views:

61

answers:

3

I have a PyMongo newbie question: If collection is a PyMongo Collection and I use it to save an object with

obj = {'foo': 'bar'}
collection.insert(obj)

then MongoDB automatically generates an _id field for obj; once can confirm this with

print obj

which yields something like

{'foo': 'bar', '_id': ObjectId('4c2fea1d289c7d837e000000')}

My question is: How do I get that _id back out in such a way that I can use it?

For instance, if I want to delete obj from the database, I would think that I would want to do something like

collection.remove(obj['_id'])

but when I try this I get the message

TypeError: 'ObjectId' object is unsubscriptable.

What's going on?

A: 

You can just pass obj.

Matthew Flaschen
+2  A: 

You just need to pass remove a dict, just like you did insert. So, to remove a document based on its _id value, do something like:

collection.remove({'_id': ObjectId('4c2fea1d289c7d837e000000')})
kristina
Ah, I see. What is the purpose of the ObjectId wrapper, anyway?
Trevor Burnham
It's a special type. It's more compact than a string and contains a timestamp for when it was created, which you can access with generation_time (http://api.mongodb.org/python/1.7%2B/api/pymongo/objectid.html)
kristina
`ObjectId` can be imported with `from pymongo.objectid import ObjectId` for those looking for it.
T. Stone
+1  A: 

insert returns the _id of the inserted document.

and remove will remove based on _id, so try something like:

doc_id = db.test.insert({"foo": 1})
db.test.remove(doc_id)
mdirolf