Summing Mongo Sub-Document Array

Question

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

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 }

EDIT

Desired output:

100 // sum of each "priceAmt"

Rafa Viotti · Accepted Answer · 2013-10-29 17:39:27Z

4

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}})

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

edited Oct 29, 2013 at 17:39

answered Oct 29, 2013 at 17:27

Rafa Viotti

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