判断IP地址和网段是否包含于另一个网段中
>>> IP('10.10.10.0/24')<IP('10.10.11.0/24')
True
>>> '10.10.12.11' in IP('10.10.11.0/24')
False
>>> '10.10.11.11' in IP('10.10.11.0/24')
True
网段两个网段是否有重叠
>>> IP('192.168.0.0/23').overlaps('192.168.1.0/24') #1表示有重叠
1
>>> IP('192.168.2.0/23').overlaps('192.168.1.0/24') #0表示没重叠
示例
根据输入的IP或子网返回网络、掩码、广播、反向解析、子网数、IP类型
[root@localhost python2.6]# cat netsimple1.py
#!/usr/bin/env python
#by lineqi
#create time 2015-09-04
from IPy import IP
ip_s = raw_input('please input an IP or net-range:')
ips = IP(ip_s)
if len(ips) >1:
print('net :%s ' % ips.net())
print('netmask :%s ' % ips.netmask())
print('broadcast :%s ' % ips.broadcast())
print('revese address :%s ' % ips.reverseNames()[0]) #输出第1个子网地址反向解析
print('subnet :%s ' % len(ips)) #输出网络子网数
else:
print('reverse address :%s' % ips.reverseNames())
print('iptype :%s' % ips.iptype())
[root@localhost python2.6]# python netsimple1.py
please input an IP or net-range:192.168.1.0/28
net :192.168.1.0
netmask :255.255.255.240
broadcast :192.168.1.15
revese address :0.1.168.192.in-addr.arpa.
subnet :16
iptype :PRIVATE
[root@localhost python2.6]# python netsimple1.py
please input an IP or net-range:192.168.1.23
reverse address :['23.1.168.192.in-addr.arpa.']
iptype :PRIVATE
