<int(bits)-1) - 1)
return ipaddr & netmask == netmask
马克的代码几乎是正确的。代码的完整版本 -
def addressInNetwork3(ip,net):
'''This function allows you to check if on IP belogs to a Network'''
ipaddr = struct.unpack('=L',socket.inet_aton(ip))[0]
netaddr,bits = net.split('/')
netmask = struct.unpack('=L',socket.inet_aton(calcDottedNetmask(int(bits))))[0]
network = struct.unpack('=L',socket.inet_aton(netaddr))[0] & netmask
return (ipaddr & netmask) == (network & netmask)
def calcDottedNetmask(mask):
bits = 0
for i in xrange(32-mask,32):
bits |= (1 << i)
return "%d.%d.%d.%d" % ((bits & 0xff000000) >> 24, (bits & 0xff0000) >> 16, (bits & 0xff00) >> 8 , (bits & 0xff))
显然,从相同的源代码上面...
一个非常重要的注意的是,第一代码有一个小故障 - IP地址255.255.255.255也显示了作为任何子网有效的IP。我有时间得到这个代码工作,并感谢马克为正确答案赫克。
依托“结构”模块会导致与字节序和类型的大小问题,恰恰是没有必要的。也不是socket.inet_aton()。 Python的工作非常好,点分IP地址:
def ip_to_u32(ip):
return int(''.join('%02x' % int(d) for d in ip.split('.')), 16)
我需要每个插座上做IP匹配接受()调用,对整个组允许的源网络的,所以我预先计算口罩和网络,为整数:
SNS_SOURCES = [
# US-EAST-1
'207.171.167.101',
'207.171.167.25',
'207.171.167.26',
'207.171.172.6',
'54.239.98.0/24',
'54.240.217.16/29',
'54.240.217.8/29',
'54.240.217.64/28',
'54.240.217.80/29',
'72.21.196.64/29',
'72.21.198.64/29',
'72.21.198.72',
'72.21.217.0/24',
]
def build_masks():
masks = [ ]
for cidr in SNS_SOURCES:
if '/' in cidr:
netstr, bits = cidr.split('/')
mask = (0xffffffff << (32 - int(bits))) & 0xffffffff
net = ip_to_u32(netstr) & mask
else:
mask = 0xffffffff
net = ip_to_u32(cidr)
masks.append((mask, net))
return masks
然后,我可以很快地看到一个给定的IP是这些网络中的一个内:
ip = ip_to_u32(ipstr)
for mask, net in cached_masks:
if ip & mask == net:
# matched!
break
else:
raise BadClientIP(ipstr)
无模块需要进口,而且代码是在匹配速度非常快。
>>> from netaddr import all_matching_cidrs
>>> all_matching_cidrs("212.11.70.34", ["192.168.0.0/24","212.11.64.0/19"] )
[IPNetwork('212.11.64.0/19')]
下面是该方法的用法:
>>> help(all_matching_cidrs)
Help on function all_matching_cidrs in module netaddr.ip:
all_matching_cidrs(ip, cidrs)
Matches an IP address or subnet against a given sequence of IP addresses and subnets.
@param ip: a single IP address or subnet.
@param cidrs: a sequence of IP addresses and/or subnets.
@return: all matching IPAddress and/or IPNetwork objects from the provided
sequence, an empty list if there was no match.
