Python请求库如何使用单个令牌传递授权标头
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Python requests library how to pass Authorization header with single token
提问by
I have a request URI and a token. If I use:
我有一个请求 URI 和一个令牌。如果我使用：
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:
等等，我得到一个200并查看相应的JSON数据。因此，我安装了请求，当我尝试访问此资源时，我得到 403，可能是因为我不知道传递该令牌的正确语法。谁能帮我弄清楚吗？这就是我所拥有的：
import sys,socket
import requests
r = requests.get('<MY_URI>','<MY_TOKEN>')
r. status_code
I already tried:
我已经尝试过：
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.
但这些都不起作用。
采纳答案by Ian Stapleton Cordasco
In python:
在蟒蛇中：
('<MY_TOKEN>')
is equivalent to
相当于
'<MY_TOKEN>'
And requests interprets
并请求解释
('TOK', '<MY_TOKEN>')
As you wanting requests to use Basic Authentication and craft an authorization header like so:
当您希望请求使用基本身份验证并制作授权标头时，如下所示：
'VE9LOjxNWV9UT0tFTj4K'
Which is the base64 representation of 'TOK:<MY_TOKEN>'
哪个是 base64 表示 'TOK:<MY_TOKEN>'
To pass your own header you pass in a dictionary like so:
要传递您自己的标题，您需要传入一个字典，如下所示：
r = requests.get('<MY_URI>', headers={'Authorization': 'TOK:<MY_TOKEN>'})
回答by BajajG
I was looking for something similar and came across this. It looks like in the first option you mentioned
我正在寻找类似的东西并遇到了这个。看起来像你提到的第一个选项
r = requests.get('<MY_URI>', auth=('<MY_TOKEN>'))
"auth" takes two parameters: username and password, so the actual statement should be
“auth”有两个参数：用户名和密码，所以实际语句应该是
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:
就我而言，没有密码，所以我将 auth 字段中的第二个参数留空，如下所示：
r=requests.get('<MY_URI', auth=('MY_USERNAME', ''))
Hope this helps somebody :)
希望这对某人有所帮助:)
回答by swatisinghi
You can try something like this
你可以试试这样的
r = requests.get(ENDPOINT, params=params, headers={'Authorization': 'Basic %s' % API_KEY})
回答by Anton Tarasenko
You can also set headers for the entire session:
您还可以为整个会话设置标题：
TOKEN = 'abcd0123'
HEADERS = {'Authorization': 'token {}'.format(TOKEN)}
with requests.Session() as s:
s.headers.update(HEADERS)
resp = s.get('http://example.com/')
回答by Edgar N
This worked for me:
这对我有用：
access_token = #yourAccessTokenHere#
result = requests.post(url,
headers={'Content-Type':'application/json',
'Authorization': 'Bearer {}'.format(access_token)})
回答by Amit Blum
Requests natively supports basic auth only with user-pass params, not with tokens.
请求本身仅支持用户传递参数的基本身份验证，而不支持令牌。
You could, if you wanted, add the following class to have requests support token based basic authentication:
如果需要，您可以添加以下类以使请求支持基于令牌的基本身份验证：
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 :
然后，要使用它运行以下请求：
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.
另一种方法是制定自定义标题，正如其他用户在这里建议的那样。
回答by Tri Tran
i founded here, its ok with me for linkedin: https://auth0.com/docs/flows/guides/auth-code/call-api-auth-codeso my code with with linkedin login here:
我在这里成立，它可以与我一起使用linkedin：https: //auth0.com/docs/flows/guides/auth-code/call-api-auth-code所以我的代码与linkedin登录：
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()
回答by Reymond Joseph
This worked for me:
这对我有用：
r = requests.get('http://127.0.0.1:8000/api/ray/musics/', headers={'Authorization': 'Token 22ec0cc4207ebead1f51dea06ff149342082b190'})
My code uses user generated token.
我的代码使用用户生成的令牌。
