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Sum with MongoDB group by multiple columns to calculate total marks with duplicate ids
For this, use aggregate() along with $group. Let us create a collection with documents −
> db.demo627.insertOne({id:101,"Name":"Chris","Marks":54});
{
"acknowledged" : true,
"insertedId" : ObjectId("5e9acb306c954c74be91e6b2")
}
> db.demo627.insertOne({id:102,"Name":"Bob","Marks":74});
{
"acknowledged" : true,
"insertedId" : ObjectId("5e9acb3c6c954c74be91e6b3")
}
> db.demo627.insertOne({id:101,"Name":"Chris","Marks":87});
{
"acknowledged" : true,
"insertedId" : ObjectId("5e9acb426c954c74be91e6b4")
}
> db.demo627.insertOne({id:102,"Name":"Mike","Marks":70});
{
"acknowledged" : true,
"insertedId" : ObjectId("5e9acb4b6c954c74be91e6b5")
}
Display all documents from a collection with the help of find() method −
> db.demo627.find();
This will produce the following output −
{ "_id" : ObjectId("5e9acb306c954c74be91e6b2"), "id" : 101, "Name" : "Chris", "Marks" : 54 }
{ "_id" : ObjectId("5e9acb3c6c954c74be91e6b3"), "id" : 102, "Name" : "Bob", "Marks" : 74 }
{ "_id" : ObjectId("5e9acb426c954c74be91e6b4"), "id" : 101, "Name" : "Chris", "Marks" : 87 }
{ "_id" : ObjectId("5e9acb4b6c954c74be91e6b5"), "id" : 102, "Name" : "Mike", "Marks" : 70 }
Following is the query to sum with MongoDB group by multiple columns to calculate total marks with duplicate ids −
> db.demo627.aggregate([
... { "$group": {
... "_id": {
... "id" : "$id",
.. . "Name":"$Name"
... },
... "Marks": { "$sum": "$Marks" }
... }},
... { "$group": {
... "_id": {
... "id" : "$_id.id",
... "Name": "$_id.Name"
... },
...
... "TotalMarks": { "$sum": "$Marks" }
... }}
... ], { "allowDiskUse": true }
... );
This will produce the following output −
{ "_id" : { "id" : 101, "Name" : "Chris" }, "TotalMarks" : 141 }
{ "_id" : { "id" : 102, "Name" : "Bob" }, "TotalMarks" : 74 }
{ "_id" : { "id" : 102, "Name" : "Mike" }, "TotalMarks" : 70 }
Updated on: 2020-05-12T07:11:54+05:30
784 Views
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