Equivalent python code for curl command for https call

Question

I have a curl command that works and gives me back the JSON. Curl command:

    curl -sS -k -L -H "Authorization: bearer <token>" -X GET https://IP:PORT/api/v1/namespaces

I tried with requestsand pycurl modules which I found in the other posts but no luck.

Can anyone help me with finding the equivalent in python???

rje · Accepted Answer · 2018-10-16 20:54:48Z

1

We can do this with requests like this:

import requests
header = {'Authorization': 'bearer <token>'}
resp = requests.get("https://IP:PORT/api/v1/namespaces", 
    headers = header, 
    verify=False)
print(resp.status_code)
print(resp.text)

The -H switch behaviour is replicated by sending a header
The -L switch behaviour is replicated by specifying verify=False
The -sS and -k are about the behaviour of curl and not relevant here
The -X GET is replicated by using requests.get()

answered Oct 16, 2018 at 20:54

rje

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1 Comment

Mohammed Over a year ago

Thanks, it works, the token I had was also not correct. Thanks again.

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