Remove a PyMongo SON Manipulator

Question

For some queries in my application, I am using a SON manipulator to conveniently render database output. In some other cases, I just want the raw data.

So I add my manipulator thus:

db.add_son_manipulator(Renderer())

And here's my Renderer:

class Renderer(pymongo.son_manipulator.SONManipulator):
    def transform_outgoing(self, son, collection):
        rendered_data = {}

        for field in son:
            try:
                rendered_data[field] = son[field]['value']
            except (KeyError, TypeError):
                rendered_data[field] = son[field]

        return rendered_data

Question: How do I remove the manipulator from the database class instance for cases when I don't want my data transformed?

I don't understand your question: what do you mean remove manipulator from database? — Rafael Barros, Sep 15 '14 at 20:59
@RafaelBarros I mean the database class instance. Updated the question. — sssilver, Sep 15 '14 at 22:47

Serge · Answer 1 · 2015-01-22T08:44:49.037

0

EDITED

Seems it is impossible to do in a normal way (no API method exists for that purpose). Instead I should have another database object with no manipulator and use it when I don't need to transform the data (probably would work even faster).

edited Jan 22 '15 at 08:44

answered Jan 21 '15 at 12:03

Serge

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@sssilver ok, so I edited answer to remove it at all. – Serge Jan 22 '15 at 08:46

Remove a PyMongo SON Manipulator

1 Answers1

EDITED