regular expression for ipaddress and mac address

Question

can anyone suggest me the regular expression for ip address and mac address ?

i am using python & django

for example , http://[ipaddress]/SaveData/127.0.0.1/00-0C-F1-56-98-AD/

for mac address i tried following but didn't work

([0-9A-F]{2}[:-]){5}([0-9A-F]{2})

^([0-9A-F]{2}[:-]){5}([0-9A-F]{2})$

Answer 1

You can use /^([0-2]?\d{0,2}\.){3}([0-2]?\d{0,2})$/ for IPv4 Address and /^([\da-fA-F]{1,4}:){7}([\da-fA-F]{1,4})$/i for IPv6 address.

You can combine these two as /^((([0-2]?\d{0,2}\.){3}([0-2]?\d{0,2}))|(([\da-fA-F]{1,4}:){7}([\da-fA-F]{1,4})))$/i. You can find a sample here.

Ref: http://snipplr.com/view/49994/ipv4-regex/, http://snipplr.com/view/49993/ipv6-regex/

For Mac Address You can use /^([0-9A-F]{2}[-:]){5}[0-9A-F]{2}$/i. You can find a sample here.

Answer 2

import re
s = "http://[ipaddress]/SaveData/127.0.0.1/00-0C-F1-56-98-AD/"

re.search(r'([0-9A-F]{2}[:-]){5}([0-9A-F]{2})', s, re.I).group()
'00-0C-F1-56-98-AD'

re.search(r'((2[0-5]|1[0-9]|[0-9])?[0-9]\.){3}((2[0-5]|1[0-9]|[0-9])?[0-9])', s, re.I).group()
'127.0.0.1'

Place this snippet in your django routing definitions file - urls.py

url(r'^SaveData/(?P<ip>((2[0-5]|1[0-9]|[0-9])?[0-9]\.){3}((2[0-5]|1[0-9]|[0-9])?[0-9]))/(?P<mac>([0-9A-F]{2}[:-]){5}([0-9A-F]{2}))', SaveDataHandler.as_view()),

Answer 3

Your regular expression only contains two capturing groups (parentheses), so it isn't storing the entire address (the first group gets "overwritten"). Try these:

# store each octet into its own group
r"([\dA-F]{2})[-:]([\dA-F]{2})[-:]([\dA-F]{2})[-:]([\dA-F]{2})[-:]([\dA-F]{2})[-:]([\dA-F]{2})"
# store entire MAC address into a single group
r"([\dA-F]{2}(?:[-:][\dA-F]{2}){5})"

IP addresses get trickier because the ranges are binary but the representation is decimal.

# store each octet into its own group
r"(\d|[1-9]\d|1\d\d|2(?:[0-4]\d|5[0-5]))\.(\d|[1-9]\d|1\d\d|2(?:[0-4]\d|5[0-5]))\.(\d|[1-9]\d|1\d\d|2(?:[0-4]\d|5[0-5]))\.(\d|[1-9]\d|1\d\d|2(?:[0-4]\d|5[0-5]))"
# store entire IP address into a single group
r"((?:\d|[1-9]\d|1\d\d|2(?:[0-4]\d|5[0-5]))(?:\.(?:\d|[1-9]\d|1\d\d|2(?:[0-4]\d|5[0-5]))){3})"

Answer 4

consider s=256.1.1.1 i'd like to make a little modification from Michal's answer:

def find_ip(s):
    part = '(2[0-4]|1[0-9]|[0-9])?[0-9]|25[0-5]'
    res =re.search(r'(^| )((%s)\.){3}(%s)' %(part,part), s,re.I )
    if res:
        return res.group().strip()
    else:
        return ''

notice '(^| )' means line start or space ahead, to avoid get '56.1.1.1' from '256.1.1.1'

Answer 5

alright so this is what I use for IPV4

([0-9]{1,3}.){3}[0-9]{1,3}

tested with

127.0.0.1 255.255.255.255

and works for all

Answer 6

This is for MAC address:

([0-9A-F]{2}[:-]){5}([0-9A-F]{2})