Question

给出以下数据库的集合：

> db.collection.find()
{ "_id" : 1, "results" : { "record" : { "price" : 100 } } }
{ "_id" : 2, "results" : { "record" : { "price" : 200 } } }
{ "_id" : 3, "results" : { "record" : { "price" : 300 } } }

如何使用聚合（或map-reduce）来获取每个results.record.price字段的总和？

> db.collection.aggregate( {$group : { _id:"", "results.record.price" : {$sum : "$results.record.price"}}}, {$project:{_id: 0, "results.record.price" : "$results.record.price"}})
Error: Printing Stack Trace
    at printStackTrace (src/mongo/shell/utils.js:37:15)
    at DBCollection.aggregate (src/mongo/shell/collection.js:897:9)
    at (shell):1:15
Mon Oct 28 20:15:24.065 aggregate failed: {
        "errmsg" : "exception: the group aggregate field name 'results.record.price' cannot be used because $group's field names cannot contain '
.'",
        "code" : 16414,
        "ok" : 0
} at src/mongo/shell/collection.js:898

Answer 1

试试这个。

db.collection.aggregate({$group : {_id: "", total : {$sum: "$results.record.price"}}}, {$project: {_id: 0, total: 1}})

总结Mongo子文档的字段

1 个答案: