Python - SimpleJSON Issue

Question

I'm working with the Mega API and Python in hope to produce a folder tree readable by Python. At the moment I'm working with the JSON responses Mega's API gives, but for some reason am having trouble parsing it. In the past I would simply use simplejson in the format below, though right now it's not working. At the moment I'm just trying to get the file name. Any help is appreciated!

import simplejson    

megaResponseToFileSearch = "(u'BMExefXbYa', {u'a': {u'n': u'A Bullet For Pretty Boy - 1 - Dial M For Murder.mp3'}, u'h': u'BMExXbYa', u'k': (5710166, 21957970, 11015946, 7749654L), u'ts': 13736999, 'iv': (7949460, 15946811, 0, 0), u'p': u'4FlnwBTb', u's': 5236864, 'meta_mac': (529642, 2979591L), u'u': u'xpz_tb-YDUg', u't': 0, 'key': (223xx15874, 642xx8505, 1571620, 26489769L, 799460, 1596811, 559642, 279591L)})"

jsonRespone = simplejson.loads(megaResponseToFileSearch)

print jsonRespone[u'a'][u'n']

ERROR:

Traceback (most recent call last):
  File "D:/Projects/Mega Sync/megasync.py", line 18, in <module>
    jsonRespone = simplejson.loads(file4)
  File "D:\Projects\Mega Sync\simplejson\__init__.py", line 453, in loads
    return _default_decoder.decode(s)
  File "D:\Projects\Mega Sync\simplejson\decoder.py", line 429, in decode
    obj, end = self.raw_decode(s)
  File "D:\Projects\Mega Sync\simplejson\decoder.py", line 451, in raw_decode
    raise JSONDecodeError("No JSON object could be decoded", s, idx)
simplejson.decoder.JSONDecodeError: No JSON object could be decoded: line 1 column 0 (char 0)

EDIT:

I was asked where I got the string from. It's a response to searching for a file using the Mega API. I'm using the module found here. https://github.com/richardasaurus/mega.py The code itself looks like this:

from mega import Mega
mega = Mega({'verbose': True})
m = mega.login(email, password)
file = m.find('A Bullet For Pretty Boy - 1 - Dial M For Murder.mp3')
print file

That doesn't look like a valid json string. Where did you get it (post the code if possible)? — alecxe
– alecxe, Commented Jul 10, 2013 at 18:41
Updated question with code. Was it too quick of me to assume it was JSON it was spitting out? — Dustin
– Dustin, Commented Jul 10, 2013 at 18:45

alecxe · Accepted Answer · 2013-07-10 19:01:46Z

1

The thing you are getting from m.find is just a python tuple, where the 1-st (next after the 0th) element is a dictionary:

(u'99M1Tazb',
 {u'a': {u'n': u'test.txt'}, 
  u'h': u'99M1Tazb', 
  u'k': (1145485578, 1435138417, 702505527, 274874292), 
  u'ts': 1373482712,
  'iv': (1883603069, 763415510, 0, 0), 
  u'p': u'9td12YaY', 
  u's': 0, 
  'meta_mac': (1091379956, 402442960),
  u'u': u'79_166PAQCA', 
  u't': 0,
  'key': (872626551, 2013967015, 1758609603, 127858020, 1883603069, 763415510, 1091379956, 402442960)})

To get the filename, just use:

print file[1]['a']['n']

So, no need to use simplejson at all.

answered Jul 10, 2013 at 19:01

alecxe

476k127 gold badges1.1k silver badges1.2k bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Dustin Over a year ago

AH! That's why I got mixed up. Thank you very much!

Collectives™ on Stack Overflow

Python - SimpleJSON Issue

1 Answer 1

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related