You need to use $group to group documents with specified _id expression. Let us first create a collection with documents −
> db.aggreagateDemo.insertOne({"Product_Id":1,"ProductPrice":50}); { "acknowledged" : true, "insertedId" : ObjectId("5e06d3c025ddae1f53b621d9") } > db.aggreagateDemo.insertOne({"Product_Id":2,"ProductPrice":100}); { "acknowledged" : true, "insertedId" : ObjectId("5e06d3c625ddae1f53b621da") } > db.aggreagateDemo.insertOne({"Product_Id":2,"ProductPrice":500}); { "acknowledged" : true, "insertedId" : ObjectId("5e06d3cb25ddae1f53b621db") } > db.aggreagateDemo.insertOne({"Product_Id":1,"ProductPrice":150}); { "acknowledged" : true, "insertedId" : ObjectId("5e06d3d125ddae1f53b621dc") }
Following is the query to display all documents from a collection with the help of find() method −
> db.aggreagateDemo.find().pretty();
This will produce the following output −
{ "_id" : ObjectId("5e06d3c025ddae1f53b621d9"), "Product_Id" : 1, "ProductPrice" : 50 } { "_id" : ObjectId("5e06d3c625ddae1f53b621da"), "Product_Id" : 2, "ProductPrice" : 100 } { "_id" : ObjectId("5e06d3cb25ddae1f53b621db"), "Product_Id" : 2, "ProductPrice" : 500 } { "_id" : ObjectId("5e06d3d125ddae1f53b621dc"), "Product_Id" : 1, "ProductPrice" : 150 }
Here is the query to perform aggregation and group with id −
> db.aggreagateDemo.aggregate([ ... { ... $group: { ... _id: "$Product_Id", ... TotalValue:{$sum: "$ProductPrice"} ... } ... } ... ] ... );
This will produce the following output −
{ "_id" : 2, "TotalValue" : 600 } { "_id" : 1, "TotalValue" : 200 }