Python：请求有效的 urllib3 错误

Question

我有以下代码

http = urllib3.PoolManager()
resp = http.request(
    "GET",
    "https://api.nhs.uk/medicines/aciclovir",
    headers={
        "subscription-key":"XXX"
    }
)
print(resp.data)

回应是：

You don't have permission to access "http://api.nhs.uk/medicines/aciclovir" on this server

为什么它去了http://地址？

但是，当我使用如下请求时，它是成功的

url = "https://api.nhs.uk/medicines/aciclovir"

payload={}
headers = {
  'subscription-key': 'XXX'
}

response = requests.request("GET", url, headers=headers, data=payload)

print(response.text)

有什么建议么？

更新 1

这是为发生错误而执行的确切代码，是的，它确实显示了http://错误

更新 2

urllib3

Listening on 0.0.0.0 7777
Connection received on *** 50054
GET / HTTP/1.1
Host: ***:7777
Accept-Encoding: identity
subscription-key: xxx
User-Agent: python-urllib3/1.26.9

要求

Listening on 0.0.0.0 7777
Connection received on *** 50095
GET / HTTP/1.1
Host: ***:7777
User-Agent: python-requests/2.28.0
Accept-Encoding: gzip, deflate
Accept: */*
Connection: keep-alive
subscription-key: xxx

Answer 1

感谢https://stackoverflow.com/users/240443/amadan ，问题是服务器不喜欢 urllib3 提供的默认用户代理字符串

更改它甚至只是分解python-urllib字符串（例如pyXthon-urllib ）使其工作

我在他们的支持下接受了这个