Python请求库如何通过单个令牌传递Authorization标头

I have a request URI and a token. If I use:

curl -s "<MY_URI>" -H "Authorization: TOK:<MY_TOKEN>"

etc., I get a 200 and view the corresponding JSON data. So, I installed requests and when I attempt to access this resource I get a 403 probably because I do not know the correct syntax to pass that token. Can anyone help me figure it out? This is what I have:

import sys,socket
import requests

r = requests.get('<MY_URI>','<MY_TOKEN>')
r. status_code

I already tried:

r = requests.get('<MY_URI>',auth=('<MY_TOKEN>'))
r = requests.get('<MY_URI>',auth=('TOK','<MY_TOKEN>'))
r = requests.get('<MY_URI>',headers=('Authorization: TOK:<MY_TOKEN>'))

But none of these work.

In python:

('<MY_TOKEN>')

is equivalent to

'<MY_TOKEN>'

And requests interprets

('TOK', '<MY_TOKEN>')

As you wanting requests to use Basic Authentication and craft an authorization header like so:

'VE9LOjxNWV9UT0tFTj4K'

Which is the base64 representation of 'TOK:<MY_TOKEN>'

To pass your own header you pass in a dictionary like so:

r = requests.get('<MY_URI>', headers={'Authorization': 'TOK:<MY_TOKEN>'})

I was looking for something similar and came across this. It looks like in the first option you mentioned

r = requests.get('<MY_URI>', auth=('<MY_TOKEN>'))

“auth” takes two parameters: username and password, so the actual statement should be

r=requests.get('<MY_URI>', auth=('<YOUR_USERNAME>', '<YOUR_PASSWORD>'))

In my case, there was no password, so I left the second parameter in auth field empty as shown below:

r=requests.get('<MY_URI', auth=('MY_USERNAME', ''))

Hope this helps somebody :)

This worked for me:

access_token = #yourAccessTokenHere#

result = requests.post(url,
      headers={'Content-Type':'application/json',
               'Authorization': 'Bearer {}'.format(access_token)})

You can also set headers for the entire session:

TOKEN = 'abcd0123'
HEADERS = {'Authorization': 'token {}'.format(TOKEN)}

with requests.Session() as s:

    s.headers.update(HEADERS)
    resp = s.get('http://example.com/')

Requests natively supports basic auth only with user-pass params, not with tokens.

You could, if you wanted, add the following class to have requests support token based basic authentication:

import requests
from base64 import b64encode

class BasicAuthToken(requests.auth.AuthBase):
    def __init__(self, token):
        self.token = token
    def __call__(self, r):
        authstr = 'Basic ' + b64encode(('token:' + self.token).encode('utf-8')).decode('utf-8')
        r.headers['Authorization'] = authstr
        return r

Then, to use it run the following request :

r = requests.get(url, auth=BasicAuthToken(api_token))

An alternative would be to formulate a custom header instead, just as was suggested by other users here.

i founded here, its ok with me for linkedin: https://auth0.com/docs/flows/guides/auth-code/call-api-auth-code so my code with with linkedin login here:

ref = 'https://api.linkedin.com/v2/me'
headers = {"content-type": "application/json; charset=UTF-8",'Authorization':'Bearer {}'.format(access_token)}
Linkedin_user_info = requests.get(ref1, headers=headers).json()

You can try something like this

r = requests.get(ENDPOINT, params=params, headers={'Authorization': 'Basic %s' %  API_KEY})

This worked for me:

r = requests.get('http://127.0.0.1:8000/api/ray/musics/', headers={'Authorization': 'Token 22ec0cc4207ebead1f51dea06ff149342082b190'})

My code uses user generated token.