Python无法通过请求发送GET请求

Question

所以我编码已经很久了。最近重新回到这里，这就是我想要做的事情。我从Google电子表格中获取VINS，然后使用Requests库向URL发送GET请求，然后URL解码该VIN并返回数据（https://vpic.nhtsa.dot.gov/api/）。

这是我到目前为止所做的，我可以从电子表格中获取VIN并显示它，但是当我尝试发送GET请求时，Python会为我返回大量错误。这是我第一次使用Python编写代码，花了一些时间将代码放到现在的位置。

def main():
store = file.Storage('token.json')
creds = store.get()
if not creds or creds.invalid:
    flow = client.flow_from_clientsecrets('credentials.json', SCOPES)
    creds = tools.run_flow(flow, store)
service = build('sheets', 'v4', http=creds.authorize(Http()))

# Call the Sheets API
SPREADSHEET_ID = 'The Sheets ID that I use'
RANGE_NAME = 'A:A'
result = service.spreadsheets().values().get(spreadsheetId=SPREADSHEET_ID,
                                            range=RANGE_NAME).execute()
values = result.get('values', [])

if not values:
    print('No data found.')
else:

    for row in values:
        # Print columns A, which correspond to indices 0.
        print('%s' % (row[0]))
        URL = "https://vpic.nhtsa.dot.gov/api/vehicles/DecodeVin/"
        VIN=row[0]
        PARAMS={VIN}
        r=requests.get(url=URL, params=PARAMS)
        data=r.json
        r.status_code
        text=r.text
        print(text)

if __name__ == '__main__':
    main()

我得到的错误如下，我确实得到了VIN（这里没有提到）：

 Traceback (most recent call last):
  File "quickstart.py", line 47, in <module>
    main()
  File "quickstart.py", line 40, in main
    r=requests.get(url=URL, params=PARAMS)
  File "/Library/Python/2.7/site-packages/requests/api.py", line 72, in get
    return request('get', url, params=params, **kwargs)
  File "/Library/Python/2.7/site-packages/requests/api.py", line 58, in request
    return session.request(method=method, url=url, **kwargs)
  File "/Library/Python/2.7/site-packages/requests/sessions.py", line 498, in request
    prep = self.prepare_request(req)
  File "/Library/Python/2.7/site-packages/requests/sessions.py", line 441, in prepare_request
    hooks=merge_hooks(request.hooks, self.hooks),
  File "/Library/Python/2.7/site-packages/requests/models.py", line 309, in prepare
    self.prepare_url(url, params)
  File "/Library/Python/2.7/site-packages/requests/models.py", line 427, in prepare_url
    enc_params = self._encode_params(params)
  File "/Library/Python/2.7/site-packages/requests/models.py", line 97, in _encode_params
    for k, vs in to_key_val_list(data):
ValueError: too many values to unpack

我怎样才能确定问题所在？我在很长一段时间后编写代码并且对Python完全陌生。

Answer 1

从他们的文档来看，你想要的

VIN=row[0]
r=requests.get(url=URL + VIN, params={'format': 'json'})

代替