问题
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'}
回答1:
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#
回答2:
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"}
}
回答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 : { $exists : true } } );
回答4:
Match more simple:
db['dvd'].find_one( {'episodes.title': "Episode 1"},{'episodes.title': true} )
Query:
coll.find( criteria, fields );
Get just specific fields from the object. E.g.:
coll.find( {}, {name:true} );
http://www.mongodb.org/display/DOCS/dbshell+Reference
回答5:
Look at db.eval:
You should do something like:
episode = connection['dvds'].eval('function(title){
var t = db.dvds.findOne({"episodes.title" : title},{episodes:true});
if (!t) return null;
for (var i in t.episodes) if (t.episodes[i].title == title) return t.episodes[i];
}', "Episode 1");
so filtering of episodes will be on a server-side.
回答6:
I also met the same problem. The way I do is use aggregate function. First unwind it and then match it.
db.dvds.aggregate([{$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.
来源:https://stackoverflow.com/questions/2268568/mongodb-query-to-return-only-embedded-document