How to get current URL in python web page?

Question

I am a noob in Python. Just installed it, and spent 2 hours googleing how to get to a simple parameter sent in the URL to a Python script

Found this

Very helpful, except I cannot for anything in the world to figure out how to replace

import urlparse
url = 'http://foo.appspot.com/abc?def=ghi'
parsed = urlparse.urlparse(url)
print urlparse.parse_qs(parsed.query)['def']

With what do I replace url = 'string' to make it work? I just want to access http://site.com/test/test.py?param=abc and see abc printed.

Final code after Alex's answer:

url = os.environ["REQUEST_URI"] 
parsed = urlparse.urlparse(url) 
print urlparse.parse_qs(parsed.query)['param']

Are you using a framework in particular? The frameworks will help abstract your problem — Bryan, Jan 22 '13 at 22:03
Final code after Alex's answer: `code` url = os.environ["REQUEST_URI"] parsed = urlparse.urlparse(url) print urlparse.parse_qs(parsed.query)['param'] `code` — Bogdan P., Jan 22 '13 at 22:50

score 12 · Accepted Answer · edited May 23 '17 at 12:10

12

If you don't have any libraries to do this for you, you can construct your current URL from the HTTP request that gets sent to your script via the browser.

The headers that interest you are Host and whatever's after the HTTP method (probably GET, in your case). Here are some more explanations (first link that seemed ok, you're free to Google some more :).

This answer shows you how to get the headers in your CGI script:

If you are running as a CGI, you can't read the HTTP header directly, but the web server put much of that information into environment variables for you. You can just pick it out of os.environ[].

If you're doing this as an exercise, then it's fine because you'll get to understand what's behind the scenes. If you're building anything reusable, I recommend you use libraries or a framework so you don't reinvent the wheel every time you need something.

edited May 23 '17 at 12:10

Community

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answered Jan 22 '13 at 22:17

Alex Ciminian

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Thanks so much, and you were right. Found what I was looking in os.environ["REQUEST_URI"]. – Bogdan P. Jan 22 '13 at 22:49
You're welcome, glad to help. :) I've pasted your working solution from the comment in the question to make it more readable. – Alex Ciminian Jan 22 '13 at 23:41
I would upvote this if there was an actual code example. – nu everest Feb 24 '17 at 14:59

score 3 · Answer 2 · edited Jun 20 '20 at 09:12

This is how I capture in Python 3 from CGI (A) URL, (B) GET parameters and (C) POST data:

=======================================================

import sys, os, io

CAPTURE URL

myDomainSelf = os.environ.get('SERVER_NAME')

myPathSelf = os.environ.get('PATH_INFO')

myURLSelf = myDomainSelf + myPathSelf

CAPTURE GET DATA

myQuerySelf = os.environ.get('QUERY_STRING')

CAPTURE POST DATA

myTotalBytesStr=(os.environ.get('HTTP_CONTENT_LENGTH'))

if (myTotalBytesStr == None):

myJSONStr = '{"error": {"value": true, "message": "No (post) data received"}}'

else:

myTotalBytes=int(os.environ.get('HTTP_CONTENT_LENGTH'))

myPostDataRaw = io.open(sys.stdin.fileno(),"rb").read(myTotalBytes)

myPostData = myPostDataRaw.decode("utf-8")

Write RAW to FILE

mySpy = "myURLSelf: [" + str(myURLSelf) + "]\n"

mySpy = mySpy + "myQuerySelf: [" + str(myQuerySelf) + "]\n"

mySpy = mySpy + "myPostData: [" + str(myPostData) + "]\n"

You need to define your own myPath here

myFilename = "spy.txt"

myFilePath = myPath + "\" + myFilename

myFile = open(myFilePath, "w")

myFile.write(mySpy)

myFile.close()

=======================================================

Here are some other useful CGI environment vars:

AUTH_TYPE

CONTENT_LENGTH

CONTENT_TYPE

GATEWAY_INTERFACE

PATH_INFO

PATH_TRANSLATED

QUERY_STRING

REMOTE_ADDR

REMOTE_HOST

REMOTE_IDENT

REMOTE_USER

REQUEST_METHOD

SCRIPT_NAME

SERVER_NAME

SERVER_PORT

SERVER_PROTOCOL

SERVER_SOFTWARE