Home/ Questions/The complete guide to json.decoder.jsondecodeerror: expecting value: line 1 column 1 (char 0)
Next
Answered
Zoé Tran
  • 7
Asked: 2022-05-11T19:02:54+00:00 In:

The complete guide to json.decoder.jsondecodeerror: expecting value: line 1 column 1 (char 0)

  • 7

. Advertisement .

..3..

. Advertisement .

..4..

I am new to programs and searching the json.decoder.jsondecode error: expecting value: line 1 column 1 (char 0) to understand it better. It seems it doesn’t work as expected when I used some suggestions before. This is the command line I use:

 jsn_res = self.getJson(inputUrl)
 jsn_resp  = jsn_res.decode('utf-8')
 return json.loads(jsn_resp)
 
 
 def getJson(self, inputUrl):
  pyCurl = pycurl.Curl()
  IObuff = StringIO()
  curl.setopt(pyCurl.WRITEFUNCTION, IObuff.write)
  curl.setopt(pyCurl.TIMEOUT, self.timeout)
  pyCurl.setopt(pyCurl.URL, inputUrl)
  pyCurl.perform()
  pyCurl.close()
  outResp = IObuff.getvalue().strip()
  return outResp

The error I’m getting is below:

 Exception Type: JSONDecodeError at /pricemodels/2/dir/
 Exception Value: Expecting value: line 1 column 1 (char 0)

Please give me the solution for this issue.

json.decoder.jsondecodeerror
Share

3 Answers

  1. Best Answer

    Two things i might recommend you to do better:

    1. Handle your exceptions i.e.: using try catch
    2. Using ‘requests’ rather than ‘pycurl’ in order to better handle your requests/responses/exceptions, in your case handle json response as well. Below is an example for you
    import requests

res = requests.get(inputUrl)
res.raise_for_status() 
if (
res.status_code != 204 and
res.headers["content-type"].strip().startswith("application/json")
):
try:
return res.json()
except ValueError:
    • 19

  2. Remember to invoke json.loads() on content of the file and not the path in that JSON. 

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

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

    This is something I believe many people do every now and again (me included). 

    contents = json.loads(json_file_path)
    • 13
  3. Expert

    You can check if your code returned an empty response body or catch the exception. You may have received a 204 no content response or a non 200-range status code. This is what you should check.

    Note:

    • You don’t need the simplejson library. The same library is available with Python as json module.
    • It is not necessary to decode the response from UTF8 into unicode. The simplejson/json .loads() method handles UTF8 encoded data natively.
    • pycurl‘s API is very primitive. There are better alternatives if you don’t have a particular requirement.

    Either the or offer much friendlier APIs including JSON support. Replace your call with:

    import requests

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

    This won’t protect against URLs that don’t conform to HTTP standards. If you use arbirary URLs, make sure you check the Content-Type header to see if the server meant to provide JSON. And, for good measure, catch the exception.

    if (
    resp.status_code != 204 and
    resp.headers["content-type"].strip().startswith("application/json")
):
    try:
        return resp.json()
    except ValueError:
        # decide how to handle a server that's misbehaving to this extent
    • 0

Leave an answer

You must login to add an answer.

Need An Account,