python - urllib.request.Request() 总是以 HTTP 错误 400 问题结束

Question

当使用 urllib 中的 Request 对象来获取 URL 时，我总是以 HTTP 错误 400：错误请求结束。

我想创建一个 watch2gether 房间，您可以在其中发布例如 YouTube URL，它会自动创建一个 watch2gether 房间，以便您可以共享视频。如果我尝试这样做：

import json
import urllib.request, urllib.parse
import urllib.response

wurl = "https://w2g.tv/rooms/create.json"

headers = {'Accept': 'application/json',
           'Content-Type': 'application/json'}

items = []
 
body_create = {"w2g_api_key": "TOKEN", "share": " ","bg_color": "#00ff00", "bg_opacity": "50" }

body_update = { "w2g_api_key": "TOKEN","item_url" : " " }

class watch2gether():
    def createroom(self, url : str): 
        body_create['share'] = url
        data = urllib.parse.urlencode(body_create).encode("utf-8")
        req = urllib.request.Request(wurl, data, headers)
        with urllib.request.urlopen(req) as response: 
            source = response.read()
            dict = json.loads(source)
            items.append(dict['streamkey'])
        return items[0]   

    def updateroom(self, url: str): 
            input_url = f"https://w2g.tv/rooms/{items[0]}/sync_update"
            body_update['item_url'] = url
            data = urllib.parse.urlencode(body_update).encode("utf-8")
            urllib.request.urlopen(input_url, data)

s1 = watch2gether()

user_url = str(input("post url to create room: "))
print("hey here is your link https://w2g.tv/rooms/" + s1.createroom(user_url))

new_url = str(input("post link to update room: " ))
s1.updateroom(new_url)

我得到这个回应：

post url to create room: https://www.youtube.com/watch?v=8RJKBNN4dTw
Traceback (most recent call last):
  File "/Users/jannitselekoglou/Desktop/Discordbot/w2g.py", line 47, in <module>
    print("hey here is your link https://w2g.tv/rooms/" + s1.createroom(user_url))
  File "/Users/jannitselekoglou/Desktop/Discordbot/w2g.py", line 27, in createroom
    with urllib.request.urlopen(req, data) as response: 
  File "/Users/jannitselekoglou/miniconda3/lib/python3.9/urllib/request.py", line 214, in urlopen
    return opener.open(url, data, timeout)
  File "/Users/jannitselekoglou/miniconda3/lib/python3.9/urllib/request.py", line 523, in open
    response = meth(req, response)
  File "/Users/jannitselekoglou/miniconda3/lib/python3.9/urllib/request.py", line 632, in http_response
    response = self.parent.error(
  File "/Users/jannitselekoglou/miniconda3/lib/python3.9/urllib/request.py", line 561, in error
    return self._call_chain(*args)
  File "/Users/jannitselekoglou/miniconda3/lib/python3.9/urllib/request.py", line 494, in _call_chain
    result = func(*args)
  File "/Users/jannitselekoglou/miniconda3/lib/python3.9/urllib/request.py", line 641, in http_error_default
    raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 400: Bad Request

但是，如果我只使用urlopen()方法而不传入标头，它就可以完美地工作。（您可以在此处阅读 Watch2Gether API 文档，非常简短）

我真的不明白我在使用urllib.request.Request()时做错了什么，以及为什么我不需要传入标头，因为 API 文档明确指出你必须这样做“确保设置您对 application/json 的请求的内容类型”。

标签： python