基本上你提供一个IP地址作为第一个参数和cidrs作为第二个参数列表。返回的命中列表。
#This works properly without the weird byte by byte handling def addressInNetwork(ip,net): '''Is an address in a network''' # Convert addresses to host order, so shifts actually make sense ip = struct.unpack('>L',socket.inet_aton(ip))[0] netaddr,bits = net.split('/') netaddr = struct.unpack('>L',socket.inet_aton(netaddr))[0] # Must shift left an all ones value, /32 = zero shift, /0 = 32 shift left netmask = (0xffffffff << (32-int(bits))) & 0xffffffff # There's no need to mask the network address, as long as its a proper network address return (ip & netmask) == netaddr
所选择的答案有一个bug。
下面是正确的代码:
def addressInNetwork(ip, net_n_bits):
ipaddr = struct.unpack('<L', socket.inet_aton(ip))[0]
net, bits = net_n_bits.split('/')
netaddr = struct.unpack('<L', socket.inet_aton(net))[0]
netmask = ((1L << int(bits)) - 1)
return ipaddr & netmask == netaddr & netmask
注:ipaddr & netmask == netaddr & netmask
代替ipaddr & netmask == netmask
。
我也((2L<<int(bits)-1) - 1)
取代((1L << int(bits)) - 1)
,因为后者似乎更容易理解。
这是一类我写的最长前缀匹配:
#!/usr/bin/env python
class Node:
def __init__(self):
self.left_child = None
self.right_child = None
self.data = "-"
def setData(self, data): self.data = data
def setLeft(self, pointer): self.left_child = pointer
def setRight(self, pointer): self.right_child = pointer
def getData(self): return self.data
def getLeft(self): return self.left_child
def getRight(self): return self.right_child
def __str__(self):
return "LC: %s RC: %s data: %s" % (self.left_child, self.right_child, self.data)
class LPMTrie:
def __init__(self):
self.nodes = [Node()]
self.curr_node_ind = 0
def addPrefix(self, prefix):
self.curr_node_ind = 0
prefix_bits = ''.join([bin(int(x)+256)[3:] for x in prefix.split('/')[0].split('.')])
prefix_length = int(prefix.split('/')[1])
for i in xrange(0, prefix_length):
if (prefix_bits[i] == '1'):
if (self.nodes[self.curr_node_ind].getRight()):
self.curr_node_ind = self.nodes[self.curr_node_ind].getRight()
else:
tmp = Node()
self.nodes[self.curr_node_ind].setRight(len(self.nodes))
tmp.setData(self.nodes[self.curr_node_ind].getData());
self.curr_node_ind = len(self.nodes)
self.nodes.append(tmp)
else:
if (self.nodes[self.curr_node_ind].getLeft()):
self.curr_node_ind = self.nodes[self.curr_node_ind].getLeft()
else:
tmp = Node()
self.nodes[self.curr_node_ind].setLeft(len(self.nodes))
tmp.setData(self.nodes[self.curr_node_ind].getData());
self.curr_node_ind = len(self.nodes)
self.nodes.append(tmp)
if i == prefix_length - 1 :
self.nodes[self.curr_node_ind].setData(prefix)
def searchPrefix(self, ip):
self.curr_node_ind = 0
ip_bits = ''.join([bin(int(x)+256)[3:] for x in ip.split('.')])
for i in xrange(0, 32):
if (ip_bits[i] == '1'):
if (self.nodes[self.curr_node_ind].getRight()):
self.curr_node_ind = self.nodes[self.curr_node_ind].getRight()
else:
return self.nodes[self.curr_node_ind].getData()
else:
if (self.nodes[self.curr_node_ind].getLeft()):
self.curr_node_ind = self.nodes[self.curr_node_ind].getLeft()
else:
return self.nodes[self.curr_node_ind].getData()
return None
def triePrint(self):
n = 1
for i in self.nodes:
print n, ':'
print i
n += 1
这里是一个测试程序:
n=LPMTrie()
n.addPrefix('10.25.63.0/24')
n.addPrefix('10.25.63.0/16')
n.addPrefix('100.25.63.2/8')
n.addPrefix('100.25.0.3/16')
print n.searchPrefix('10.25.63.152')
print n.searchPrefix('100.25.63.200')
#10.25.63.0/24
#100.25.0.3/16
谢谢您的脚本! 我有工作相当长的就可以了,让一切工作...所以我在这里分享它
#!/usr/bin/python
>>> calcDottedNetmask(21)
>>> '255.255.248.0'
#!/usr/bin/python
>>> addressInNetwork('188.104.8.64','172.16.0.0/12')
>>>True which is completely WRONG!!
