Can't read JSON file in Python

Question

I tried to read data from a JSON file, but I encountered weird error and have no idea what it means. I tried googling it, but it didn't help. I got the following error:

Traceback (most recent call last):
  File "items_uploader.py", line 40, in <module>
    main()
  File "items_uploader.py", line 16, in main
    LoadItemsData(settings['items_filename'])
  File "items_uploader.py", line 36, in LoadItemsData
    data = json.load(json_data)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/json/__init__.py", line 278, in load
    **kw)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/json/__init__.py", line 326, in loads
    return _default_decoder.decode(s)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/json/decoder.py", line 366, in decode
    obj, end = self.raw_decode(s, idx=_w(s, 0).end())
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/json/decoder.py", line 382, in raw_decode
    obj, end = self.scan_once(s, idx)
ValueError: Expecting , delimiter: line 8 column 397 (char 3064)

The code itself is quite simple:

import socket
import MySQLdb
from ConfigParser import SafeConfigParser
import json
from pprint import pprint

def main():
    settings = GetSettings()
    LoadItemsData(settings['items_filename'])
    return

def GetSettings():
    settings = {}
    parser = SafeConfigParser()
    parser.read('settings.yaml')
    settings['items_filename'] =  parser.get('files', 'items_filename')
    return settings

def LoadItemsData(filename):
    json_data=open(filename)
    data = json.load(json_data)
    return data

if __name__ == '__main__':
    main()

Any help would be appreciated!

Are you sure your JSON data is valid? The error appears to be a syntax error in your input data. — Jesse the Game
– Jesse the Game, Commented Dec 3, 2012 at 0:05
Please include the portion of the file around where the JSON error occurs (line 8 column 397, char 3064). — Andrew Clark
– Andrew Clark, Commented Dec 3, 2012 at 0:05
Basically none of the python code you posted (other than the traceback) is relevant here, as the only important thing is the contents of settings['items_filename'] file. — jdi
– jdi, Commented Dec 3, 2012 at 0:14
Completely unrelated, but using CamelCase in function names in Python is frowned upon, this style is reserved for classes. python.org/dev/peps/pep-0008/#naming-conventions — Sergey
– Sergey, Commented Dec 3, 2012 at 0:48

user1786283 · Accepted Answer · 2012-12-03 03:01:02Z

3

Make sure your JSON data is in a valid format, one extra character will mess up the python parser. To test your JSON data go here, make sure you can see it in a correct format.

For example, if I had

JSON_data ='{"c":[{"xy":{"xstart":0,"xend":5,"ystart":1,"yend":5},"names":["D","T","O","H","L","C",],"co":["rgb(0,0,128)"]}],"Values":{"D":["11/30/2012"],"T":["09:44:00"],"O":["5848.40"],"H":["5848.40"],"L":["5847.45"],"C":["5848.40"]}}'

The , after C (here ["D","T","O","H","L","C",]) will show an error. So make sure that your data is in correct format and there are no unnecessary characters.

Hope this helps.

answered Dec 3, 2012 at 3:01

user1786283

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1 Comment

pisarzp Over a year ago

Thanks, it turns out it was a bug in JSON file!

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