将curl命令转换为Python请求

Question

The curl command that I have that works properly is -

curl -X GET -H "Authorization: Basic <base64userpass>" -H "Content-Type: application/json" "http://<host>/bamboo/rest/api/latest/result/<plankey>.json?expand=results.result&os_authType=basic"

In Python, this is what I currently have -

   headers = {'Authorization': 'Basic <base64userpass>', 'Content-Type': 'application/json'}
   datapoints = {'expand': 'results.result', 'os_authType': 'basic'}
   url = "http://<host>/bamboo/rest/api/latest/result/<plankey>.json"
   r = requests.get(url, headers=headers, data=datapoints)

The response I get when using the Python request is <Response [403]> , but when using curl I get back the expected data.

What am I missing here?

Thanks.

Answer 1

You should use the auth option of requests to do basic authentication. There are more headers that the CURL command-line handle for you (and requests will not handle them unless you use the auth ):

>>> from requests.auth import HTTPBasicAuth
>>> requests.get('https://api.github.com/user', auth=HTTPBasicAuth('user', 'pass'))

Or just use:

>>> requests.get('https://api.github.com/user', auth=('user', 'pass'))

(Change the URL and everything).

Also note that requests should get the params= (and not data= ).