所以我的新addressInNetwork函数看起来类似:
#!/usr/bin/python
import socket,struct
def addressInNetwork(ip,net):
'''This function allows you to check if on IP belogs to a Network'''
ipaddr = struct.unpack('=L',socket.inet_aton(ip))[0]
netaddr,bits = net.split('/')
netmask = struct.unpack('=L',socket.inet_aton(calcDottedNetmask(bits)))[0]
network = struct.unpack('=L',socket.inet_aton(netaddr))[0] & netmask
return (ipaddr & netmask) == (network & netmask)
def calcDottedNetmask(mask):
bits = 0
for i in xrange(32-int(mask),32):
bits |= (1 > 24, (bits & 0xff0000) >> 16, (bits & 0xff00) >> 8 , (bits & 0xff))
而现在,答案是正确的!
#!/usr/bin/python
>>> addressInNetwork('188.104.8.64','172.16.0.0/12')
False
我希望这将帮助其他人,节省时间让他们!
与所有上述情况,我认为socket.inet_aton()返回在网络顺序字节,所以解压它们正确的做法是可能
struct.unpack('!L', ... )
以前的解决方案在IP和网==净的错误。正确的IP查找是IP&网络掩码=网
bugfixed代码:
import socket
import struct
def makeMask(n):
"return a mask of n bits as a long integer"
return (2L<<n-1) - 1
def dottedQuadToNum(ip):
"convert decimal dotted quad string to long integer"
return struct.unpack('L',socket.inet_aton(ip))[0]
def addressInNetwork(ip,net,netmask):
"Is an address in a network"
print "IP "+str(ip) + " NET "+str(net) + " MASK "+str(netmask)+" AND "+str(ip & netmask)
return ip & netmask == net
def humannetcheck(ip,net):
address=dottedQuadToNum(ip)
netaddr=dottedQuadToNum(net.split("/")[0])
netmask=makeMask(long(net.split("/")[1]))
return addressInNetwork(address,netaddr,netmask)
print humannetcheck("192.168.0.1","192.168.0.0/24");
print humannetcheck("192.169.0.1","192.168.0.0/24");
有这就是所谓的SubnetTree提供蟒蛇是做好这项工作非常出色的API。这是一个简单的例子:
import SubnetTree
t = SubnetTree.SubnetTree()
t.insert("10.0.1.3/32")
print("10.0.1.3" in t)
我喜欢用netaddr为:
from netaddr import CIDR, IP
if IP("192.168.0.1") in CIDR("192.168.0.0/24"):
print "Yay!"
作为arno_v在评论中指出,netaddr中的新版本,它是这样的:
from netaddr import IPNetwork, IPAddress
if IPAddress("192.168.0.1") in IPNetwork("192.168.0.0/24"):
print "Yay!"
import socket,struct
def addressInNetwork(ip,net):
"Is an address in a network"
ipaddr = struct.unpack('!L',socket.inet_aton(ip))[0]
netaddr,bits = net.split('/')
netaddr = struct.unpack('!L',socket.inet_aton(netaddr))[0]
netmask = ((1<<(32-int(bits))) - 1)^0xffffffff
return ipaddr & netmask == netaddr & netmask
print addressInNetwork('10.10.10.110','10.10.10.128/25')
print addressInNetwork('10.10.10.110','10.10.10.0/25')
print addressInNetwork('10.10.10.110','10.20.10.128/25')
$蟒蛇check-subnet.py 假 真正 假
我不知道在标准库中任何东西,但PySubnetTree是一个Python库,将做子网匹配。
从以上各种来源,从我自己的研究,我这是怎么了子网和地址计算工作。这些作品都不足以解决问题和其他相关问题。
class iptools:
@staticmethod
def dottedQuadToNum(ip):
"convert decimal dotted quad string to long integer"
return struct.unpack('>L', socket.inet_aton(ip))[0]
@staticmethod
def numToDottedQuad(n):
"convert long int to dotted quad string"
return socket.inet_ntoa(struct.pack('>L', n))
@staticmethod
def makeNetmask(mask):
bits = 0
for i in xrange(32-int(mask), 32):
