mongodb - Mongo group分组后，怎么对分组后的结果进行从大到小排序，并取前5个

Question

{代码...} 最后的结果，我想要count数量从大到小排列，并且只取前五个，能用finalize实现吗？具体怎么写函数呢？？或者能用什么方法实现？？？？

迷茫 · Answer

To be honest, group-by under mongo is quite troublesome. Native mongo statement, aggregate query through pipeline

sqldb.term.aggregate([
     {$match:{library_id:3607}},
     {$limit:5},
     {$group:{_id:"$version", count: {$sum:1}}},
     {$sort:{count:-1}}
])

term: The name of the collection you want to query
$match: Matching conditions, optional
$limit: Number of results, optional
$group: Aggregation rules
$sort: Query The results are sorted, -1 means descending order

http://docs.mongoing.com/manual-zh/reference/operator/aggregation.html#aggregation-pipeline-operator-reference

高洛峰 · Answer

Use aggregate for group calculation and sorting.
Simple example

db.collection.aggregate([  {$group:{_id:"$source",total:{$sum:"$changeValue"}}},{$limit:1}])

See http://docs.mongodb.org/manual/reference/command/aggregate/