Unable to deserialize PyMongo ObjectId from JSON

Question

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?

Accepted Answer

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.

Unable to deserialize PyMongo ObjectId from JSON

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