假设我有一个BlogPost包含零到多嵌入式Comment文档的模型.我可以查询并让MongoDB 只 返回Comment与我的查询规范匹配的对象吗?
例如,db.blog_posts.find({"comment.submitter": "some_name"})仅返回评论列表.
编辑:一个例子:
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'])
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在这个例子中,episode是:
{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'}]}
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但我只是想:
{u'desc': u'...', u'title': u'Episode 1'}
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我想你想要的是:
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": "$title"} # Just take any title
}
},
])["result"]
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输出(除了空格)是:
[ { u'episodes': [ { u'title': u'Episode 1',
u'desc': u'...'
}
],
u'_id': ObjectId('51542645a0c6dc4da77a65b6'),
u'title': u'The Hitchhikers Guide to the Galaxy'
}
]
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如果你想摆脱它u"_id",附加管道:
{"$project": {"_id": 0,
"episodes": "$episodes",
"title": "$title"}
}
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在Mongo DB Google Groups页面上询问了同样的问题.显然它目前不可能,但计划在未来.
http://groups.google.com/group/mongodb-user/browse_thread/thread/4e6f5a0bac1abccc#
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