Group distinct count() on a single field in a single request - mongodb

Question

Is there a way to get different counts on a single field from a single document ?
Here is a schema for a document User
UserSchema: {
name: { type: String, required: true },
created_at: { type: Date, default: now }
}
I would like to get every Users created the 01/05/2013 and the 06/08/2013, maybe i'll need to count more different dates.
Can i get these datas on a sigle count() or should i get all the Users with a find() then count it using javascript ?

Answer 1

You can use the collection.count() form which accepts a query, along with the use of $or and ranges:
db.collection.count({ "$or": [
{ "created_at":
{"$gte": new Date("2014-05-01"), "$lt": new Date("2014-05-02") }
},
{ "created_at":
{"$gte": new Date("2013-08-06"), "$lt": new Date("2013-08-07") }
}
]})
Or you can pass that query to .find() and use the cursor count from there if that suits your taste.
But then, I read your title again, and distinct count would be different, and best to use aggregate to get the distinct days:
db.collection.aggregate([
// Match the dates you want to filter
{ "$match": {
{ "$or": [
{ "created_at": {
"$gte": new Date("2014-05-01"),
"$lt": new Date("2014-05-02")
}},
{ "created_at": {
"$gte": new Date("2013-08-06"),
"$lt": new Date("2013-08-07")
}}
]}
}},
// Group on the *whole* day and sum the count
{ "$group": {
"_id": {
"year": { "$year": "$created_at" },
"month": { "$month": "$created_at" },
"day": { "$dayOfMonth": "$created_at" }
},
"count": { "$sum": 1 }
}}
])
And that would give you a distinct count of the documents for each selected day you had added in your $or clause.
No need for looping in code.