JSONDecodeError: Expecting value: line 1 column 1 (char 0)

Be sure to remember to invoke json.loads() on the contents of the file, as opposed to the file path of that JSON:

json_file_path = "/path/to/example.json"

with open(json_file_path, 'r') as j:
     contents = json.loads(j.read())

I think a lot of people are guilty of doing this every once in a while (myself included):

contents = json.loads(json_file_path)

Your code produced an empty response body, you'd want to check for that or catch the exception raised. It is possible the server responded with a 204 No Content response, or a non-200-range status code was returned (404 Not Found, etc.). Check for this.

Note:

• There is no need to use simplejson library, the same library is included with Python as the json module.

• There is no need to decode a response from UTF8 to unicode, the simplejson / json .loads() method can handle UTF8 encoded data natively.

• pycurl has a very archaic API. Unless you have a specific requirement for using it, there are better choices.

Either the requests or httpx offers much friendlier APIs, including JSON support. If you can, replace your call with:

import requests

response = requests.get(url)
response.raise_for_status()  # raises exception when not a 2xx response
if response.status_code != 204:
    return response.json()

Of course, this won't protect you from a URL that doesn't comply with HTTP standards; when using arbirary URLs where this is a possibility, check if the server intended to give you JSON by checking the Content-Type header, and for good measure catch the exception:

if (
    response.status_code != 204 and
    response.headers["content-type"].strip().startswith("application/json")
):
    try:
        return response.json()
    except ValueError:
        # decide how to handle a server that's misbehaving to this extent