bits |= (1 << i)
return bits
@staticmethod
def ipToNetAndHost(ip, maskbits):
"returns tuple (network, host) dotted-quad addresses given"
" IP and mask size"
# (by Greg Jorgensen)
n = iptools.dottedQuadToNum(ip)
m = iptools.makeMask(maskbits)
net = n & m
host = n - mask
return iptools.numToDottedQuad(net), iptools.numToDottedQuad(host)
这里是我的代码
# -*- coding: utf-8 -*-
import socket
class SubnetTest(object):
def __init__(self, network):
self.network, self.netmask = network.split('/')
self._network_int = int(socket.inet_aton(self.network).encode('hex'), 16)
self._mask = ((1L << int(self.netmask)) - 1) << (32 - int(self.netmask))
self._net_prefix = self._network_int & self._mask
def match(self, ip):
'''
判断传入的 IP 是不是本 Network 内的 IP
'''
ip_int = int(socket.inet_aton(ip).encode('hex'), 16)
return (ip_int & self._mask) == self._net_prefix
st = SubnetTest('100.98.21.0/24')
print st.match('100.98.23.32')
如果你不希望导入其他模块,你可以一起去:
def ip_matches_network(self, network, ip):
"""
'{:08b}'.format(254): Converts 254 in a string of its binary representation
ip_bits[:net_mask] == net_ip_bits[:net_mask]: compare the ip bit streams
:param network: string like '192.168.33.0/24'
:param ip: string like '192.168.33.1'
:return: if ip matches network
"""
net_ip, net_mask = network.split('/')
net_mask = int(net_mask)
ip_bits = ''.join('{:08b}'.format(int(x)) for x in ip.split('.'))
net_ip_bits = ''.join('{:08b}'.format(int(x)) for x in net_ip.split('.'))
# example: net_mask=24 -> compare strings at position 0 to 23
return ip_bits[:net_mask] == net_ip_bits[:net_mask]
我想提出的解决方案的一个子集,这些答案..没有成功,我终于适配和固定提出的代码,并写了我的固定功能。
我测试了它,并至少工作在小端架构 - e.g.x86--如果有人喜欢尝试在大端架构,请给我反馈。
IP2Int
代码来自this post,另一种方法是完全(我的测试用例)在这个问题以前建议的工作定位。
编码:
def IP2Int(ip):
o = map(int, ip.split('.'))
res = (16777216 * o[0]) + (65536 * o[1]) + (256 * o[2]) + o[3]
return res
def addressInNetwork(ip, net_n_bits):
ipaddr = IP2Int(ip)
net, bits = net_n_bits.split('/')
netaddr = IP2Int(net)
bits_num = int(bits)
netmask = ((1L << bits_num) - 1) << (32 - bits_num)
return ipaddr & netmask == netaddr & netmask
希望对大家有用,
使用ipaddress(in the stdlib since 3.3,at PyPi for 2.6/2.7):
>>> import ipaddress
>>> ipaddress.ip_address('192.168.0.1') in ipaddress.ip_network('192.168.0.0/24')
True
如果你要评估很多IP地址这种方式,你可能要计算的网络掩码前期,像
n = ipaddress.ip_network('192.0.0.0/16')
netw = int(n.network_address)
mask = int(n.netmask)
然后,对于每个地址,计算所述二进制表示与一个
a = int(ipaddress.ip_address('192.0.43.10'))
a = struct.unpack('!I', socket.inet_pton(socket.AF_INET, '192.0.43.10'))[0]
a = struct.unpack('!I', socket.inet_aton('192.0.43.10'))[0] # IPv4 only
最后,你可以简单地检查:
in_network = (a & mask) == netw
对于python3
In [64]: ipaddress.IPv4Address('192.168.1.1') in ipaddress.IPv4Network('192.168.0.0/24')
Out[64]: False
此代码为我工作在Linux x86。我真的没有考虑过ENDIANESS的问题,但我一直在使用超过对8个不同的网络串测试200K IP地址测试它反对“IPADDR”模块,并IPADDR的结果是相同的,因为这代码。
def addressInNetwork(ip, net):
import socket,struct
