汇总Mongo子文档数组 [英] Summing Mongo Sub-Document Array

问题描述

 db.test3.find() 
{ "_id" : 1, "results" : [{"result" : {"cost" : [ { "priceAmt" : 100 } ] } } ] }

我未成功尝试以下操作:

I tried the following unsucessfully:

db.test3.aggregate({$group : {_id: "", total : {$sum: 
$results.result.cost.priceAmt"}}}, {$project: {_id: 0, total: 1}})
{ "result" : [ { "total" : 0 } ], "ok" : 1 }

编辑

所需的输出:

100 // sum of each "priceAmt"

推荐答案

您必须使用$unwind运算符将数组项转换为单个文档.

You'll have to use the $unwind operator to turn array items into individual documents.

db.test3.aggregate({$unwind: "$results"}, {$unwind: "$results.result.cost"}, {$group : {_id: "", total : {$sum: "$results.result.cost.priceAmt"}}}, {$project: {_id: 0, total: 1}})

$unwind需要应用两次，因为您有一个嵌套数组.

The $unwind needs to be applied twice because you have a nested array.

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