Question

我正在尝试从thetvdb.com（https://api.thetvdb.com）访问API v2。不幸的是我总是得到403错误。

这就是我所拥有的：

#!/usr/bin/python3
import requests

url = "https://api.thetvdb.com/login"
headers = {'content-type': 'application/json'}
payload = {"apikey":"123","username":"secretusername","userkey":"123"}
post = requests.post(url, data = payload, headers = headers)
print(post.status_code, post.reason)

根据API文档，我必须进行身份验证才能获得令牌。但我得到了403 Forbidden。

现在我尝试使用curl：

curl -X POST --header 'Content-Type: application/json' --header 'Accept: 
application/json' -d  
{"apikey":"123","username":"secretusername","userkey":"123"}' 
'https://api.thetvdb.com/login'

这完美无缺。任何人都可以解释我错过了什么吗？这让我疯了。

我也尝试了

post = requests.post(url, data = json.dumps(payload), headers = headers)

同样的错误。

Answer 1

您必须explicitly将payload转换为json字符串并传递为data。看起来您已经这样做了，您也可以尝试将用户代理设置为curl/7.47.1

headers = {'content-type': 'application/json', 'User-Agent': 'curl/7.47.1'}
post = requests.post(url, data = json.dumps(payload), headers = headers)

程序看起来像

#!/usr/bin/python3
import requests
import json    

url = "https://api.thetvdb.com/login"
headers = {'content-type': 'application/json', 'User-Agent': 'curl/7.47.1'}
payload = {"apikey":"123","username":"secretusername","userkey":"123"}
post = requests.post(url, data = json.dumps(payload), headers = headers)
print(post.status_code, post.reason)

Answer 2

我认为您需要在python请求中传递Accept标头。像这样的东西 header = { 'Accept' : 'application/json', 'Content-Type' : 'application/json' "Accept-Encoding": "gzip, deflate, sdch, br", "Accept-Language": "en-US,en;q=0.8", "User-Agent": "some user-agent", }

Python请求。错误403

2 个答案: