Curl POST request into pycurl code

In your bash example, the property target is an array, in your Python example it is a string.

Try this:

data = json.dumps({"name": "abc", "path": "def", "target": ["ghi"]})

I also strongly advise you to check out the requests library which has a much nicer API:

import requestsdata = {"name": "abc", "path": "def", "target": ["ghi"]}response = requests.post('http://some-url', json=data)print response.status_code

PycURL is a wrapper on the libcurl library written in C language so its Python API can be bit puzzling. As some people are advocating use of python requests instead I just want to point out that it isn't a perfect replacement. For me, its lack of DNS resolution timeout was a deal breaker. I also find it much slower on my Raspberry Pi. This comparison may be relevant:Python Requests vs PyCurl Performance

So here's something that doesn't evade OP's question:

import pycurlimport jsonfrom cStringIO import StringIOcurl = pycurl.Curl()curl.setopt(pycurl.URL, 'http://some-url')curl.setopt(pycurl.HTTPHEADER, ['Accept: application/json',                                'Content-Type: application/json'])curl.setopt(pycurl.POST, 1)# If you want to set a total timeout, say, 3 secondscurl.setopt(pycurl.TIMEOUT_MS, 3000)## depending on whether you want to print details on stdout, uncomment either# curl.setopt(pycurl.VERBOSE, 1) # to print entire request flow## or# curl.setopt(pycurl.WRITEFUNCTION, lambda x: None) # to keep stdout clean# preparing body the way pycurl.READDATA wants it# NOTE: you may reuse curl object setup at this point#  if sending POST repeatedly to the url. It will reuse#  the connection.body_as_dict = {"name": "abc", "path": "def", "target": "ghi"}body_as_json_string = json.dumps(body_as_dict) # dict to jsonbody_as_file_object = StringIO(body_as_json_string)# prepare and send. See also: pycurl.READFUNCTION to pass function insteadcurl.setopt(pycurl.READDATA, body_as_file_object) curl.setopt(pycurl.POSTFIELDSIZE, len(body_as_json_string))curl.perform()# you may want to check HTTP response code, e.g.status_code = curl.getinfo(pycurl.RESPONSE_CODE)if status_code != 200:    print "Aww Snap :( Server returned HTTP status code {}".format(status_code)# don't forget to release connection when finishedcurl.close()

There are some more interesting features worth checking out in the libcurl curleasy setopts documentation

I know this is over a year old now, but please try removing the whitespace in your header value.

c.setopt(pycurl.HTTPHEADER, ['Accept:application/json'])

I also prefer using the requests module as well because the APIs/methods are clean and easy to use.