Convert PHP Curl To Python - php

Question

I've been trying to translate some PHP code to Python 3 but can't quite get it to work. In PHP I have the following:
$request = "https://api.example.com/token";
$developerKey = "Basic VVVfdFdfsjkUIHDfdsjYTpMX3JQSDNJKSFQUkxCM0p0WWFpRklh";
$data = array('grant_type'=>'password',
'username'=>'name',
'password'=>'pass',
'scope'=>'2346323');
$cjconn = curl_init($request);
curl_setopt($cjconn, CURLOPT_POST, TRUE);
curl_setopt($cjconn, CURLOPT_HTTPHEADER, array('Authorization: '.$developerKey));
curl_setopt($cjconn, CURLOPT_SSL_VERIFYPEER, FALSE);
curl_setopt($cjconn, CURLOPT_RETURNTRANSFER, TRUE);
curl_setopt($cjconn, CURLOPT_POSTFIELDS,http_build_query($data));
$result = curl_exec($cjconn);
curl_close($cjconn);
$tokens = json_decode($result,true);
$accesstoken = $tokens['access_token'];
echo $accesstoken."\n";
I tried converting it to the following in Python:
import pycurl, json
url = 'https://api.example.com/token'
data = json.dumps({"grant_type":"password",
"username":"name",
"password":"pass",
"scope":"2346323"})
key = 'Basic VVVfdFdfsjkUIHDfdsjYTpMX3JQSDNJKSFQUkxCM0p0WWFpRklh'
c = pycurl.Curl()
c.setopt(pycurl.URL,url)
c.setopt(pycurl.HTTPHEADER,['Authorization: {}'.format(key)])
c.setopt(pycurl.POST,1)
c.setopt(pycurl.POSTFIELDS,data)
c.perform()
But I get the following error:
<faultstring>String index out of range: -1</faultstring>
How can I correct this, or is there a more pythonic solution?

Answer 1

If anyone is interested in the solution, I came up with the following which worked:
def getToken(self):
"""Retrieves the token from provider"""
#The data to be passed to retrieve the token
tokenData = {'grant_type':'password',
'username':TOKENUSERNAME,
'password':TOKENPASSWORD,
'scope':TOKENSCOPE}
#The header parameters
header_params = {'Authorization':KEY}
#Make the request for the token
r = requests.post(TOKENURL,data=tokenData,headers=header_params)
#Check the status code
if r.status_code not in [200,203]:
self.log.logentry("There was an error retrieving the data from Linkshare: {}:{}".format(r.status_code,r.text))
sys.exit()
#Extract the data from the response
data = r.json()
#Parse the access token
token = {'token':data['access_token'],
'type':data['bearer']}
return token