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

Question

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

I'd agree with this post as a duplicate -- same answers anyway — Alvin, May 25 '12 at 07:40
possible duplicate of [Finding a public facing IP address in Python?](http://stackoverflow.com/questions/166545/finding-a-public-facing-ip-address-in-python) — jfs, Feb 24 '14 at 21:49

score 109 · Accepted Answer · edited Jul 13 '22 at 14:39

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.

I like icanhazip.com, it's a whole website with just your raw IP; no need to add arrays, formats, etc. — anonymous, Aug 09 '17 at 01:01
Just to add for ipify.org: With the url `https://api.ipify.org/?format=raw` you'll get the IP in plaintext, no JSON nonsense. — pepoluan, Jul 17 '20 at 03:37
https://api.ident.me works too, please fallback to https://tnedi.me if you want to write a complete code example ;) — Pierre Carrier, Feb 22 '22 at 22:35

score 15 · Answer 2 · answered Feb 04 '17 at 16:41

15

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

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

answered Feb 04 '17 at 16:41

HyperActive

1,129
12
12

score 4 · Answer 3 · answered Mar 04 '19 at 18:50

4

With requests module

import requests

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

answered Mar 04 '19 at 18:50

Ahmed

414
4
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1

nice! :) i like this over pinging some random site. i want to say it's more fragile, but i guess it's a bigger name than some of the other sites. As long as they don't change how their headers operate... – digitalfoo Feb 24 '20 at 00:15

score 2 · Answer 4 · answered Jan 15 '17 at 22:58

2

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'])

answered Jan 15 '17 at 22:58

Xavi Martínez

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1
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import ipgetter; IP = ipgetter.myip() is enough for the job. – Niko Föhr Feb 28 '17 at 21:18
ipgetter does not (yet) support IPv6, although IPv6 is already celebrating its twentieth birthday. :( – qräbnö Jun 06 '18 at 20:19
Dead as of 2022-01-24, try this: http://freegeoip.net/json/1.1.1.1 and it redirects to http://freegeoip.net/shutdown. – Contango Jan 24 '22 at 18:37

score 1 · Answer 5 · edited Aug 03 '17 at 03:21

1

You can just do this:

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

Produces:

XX.XX.XXX.XXX

edited Aug 03 '17 at 03:21

Rob

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answered Aug 03 '17 at 02:33

Isaac Chabot

31
5

score 1 · Answer 6 · edited Feb 25 '22 at 20:22

1

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

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

edited Feb 25 '22 at 20:22

OneCricketeer

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answered Apr 19 '20 at 20:16

Bella

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Please don't reply to duplicate questions. Mark them as duplicate. It makes SO more organized. – toliveira Apr 19 '20 at 23:37
It's actually three lines if you include `pip install requests` – OneCricketeer Feb 25 '22 at 20:22

score 0 · Answer 7 · answered May 17 '18 at 16:47

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"]

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

7 Answers7

ip.42.pl

jsonip.com

httpbin.org

ipify.org

Linked

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