MongoDB query to return only embedded document

Question

assume that i have a BlogPost model with zero-to-many embedded Comment documents. can i query for and have MongoDB return only Comment objects matching my query spec?

eg, db.blog_posts.find({"comment.submitter": "some_name"}) returns only a list of comments.

edit: an example:

import pymongo

connection = pymongo.Connection()
db = connection['dvds']

db['dvds'].insert({'title': "The Hitchhikers Guide to the Galaxy",
                   'episodes': [{'title': "Episode 1", 'desc': "..."},
                                {'title': "Episode 2", 'desc': "..."},
                                {'title': "Episode 3", 'desc': "..."},
                                {'title': "Episode 4", 'desc': "..."},
                                {'title': "Episode 5", 'desc': "..."},
                                {'title': "Episode 6", 'desc': "..."}]})

episode = db['dvds'].find_one({'episodes.title': "Episode 1"}, 
                              fields=['episodes'])

in this example, episode is:

{u'_id': ObjectId('...'),
 u'episodes': [{u'desc': u'...', u'title': u'Episode 1'},
               {u'desc': u'...', u'title': u'Episode 2'},
               {u'desc': u'...', u'title': u'Episode 3'},
               {u'desc': u'...', u'title': u'Episode 4'},
               {u'desc': u'...', u'title': u'Episode 5'},
               {u'desc': u'...', u'title': u'Episode 6'}]}

but i just want:

{u'desc': u'...', u'title': u'Episode 1'}

"The future" of the accepted answer is now the past, as aggregation framework already exists. I answered how to do it. — H.D.
– H.D., Commented Aug 31, 2013 at 12:27

H.D. · Accepted Answer · 2013-03-28 11:25:44Z

I think what you wanted is this:

print db.dvds.aggregate([
  {"$unwind": "$episodes"}, # One document per episode
  {"$match": {"episodes.title": "Episode 1"} }, # Selects (filters)
  {"$group": {"_id": "$_id", # Put documents together again
              "episodes": {"$push": "$episodes"},
              "title": {"$first": "$title"} # Just take any title
             }
  },
])["result"]

The output (besides the whitespaces) is:

[ { u'episodes': [ { u'title': u'Episode 1',
                     u'desc': u'...'
                   }
                 ],
    u'_id': ObjectId('51542645a0c6dc4da77a65b6'),
    u'title': u'The Hitchhikers Guide to the Galaxy'
  }
]

If you want to get rid from the u"_id", append the pipeline with:

  {"$project": {"_id": 0,
                "episodes": "$episodes",
                "title": "$title"}
               }

Stephen Curran · Accepted Answer · 2010-02-15 22:34:20Z

6

This same question was asked over on the Mongo DB Google Groups page. Apparently its not currently possible but it is planned for the future.

http://groups.google.com/group/mongodb-user/browse_thread/thread/4e6f5a0bac1abccc#

answered Feb 15, 2010 at 22:34

Stephen Curran

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Comments

Alex Voikov · Accepted Answer · 2014-08-01 08:32:28Z

3

The mongodb javascript shell is documented at http://docs.mongodb.org/manual/reference/method/

If you want to get back only specific fields of an object, you can use

db.collection.find( { }, {fieldName:true});

If, on the other hand, you are looking for objects which contain a specific field, you can sue

db.collection.find( { fieldName : { $exists : true } } );

edited Aug 1, 2014 at 8:32

Alex Voikov

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answered Feb 15, 2010 at 20:14

Dominik

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1 Comment

Carson Over a year ago

i've tried to clarify my question with an example in the original post. thanks.

franzlorenzon · Accepted Answer · 2013-02-27 09:57:51Z

1

Match more simple:

db['dvd'].find_one( {'episodes.title': "Episode 1"},{'episodes.title': true} )

Query:

coll.find( criteria, fields );

Get just specific fields from the object. E.g.:

coll.find( {}, {name:true} );

http://www.mongodb.org/display/DOCS/dbshell+Reference

edited Feb 27, 2013 at 9:57

franzlorenzon

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answered May 10, 2010 at 4:11

Derlok

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Comments

pirho · Accepted Answer · 2017-11-20 10:56:42Z

1

I also met the same problem. The way I do is use aggregate function. First unwind it and then match it.

db.dvds.aggregate([{$unwind:"$episodes"},{$match:{"episodes.title":"Episode 1"}}])

The results will be like

{ "_id" : ObjectId("5a129c9e6944555b122c8511"),
   "title" : "The Hitchhikers Guide to the Galaxy",
   "episodes" : { "title" : "Episode 1", "desc" : "..." } }

It's not perfect, but then you can edit it by python.

edited Nov 20, 2017 at 10:56

pirho

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answered Nov 20, 2017 at 10:31

Rachel.yuq

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Comments

poiuyttr · Accepted Answer · 2010-02-15 22:59:39Z

0

Look at db.eval:

You should do something like:

episode = connection['dvds'].eval('function(title){
  var t = db.dvds.findOne({"episodes.title" : title},{episodes:true});
  if (!t) return null;
  for (var i in t.episodes) if (t.episodes[i].title == title) return t.episodes[i];
}', "Episode 1");

so filtering of episodes will be on a server-side.

answered Feb 15, 2010 at 22:59

poiuyttr

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