Mongodb 汇总Mongo子文档'；领域

给定以下数据库的集合：

> 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

结果.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

试试这个

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

---

\u id:“

是什么意思？谢谢你help@kevin

\u id

是组标识符。在这种情况下，由于空字符串是常量，因此处理的每个记录都将合并到一个由该空字符串标识的组中。您可以使用其他常量，如

null

，

“a”

等。您能检查一下吗-