MongoDB (Mongoose) Get ranking by key value [duplicate] - mongodb

Question

I am playing around with MongoDB trying to figure out how to do a simple
SELECT province, COUNT(*) FROM contest GROUP BY province
But I can't seem to figure it out using the aggregate function. I can do it using some really weird group syntax
db.user.group({
"key": {
"province": true
},
"initial": {
"count": 0
},
"reduce": function(obj, prev) {
if (true != null) if (true instanceof Array) prev.count += true.length;
else prev.count++;
}
});
But is there an easier/faster way using the aggregate function?

Answer 1

This would be the easier way to do it using aggregate:
db.contest.aggregate([
{"$group" : {_id:"$province", count:{$sum:1}}}
])

Answer 2

I need some extra operation based on the result of aggregate function. Finally I've found some solution for aggregate function and the operation based on the result in MongoDB. I've a collection Request with field request, source, status, requestDate.
Single Field Group By & Count:
db.Request.aggregate([
{"$group" : {_id:"$source", count:{$sum:1}}}
])
Multiple Fields Group By & Count:
db.Request.aggregate([
{"$group" : {_id:{source:"$source",status:"$status"}, count:{$sum:1}}}
])
Multiple Fields Group By & Count with Sort using Field:
db.Request.aggregate([
{"$group" : {_id:{source:"$source",status:"$status"}, count:{$sum:1}}},
{$sort:{"_id.source":1}}
])
Multiple Fields Group By & Count with Sort using Count:
db.Request.aggregate([
{"$group" : {_id:{source:"$source",status:"$status"}, count:{$sum:1}}},
{$sort:{"count":-1}}
])

Answer 3

If you need multiple columns to group by, follow this model. Here I am conducting a count by status and type:
db.BusinessProcess.aggregate({
"$group": {
_id: {
status: "$status",
type: "$type"
},
count: {
$sum: 1
}
}
})

Answer 4

Starting in MongoDB 3.4, you can use the $sortByCount aggregation.
Groups incoming documents based on the value of a specified expression, then computes the count of documents in each distinct group.
https://docs.mongodb.com/manual/reference/operator/aggregation/sortByCount/
For example:
db.contest.aggregate([
{ $sortByCount: "$province" }
]);

Answer 5

Additionally if you need to restrict the grouping you can use:
db.events.aggregate(
{$match: {province: "ON"}},
{$group: {_id: "$date", number: {$sum: 1}}}
)

Answer 6

This type of query worked for me:
db.events.aggregate({$group: {_id : "$date", number: { $sum : 1} }} )
See http://docs.mongodb.org/manual/tutorial/aggregation-with-user-preference-data/

Answer 7

Starting in Mongo 5.0, we can also use { $count: { } } as an alias for { $sum : 1 }:
// { "province" : "Champagne-Ardenne" }
// { "province" : "Champagne-Ardenne" }
// { "province" : "Haute-Normandie" }
db.collection.aggregate([
{ $group: { _id: "$province", count: { $count: {} } } }
])
// { "_id" : "Champagne-Ardenne", "count" : 2 }
// { "_id" : "Haute-Normandie", "count" : 1 }

Answer 8

db.contest.aggregate([
{ $match:{.....May be some match criteria...}},
{ $project: {"province":1,_id:0}},
{ $sortByCount: "$province" }
],{allowDiskUse:true});
MongoDB have 32 MB limitation of sorting operation on memory, use allowDiskUse : true this option, when you expose this query upfront of millions of data, it will sort at disk level not in memory. MongoDB aggregation pipeline has 100MB limitation, so use $project to reduce the data flowing to next pipeline.
If you are using small data then no need to use allowDiskUse option.

Answer 9

Mongo shell command that worked for me:
db.getCollection(<collection_name>).aggregate([{"$match": {'<key>': '<value to match>'}}, {"$group": {'_id': {'<group_by_attribute>': "$group_by_attribute"}}}])