Unable to deserialize PyMongo ObjectId from JSON

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小蘑菇
小蘑菇 2020-12-05 20:34

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 h

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  • 2020-12-05 21:17

    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.

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  • 2020-12-05 21:20

    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.

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