How can I get the public IP using python2.7?

Question

How can I get the public IP using python2.7? Not private IP.

Answer 1

Currently there are several options:

• ip.42.pl
• jsonip.com
• httpbin.org
• ipify.org

Below are exact ways you can utilize each of the above.

ip.42.pl

from urllib2 import urlopen
my_ip = urlopen('http://ip.42.pl/raw').read()

This is the first option I have found. It is very convenient for scripts, you don't need JSON parsing here.

jsonip.com

from json import load
from urllib2 import urlopen

my_ip = load(urlopen('https://ipv4.jsonip.com'))['ip']

Seemingly the sole purpose of this domain is to return IP address in JSON.

httpbin.org

from json import load
from urllib2 import urlopen

my_ip = load(urlopen('http://httpbin.org/ip'))['origin']

httpbin.org is service I often recommend to junior developers to use for testing their scripts / applications.

ipify.org

from json import load
from urllib2 import urlopen

my_ip = load(urlopen('https://api.ipify.org/?format=json'))['ip']

Power of this service results from lack of limits (there is no rate limiting), infrastructure (placed on Heroku, with high availability in mind) and flexibility (works for both IPv4 and IPv6).

EDIT : Added httpbin.org to the list of available options.

EDIT : Added ipify.org thanks to kert's note .

Answer 2

I like the requests package with http://ip.42.pl/raw

import requests
requests.get('http://ip.42.pl/raw').text

Answer 3

With requests module

import requests

public_IP = requests.get("https://www.wikipedia.org").headers["X-Client-IP"]
print public_IP

Answer 4

Try this:

import ipgetter
import requests

IP = ipgetter.myip()
url = 'http://freegeoip.net/json/'+IP
r = requests.get(url)
js = r.json()
print 'IP Adress: '         +   js['ip']
print 'Country Code: '      +   js['country_code']
print 'Country Name: '      +   js['country_name']
print 'Region Code: '       +   js['region_code']
print 'Region Name: '       +   js['region_name']
print 'City Name: '         +   js['city']
print 'Zip code: '          +   js['zip_code']
print 'Time Zone: '         +   js['time_zone']
print 'Latitude: '          +   str(js['latitude'])
print 'Longitude: '         +   str(js['longitude'])

Answer 5

You can just do this:

import requests
print requests.get("http://ipecho.net/plain?").text

Produces:

XX.XX.XXX.XXX

Answer 6

in python 2.7 it's just a code of 2 lines.

>>> import requests
>>print requests.get("http://ipconfig.in/ip").text

Answer 7

Getip is a small module which returns public IP address from a random server.

Install:

~$ pip install getip2

Use:

>> import getip
>> ip = getip.get()
>>
>> ip
'66.249.76.109'

Answer 8

This is a way not to have to make a call to the internet:

Please let me know if this doesn't work, then I can update the answer (it works for ~10 servers of mine)

from subprocess import check_output
out = check_output("/sbin/ifconfig | awk '/inet / { print $2 }' | sed 's/addr://'", shell=True)
[x for x in out.decode().split() if not x == "127.0.0.1" and 
                                    not (x.startswith("172") and x.endswith("0.1"))]
["x.x.x.x.x"]