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 }