CODEWITHSUNDEEP
mongodbmongodb-queryaggregation-framework
Gatmando
Gatmando

Reputation: 2269

$sum values within array in mongodb

We have a Mongo collection that has this form:

[
    {
        "_id" : "34210db0-6g91-83e8-ae8c-659f064f503f",
        "dateReceived" : ISODate("2018-01-01T00:00:00.000Z"),
        "group" : null,
        "clientShortName" : "My Client Name",
        "sourceShortName" : "Datasource Name",
        "files" : [ 
            {
                "_id" : "807061f0-2d77-87e8-8610-9ff3cbc9c774"
                "status" : 1,
                "fileName" : "filename1.csv",
                "numRows" : 15,

            }, 
            {
                "_id" : "587036f0-2n65-55e8-8610-3ee3cbc9c814"
                "status" : 8,
                "fileName" : "filename2.csv",
                "numRows" : 30,
            }
        ]
    }
]

We have a Mongo query using "find" in conjunction with "map" in order to produce a transformed output. The find/map command looks like this:

db.getCollection('batches')
.find({_id: "34210db0-6g91-83e8-ae8c-659f064f503f"}, {"__v": false, "files.diffHistory": false})
.map( doc =>
    {
        doc.id = doc._id;
        doc.clientName = doc.clientShortName;
        doc.dataSourceName = doc.sourceShortName;
        delete doc._id;
        delete doc.clientShortName;
        delete doc.sourceShortName;

        doc.numFiles = NumberInt(doc.files.length);

        doc.files = doc.files.map( file =>
            {
                file.id = file._id;
                delete file._id;
                delete file.__v;
                delete file.edits;

                return file;
            }
        );

        // broken....how should this be formatted?    
        doc.totalNumRows = {$sum: doc.files.numRows};

        return doc;
    }
)

This find/map works and produces the expected output EXCEPT for the "totalNumRows" summation. What we are trying to accomplish is to add up all of the "files.numRows" fields so that we get a single summarised entry at the top level of the returned data set. Ie we'd see a result set that looks like:

[
        {
            "id" : "34210db0-6g91-83e8-ae8c-659f064f503f",
            "dateReceived" : ISODate("2018-01-01T00:00:00.000Z"),
            "group" : null,
            "clientName" : "My Client Name",
            "dataSourceName" : "Datasource Name",
            "files" : [ 
                {
                    "id" : "807061f0-2d77-87e8-8610-9ff3cbc9c774"
                    "status" : 1,
                    "fileName" : "filename1.csv",
                    "numRows" : 15,

                }, 
                {
                    "id" : "587036f0-2n65-55e8-8610-3ee3cbc9c814"
                    "status" : 8,
                    "fileName" : "filename2.csv",
                    "numRows" : 30,
                }
            ],

            "totalNumRows": 45

        }
    ]

Everything tried so far produces a broken query. Does anyone know the correct command/format for summing up the "numRows" field of our "file" sub-doc?

Upvotes: 1

Views: 34

Answers (1)

Ashh
Ashh

Reputation: 46441

Well you can do something better and faster approach then that... Try $project to rename your actual field name and then $sum to get total count of numRows

db.collection.aggregate([
  { "$match": { _id: "34210db0-6g91-83e8-ae8c-659f064f503f" }},
  { "$project": {
    "totalNumRows": {
      "$sum": "$files.numRows"
    },
    "clientName": "$clientShortName",
    "dataSourceName": "$sourceShortName",
    "files": "$files"
  }}
])

It will give following output

[
  {
    "_id": 1111,
    "clientName": "My Client Name",
    "dataSourceName": "Datasource Name",
    "files": [
      {
        "fileName": "filename1.csv",
        "id": 2222,
        "numRows": 15,
        "status": 1
      },
      {
        "fileName": "filename2.csv",
        "id": 3333,
        "numRows": 30,
        "status": 8
      }
    ],
    "totalNumRows": 45
  }
]

Upvotes: 1

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