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To Aggregate Totals in One Group with MongoDB
To aggregate totals, use $sum in MongoDB. Let us create a collection with documents −
> db.demo406.insertOne({"Score":35});
{
"acknowledged" : true,
"insertedId" : ObjectId("5e6f99d5fac4d418a0178599")
}
> db.demo406.insertOne({"Score":55});
{
"acknowledged" : true,
"insertedId" : ObjectId("5e6f99d8fac4d418a017859a")
}
> db.demo406.insertOne({"Score":35});
{
"acknowledged" : true,
"insertedId" : ObjectId("5e6f99dcfac4d418a017859b")
}
> db.demo406.insertOne({"Score":45});
{
"acknowledged" : true,
"insertedId" : ObjectId("5e6f99defac4d418a017859c")
}
> db.demo406.insertOne({"Score":65});
{
"acknowledged" : true,
"insertedId" : ObjectId("5e6f99e3fac4d418a017859d")
}
> db.demo406.insertOne({"Score":45});
{
"acknowledged" : true,
"insertedId" : ObjectId("5e6f99e6fac4d418a017859e")
}
Display all documents from a collection with the help of find() method −
> db.demo406.find();
This will produce the following output −
{ "_id" : ObjectId("5e6f99d5fac4d418a0178599"), "Score" : 35 }
{ "_id" : ObjectId("5e6f99d8fac4d418a017859a"), "Score" : 55 }
{ "_id" : ObjectId("5e6f99dcfac4d418a017859b"), "Score" : 35 }
{ "_id" : ObjectId("5e6f99defac4d418a017859c"), "Score" : 45 }
{ "_id" : ObjectId("5e6f99e3fac4d418a017859d"), "Score" : 65 }
{ "_id" : ObjectId("5e6f99e6fac4d418a017859e"), "Score" : 45 }
Following is the query to aggregate totals in one group −
> db.demo406.aggregate([
... { "$group": {
... "_id": null,
... "Score1": {
... "$sum": {
... "$cond": [{ "$eq": [ "$Score", 35 ] }, 1, 0 ]
... }
... },
... "Score2": {
... "$sum": {
... "$cond": [{ "$ne": [ "$Score", 35 ] }, 1, 0 ]
... }
... },
... "Score3": {
... "$sum": {
... "$cond": [{ "$ne": [ "$Score", 59 ] }, "$Score", 0 ]
... }
... }
... }}
... ])
This will produce the following output −
{ "_id" : null, "Score1" : 2, "Score2" : 4, "Score3" : 280 }Advertisements