Curl POST request into pycurl code

Question

I'm trying to convert following curl request into pycurl:

curl -v
-H Accept:application/json \
-H Content-Type:application/json \
-d "{
    name: 'abc',
    path: 'def',
    target: [ 'ghi' ]
}" \
-X POST http://some-url

I have following python code:

import pycurl, json

c = pycurl.Curl()
c.setopt(pycurl.URL, 'http://some-url')
c.setopt(pycurl.HTTPHEADER, ['Accept: application/json'])
data = json.dumps({"name": "abc", "path": "def", "target": "ghi"})
c.setopt(pycurl.POST, 1)
c.setopt(pycurl.POSTFIELDS, data)
c.setopt(pycurl.VERBOSE, 1)
c.perform()
print curl_agent.getinfo(pycurl.RESPONSE_CODE)
c.close()

Executing this I had an error 415: Unsupported media type, so I have changed:

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

into:

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

This time I have 400: Bad request. But bash code with curl works. Do you have any idea what should I fix in python code?

Answer 1

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