ipaddr = int(''.join([ '%02x' % int(x) for x in ip.split('.') ]), 16)
netstr, bits = net.split('/')
netaddr = int(''.join([ '%02x' % int(x) for x in netstr.split('.') ]), 16)
mask = (0xffffffff << (32 - int(bits))) & 0xffffffff
return (ipaddr & mask) == (netaddr & mask)
例:
>>> print addressInNetwork('10.9.8.7', '10.9.1.0/16')
True
>>> print addressInNetwork('10.9.8.7', '10.9.1.0/24')
False
我想戴夫·韦伯的解决方案,但击出了一些问题:
最根本的是 - 匹配应该通过取与检查与面罩的IP地址,然后检查结果精确匹配的网络地址。如已完成与网络地址不安定的IP地址。
我还注意到,只是忽略了尾段行为假设的一致性会救你只会对八位边界口罩工作(/ 24/16)。为了得到其它的掩模(/ 23,/ 21)正常工作我添加了一个“大于”的struct命令和改变了代码创建二进制掩模全“1”来启动和换档由(32掩模左)。
最后,我添加了一个简单的检查,网络地址是有效的面具,只打印一个警告,如果事实并非如此。
这里的结果:
def addressInNetwork(ip,net):
"Is an address in a network"
ipaddr = struct.unpack('>L',socket.inet_aton(ip))[0]
netaddr,bits = net.split('/')
netmask = struct.unpack('>L',socket.inet_aton(netaddr))[0]
ipaddr_masked = ipaddr & (4294967295<<(32-int(bits))) # Logical AND of IP address and mask will equal the network address if it matches
if netmask == netmask & (4294967295<<(32-int(bits))): # Validate network address is valid for mask
return ipaddr_masked == netmask
else:
print "***WARNING*** Network",netaddr,"not valid with mask /"+bits
return ipaddr_masked == netmask
我不使用模块不需要的时候,他们的粉丝。这项工作只需要简单的数学,所以这里是我的简单的函数来完成这项工作:
def ipToInt(ip):
o = map(int, ip.split('.'))
res = (16777216 * o[0]) + (65536 * o[1]) + (256 * o[2]) + o[3]
return res
def isIpInSubnet(ip, ipNetwork, maskLength):
ipInt = ipToInt(ip)#my test ip, in int form
maskLengthFromRight = 32 - maskLength
ipNetworkInt = ipToInt(ipNetwork) #convert the ip network into integer form
binString = "{0:b}".format(ipNetworkInt) #convert that into into binary (string format)
chopAmount = 0 #find out how much of that int I need to cut off
for i in range(maskLengthFromRight):
if i < len(binString):
chopAmount += int(binString[len(binString)-1-i]) * 2**i
minVal = ipNetworkInt-chopAmount
maxVal = minVal+2**maskLengthFromRight -1
return minVal <= ipInt and ipInt <= maxVal
然后使用它:
>>> print isIpInSubnet('66.151.97.0', '66.151.97.192',24)
True
>>> print isIpInSubnet('66.151.97.193', '66.151.97.192',29)
True
>>> print isIpInSubnet('66.151.96.0', '66.151.97.192',24)
False
>>> print isIpInSubnet('66.151.97.0', '66.151.97.192',29)
就是这样,这是比上面附带的模块解决方案快得多。
接受的答案不工作......这让我很生气。掩模是向后,并且不与不是一个简单的8比特块(例如,/ 24)中的任何位工作。我适应了答案,它工作得很好。
import socket,struct
def addressInNetwork(ip, net_n_bits):
ipaddr = struct.unpack('!L', socket.inet_aton(ip))[0]
net, bits = net_n_bits.split('/')
netaddr = struct.unpack('!L', socket.inet_aton(net))[0]
netmask = (0xFFFFFFFF >> int(bits)) ^ 0xFFFFFFFF
return ipaddr & netmask == netaddr
这里是一个返回点二进制字符串,以帮助可视掩蔽..有点像ipcalc
输出的功能。
def bb(i):
def s = '{:032b}'.format(i)
def return s[0:8]+"."+s[8:16]+"."+s[16:24]+"."+s[24:32]
例如:
没有2.5的标准库,但IPADDR使这很容易。我相信这是在名称ip地址下3.3。
import ipaddr
a = ipaddr.IPAddress('192.168.0.1')
n = ipaddr.IPNetwork('192.168.0.0/24')
#This will return True
n.Contains(a)