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Python请求库如何通过单个令牌传递Authorization标头

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问题：Python请求库如何通过单个令牌传递Authorization标头
回答 0
回答 1
回答 2
回答 3
回答 4
回答 5
回答 6
回答 7

问题：Python请求库如何通过单个令牌传递Authorization标头

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>"

等等，我得到200并查看相应的JSON数据。因此，我安装了请求，并且当我尝试访问该资源时，我得到了403，这可能是因为我不知道传递该令牌的正确语法。谁能帮我解决这个问题？这就是我所拥有的：

import sys,socket
import requests

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

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>'))

但是这些都不起作用。

点击查看英文原文

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.

回答 0

在python中：

('<MY_TOKEN>')

相当于

'<MY_TOKEN>'

并要求翻译

('TOK', '<MY_TOKEN>')

当您希望请求使用基本身份验证并设计一个授权标头时，如下所示：

'VE9LOjxNWV9UT0tFTj4K'

这是base64的表示形式 'TOK:<MY_TOKEN>'

要传递自己的标头，您需要像这样传递字典：

r = requests.get('<MY_URI>', headers={'Authorization': 'TOK:<MY_TOKEN>'})
点击查看英文原文

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>'})

回答 1

我一直在寻找类似的东西，并且遇到了这个问题。看来您提到的第一个选项

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

“ auth”具有两个参数：用户名和密码，因此实际语句应为

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

在我的情况下，没有密码，因此我将auth字段中的第二个参数留空，如下所示：

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

希望这对某人有帮助:)

点击查看英文原文

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 :)

回答 2

这为我工作：

access_token = #yourAccessTokenHere#

result = requests.post(url,
      headers={'Content-Type':'application/json',
               'Authorization': 'Bearer {}'.format(access_token)})
点击查看英文原文

This worked for me:

access_token = #yourAccessTokenHere#

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

回答 3

您还可以设置整个会话的标题：

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

with requests.Session() as s:

    s.headers.update(HEADERS)
    resp = s.get('http://example.com/')
点击查看英文原文

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/')

回答 4

请求本身仅通过用户传递参数而不是令牌支持基本身份验证。

如果需要，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

然后，要使用它，请运行以下请求：

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

一种替代方法是改为编写自定义标头，如此处其他用户所建议的那样。

点击查看英文原文

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.

回答 5

我在这里建立了，我可以 在这里登录：https : //auth0.com/docs/flows/guides/auth-code/call-api-auth-code，所以我在linkedin上的代码在这里登录：

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()
点击查看英文原文

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()

回答 6

您可以尝试这样的事情

r = requests.get(ENDPOINT, params=params, headers={'Authorization': 'Basic %s' %  API_KEY})
点击查看英文原文

You can try something like this

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

回答 7

这为我工作： 

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

我的代码使用用户生成的令牌。

点击查看英文原文

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.