获取错误JSONDecodeError:从API导入时应为“，”分隔符

Question

我正试图解析来自API的响应数据，但一直出现此错误

这是我的代码：

import requests
import json

url = "https://ratings.food.gov.uk/OpenDataFiles/FHRS776en-GB.json"

r = requests.get(url)
json_data = r.json()

这就是错误所在

File "/usr/local/lib/python3.8/site-packages/requests/models.py", line 889, in json
   return complexjson.loads(
 File "/usr/local/Cellar/python@3.8/3.8.5/Frameworks/Python.framework/Versions/3.8/lib/python3.8/json/__init__.py", line 357, in loads
   return _default_decoder.decode(s)
 File "/usr/local/Cellar/python@3.8/3.8.5/Frameworks/Python.framework/Versions/3.8/lib/python3.8/json/decoder.py", line 337, in decode
   obj, end = self.raw_decode(s, idx=_w(s, 0).end())
 File "/usr/local/Cellar/python@3.8/3.8.5/Frameworks/Python.framework/Versions/3.8/lib/python3.8/json/decoder.py", line 353, in raw_decode
   obj, end = self.scan_once(s, idx)
json.decoder.JSONDecodeError: Expecting ',' delimiter: line 1 column 233384 (char 233383)

我已经确认这是一个有效的JSON，这是一个公共API，所以我无法控制格式。我怎样才能克服这个错误

Answer 1

问题不在于代码，而在于json

您可以验证jsonhere。第7669行和第7670行各缺少一个,

Answer 2

你可以用

import requests
import json

url = "https://ratings.food.gov.uk/OpenDataFiles/FHRS776en-GB.json"

r = requests.get(url)
json_data = json.loads(r.text)