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Unable to deserialize PyMongo ObjectId from JSON

I'm seemingly unable to deserialize my MongoDB JSON document with the BSON json_util.

The json.loads function is choking on the ObjectId() string. I had understood json_util capable of handling MongoDB's ObjectId format and transforming into usable JSON.

Python code:

import json    
from bson import json_util

s = "{u'_id': ObjectId('4ed559abf047050c58000000')}"
u = json.loads(s, object_hook=json_util.object_hook)

I get the decoder exception:

...
    u = json.loads(s, object_hook=json_util.object_hook)
  File "\python27\lib\json\__init__.py", line 339, in loads
    return cls(encoding=encoding, **kw).decode(s)
  File "\python27\lib\json\decoder.py", line 366, in decode
    obj, end = self.raw_decode(s, idx=_w(s, 0).end())
  File "\python27\lib\json\decoder.py", line 382, in raw_decode
    obj, end = self.scan_once(s, idx)
ValueError: Expecting property name: line 1 column 1 (char 1)

Am I missing something?

like image 616
jdev Avatar asked Dec 07 '11 01:12

jdev


2 Answers

I think your string form actually looks like the python representation...

s = '{"_id": {"$oid": "4edebd262ae5e93b41000000"}}'
u = json.loads(s, object_hook=json_util.object_hook)

print u  # Result:  {u'_id': ObjectId('4edebd262ae5e93b41000000')}

s = json.dumps(u, default=json_util.default)

print s  # Result:  {"_id": {"$oid": "4edebd262ae5e93b41000000"}}

The bson.json_util.object_hook function does not seem to have any type of handling for there being ObjectId() in the actual json string representation.

like image 168
jdi Avatar answered Nov 19 '22 23:11

jdi


There are two problems here:

  1. The string you're attempting to JSON-decode is not JSON, it's the string representation of a Python dictionary. In particular, the problem is that u'_id' is not a valid JSON key (JSON keys are quoted strings; the "u" here indicates a Python unicode string, which is meaningless in JSON)

  2. json_util.object_hook doesn't make ObjectId available to JSON; the json module will decode the JSON, and then call the object_hook callback with each decoded object. json_util.object_hook will look for certain patterns as defined in the strict mode of MongoDB Extended JSON.

See @jdi's answer for examples of how to properly use json_util.

like image 4
dcrosta Avatar answered Nov 19 '22 22:11

dcrosta