Pyhon: HTTP POST with requests, pycurl module

Here is sample Python script when achieving the same “curl ..” POST command as below with python requests, pycurl module.

# curl -v -i -w "\n" -H "Accept: application/json" -H "Content-type: application/json" -u "admin:admin" -X POST -d '{"name":”hello"}' http://127.0.0.1:8080/list

[ requests module ]

# cat requests_create_empty_shared_list.py #!/bin/env python import json import requests url = 'http://127.0.0.1:8080/list' def creatList():        headers = {'Content-Type': 'application/json', 'Accept': 'application/json'}        values = {                'name':'foobar01'        }        data = json.dumps(values)        res = requests.post(url,data=data,headers=headers,auth=requests.auth.HTTPBasicAuth('admin','admin')) creatList()

[ pycurl module ]

# cat pycurl_create_empty_shared_list.py #!/bin/env python import pycurl import json url = 'http://127.0.0.1:8080/list' headers = {'Content-Type': 'application/json'} payload = {'name':'hi'} post = json.dumps(payload) def createList():        c = pycurl.Curl()        c.setopt(pycurl.URL, '%s' % url)        c.setopt(pycurl.HTTPHEADER, ['Accept: application/json', 'Content-Type: application/json'])        c.setopt(pycurl.VERBOSE, 0)        c.setopt(pycurl.USERPWD, 'admin:admin')        c.setopt(pycurl.POST, 1)        c.setopt(pycurl.POSTFIELDS, post)        c.perform()