There are a couple of ways to do this. The first is with Date Aggregation Operators, which allow you to dissect the “date” values in documents. Specifically for “grouping” as the primary intent: db.collection.aggregate([ { “$group”: { “_id”: { “year”: { “$year”: “$created_at” }, “dayOfYear”: { “$dayOfYear”: “$created_at” }, “hour”: { “$hour”: “$created_at” }, “interval”: … Read more
Sum To get the sum of a grouped field when using the Aggregation Framework of MongoDB, you’ll need to use $group and $sum: db.characters.aggregate([ { $group: { _id: null, total: { $sum: “$wins” } } } ] ) In this case, if you want to get the sum of all of the wins, you need … Read more
The thing to remember is that MongoDB employs an “NoSQL” approach to data storage, so perish the thoughts of selects, joins, etc. from your mind. The way that it stores your data is in the form of documents and collections, which allows for a dynamic means of adding and obtaining the data from your storage … Read more
You can use “$$REMOVE”: This system variable was added in version 3.6 (mongodb docs) db.collection.aggregate([{ $group:{ field: { $push: { $cond:[ { $gt: [“$A”, 0] }, { id: “$_id”, A:”$A” }, “$$REMOVE” ] } }, secondField:{ $push: “$B” } } ]) In this way you don’t have to filter nulls.
You can use below aggregation query. $$ROOT to keep the whole document per each name followed by $replaceRoot to promote the document to the top. db.col.aggregate([ {“$group”:{“_id”:”$name”,”doc”:{“$first”:”$$ROOT”}}}, {“$replaceRoot”:{“newRoot”:”$doc”}} ])
To get the keyword count you’d need to group the documents by the keyword field, then use the accumulator operator $sum to get the documents count. As for the other field values, since you are grouping all the documents by the keyword value, the best you can do to get the other fields is use … Read more
To accommodate the potential duplicates, you need to use two $group operations: db.test.aggregate([ { $group: { _id: { owner_id: ‘$owner_id’, car_id: ‘$car_id’ } }}, { $group: { _id: ‘$_id.owner_id’, cars_owned: { $sum: 1 } }}, { $project: { _id: 0, owner_id: ‘$_id’, cars_owned: 1 }}] , function(err, result){ console.log(result); } ); Gives a result with … Read more
You can do this by $unwinding the updates array, sorting the resulting docs by date, and then $grouping them back together on _id using the sorted order. db.servers.aggregate( {$unwind: ‘$service.apps.updates’}, {$sort: {‘service.apps.updates.date’: 1}}, {$group: {_id: ‘$_id’, ‘updates’: {$push: ‘$service.apps.updates’}}}, {$project: {‘service.apps.updates’: ‘$updates’}})
In the simplest sense this just follows the basic form of “dot notation” as used by MongoDB. That will work regardless of which array member the inner array member is in, as long as it matches a value: db.mycollection.find({ “someArray.someNestedArray.name”: “value” }) That is fine for a “single field” value, for matching multiple-fields you would … Read more
We can do multiple join conditions with the $lookup aggregation pipeline operator in version 3.6 and newer. We need to assign the fields’s values to variable using the let optional field; you then access those variables in the pipeline field stages where you specify the pipeline to run on the collections. Note that in the … Read more