Convert a curl POST request to Python only using standard library

Question

I would like to convert this curl command to something that I can use in Python for an existing script.

curl -u 7898678:X -H \'Content-Type: application/jso         


        

Answer 1

I would like this to work with the standard library if possible.

The standard library provides urllib and httplib for working with URLs:

>>> import httplib, urllib
>>> params = urllib.urlencode({'apple': 1, 'banana': 2, 'coconut': 'yummy'})
>>> headers = {"Content-type": "application/x-www-form-urlencoded",
...            "Accept": "text/plain"}
>>> conn = httplib.HTTPConnection("example.com:80")
>>> conn.request("POST", "/some/path/to/site", params, headers)
>>> response = conn.getresponse()
>>> print response.status, response.reason
200 OK

If you want to execute curl itself, though, you can just invoke os.system():

import os
TEXT = ...
cmd = """curl -u 7898678:X -H 'Content-Type: application/json'""" \
  """-d '{"message":{"body":"%{t}"}}' http://sample.com/36576/speak.json""" % \
  {'t': TEXT}

If you're willing to relax the standard-library-only restriction, you can use PycURL. Beware that it isn't very Pythonic (it's pretty much just a thin veneer over libcurl), and I'm not sure how compatible it is with Python 3.