Python One-liner to get your site's Alexa Rank

alexa.py

#!/usr/bin/env python
import urllib, sys, bs4
print bs4.BeautifulSoup(urllib.urlopen("http://data.alexa.com/data?cli=10&dat=s&url="+ sys.argv[1]).read(), "xml").find("REACH")['RANK']

First make it executable and then invoke with your domain name.

chmod +x alexa.py
./alexa.py masnun.me

Hi there.

First of all amazing and simple script, many thanks.

Just wondering if what we want here is .find("POPULARITY")['TEXT'] instead of find("REACH")['RANK'], since the figures are more consistent with information displayed in:

a) URLs like
http://www.alexa.com/siteinfo/example.com

b) PageRank plugin for Mozilla Firefox

Cheerss.

Hi there.

First of all many thanks for your one liner.

And even when it will break the "oneliner" concept just in case you want a humble improvement, below a version that catches the exception caused, probably, by an attempt of getting alexa rank for a non enough relevant domain:

!/usr/bin/env python

import urllib, sys, bs4
try:
print bs4.BeautifulSoup(urllib.urlopen("http://data.alexa.com/data?cli=10&dat=s&url="+ sys.argv[1]).read(), "xml").find("REACH")['RANK']
except TypeError:
print 'Unable to get "%s" Alexa rank. Not enough relevance?' %(sys.argv[1])`

PD: I was unable to properly format the code, sorry.

Cheers.

Thanks for the code!

I wrote something that doesn't involve any additional packages, uses only regex:

#!/usr/bin/env python
import urllib, sys, re
xml = urllib.urlopen('http://data.alexa.com/data?cli=10&dat=s&url=%s'%sys.argv[1]).read()
try: rank = int(re.search(r'<POPULARITY[^>]*TEXT="(\d+)"', xml).groups()[0])
except: rank = -1
print 'Your rank for %s is %d!\n' % (sys.argv[1], rank)

error is popping that index is out of bound...help!!!

:) @risheek20

import urllib.request, sys, re
import xmltodict, json

xml = urllib.request.urlopen('http://data.alexa.com/data?cli=10&dat=s&url={}'.format("www.example.com")).read()
 
result= xmltodict.parse(xml)
 
data = json.dumps(result).replace("@","")
data_tojson = json.loads(data)
url = data_tojson["ALEXA"]["SD"][1]["POPULARITY"]["URL"]
rank= data_tojson["ALEXA"]["SD"][1]["POPULARITY"]["TEXT"]
 
print("site {site}, rank {rank}".format(site=url,rank=rank))

I made one here: alexa_check.py

Not a one liner but it gets all ranks available, global and national. It will also show other countries rank (sometimes only one country will be shown on the website-besides global)...

For instance facebook.com will show only Global and US rank on the Alexa page. However my script finds that they are number 1 in several countries

:) @risheek20

import urllib.request, sys, re
import xmltodict, json

xml = urllib.request.urlopen('http://data.alexa.com/data?cli=10&dat=s&url={}'.format("www.example.com")).read()
 
result= xmltodict.parse(xml)
 
data = json.dumps(result).replace("@","")
data_tojson = json.loads(data)
url = data_tojson["ALEXA"]["SD"][1]["POPULARITY"]["URL"]
rank= data_tojson["ALEXA"]["SD"][1]["POPULARITY"]["TEXT"]
 
print("site {site}, rank {rank}".format(site=url,rank=rank))

hey buddy, ther's problem with parser.Parse(xml_input, True) ..... can you please edited???
I would rate it very much
Thanks in advance!