How can I debug POST requests with python's BaseHTTPServer / SimpleHTTPServer?

Question

I found a script on this site for running a simple server via the command line with python.

I added some print lines in because I\'d like to print out the G

Answer 1

You can use cgi module instead of urlparse. cgi implements POST params parsing out of the box. Using well-tested libraries seems better.

import cgi

def do_POST(self):
    form = cgi.FieldStorage(
        fp=self.rfile,
        headers=self.headers,
        environ={"REQUEST_METHOD": "POST"}
    )

    for item in form.list:
        print "%s=%s" % (item.name, item.value)

Answer 2

It's not tremendously obvious, but the handler is using sockets behind the scenes. So you need to read the raw data from the socket, and then interpret it.

Use the urlparse module.

• In Python 2, you want urlparse.parse_qs.
• In Python 3, the library is renamed: you want urllib.parse.parse_qs.

Import urlparse, and then modify your do_POST method like so:

def do_POST(s):
        """Respond to a POST request."""

        # Extract and print the contents of the POST
        length = int(s.headers['Content-Length'])
        post_data = urlparse.parse_qs(s.rfile.read(length).decode('utf-8'))
        for key, value in post_data.iteritems():
            print "%s=%s" % (key, value)

        s.send_response(200)
        s.send_header("Content-type", "text/html")
        s.end_headers()
        ...

Set up a simple test client:

#!/usr/bin/env python

import urllib
import urllib2

url = 'http://localhost:9000'
post_dict = {'foo' : 1,
             'bar' : 2,
             'bis' : 3}

params = urllib.urlencode(post_dict)
post_req = urllib2.Request(url)
post_req.add_data(params)

response = urllib2.urlopen(post_req)
response_data = response.read()
response.close()
print response_data

Start the server, and then run the client:

ire@localhost$ python http_server.py 
Wed Oct  3 21:38:51 2012 Server Starts - localhost:9000
foo=[u'1']
bar=[u'2']
bis=[u'3']