MongoDB: How can I calculate total count without using $group operator

Question

I have following result:

"result" : [ 
    {
        "_id" : "London",
        "count" : 499
    }, 
    {
        "_id" : "Paris",
        "count" : 135
    }, 
    {
        "_id" : "Lviv",
        "count" : 95
    }
]

And here is query:

 {"$group":{
        _id: "$city",
        "count" : {"$sum":1}
    }
 }

So, I want some how to calculate all fields not only grouped. I think it would better to show expected result:

"result" : [ 
    {
        "_id" : "London",
        "count" : 499,
        "total" : 729
    }, 
    {
        "_id" : "Paris",
        "count" : 135,
        "total" : 729
    }, 
    {
        "_id" : "Lviv",
        "count" : 95,
        "total" : 729
    }
]

Expected result has "total" field which calculated as amount of "count" field (499+135+95 = 729). EDITED: I must use only aggregation framework!

Can someone help me with this?

Why not just add that on the client rather than making the DB do extra work, and send the same data to the client many times? — WiredPrairie
– WiredPrairie, Commented Feb 19, 2014 at 13:10
Because client side doesn't have such functionality, I should do this on the server side. — Andrii Furmanets
– Andrii Furmanets, Commented Feb 19, 2014 at 15:45
What client are you using that can't iterate through the results to produce a total? — WiredPrairie
– WiredPrairie, Commented Feb 19, 2014 at 16:10

heinob · Accepted Answer · 2014-02-19 13:26:36Z

1

You have to count the total number before:

db.coll.count( ..., function( err, total ) {

and then use that result in your aggregation command:

{
    "$group": {
        _id: "$city",
        count: { "$sum": 1 },
        total: total
}

EDIT:

If you only want to use aggregation framework, try this instead of db.coll.count():

{
    "$group": {
        _id: 1,
        count: { "$sum": 1 }
    }
}

edited Feb 19, 2014 at 13:26

answered Feb 19, 2014 at 12:03

heinob

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4 Comments

Andrii Furmanets Over a year ago

it doesn't work for me. And probably you made mistake in second group, there is should be "count": {"$first" : "$count"}.

heinob Over a year ago

I must correct myself. It is not possible with one call. You need two calls.

Andrii Furmanets Over a year ago

it still doesn't work. This one calculate total of all fields: { "$group": { _id: 1, count: { "$sum": 1 } } } But how can I pass "$city" field to the next group operator?

heinob Over a year ago

You cannot calculate the total sum of all fields and the sum per city in one (1) call. You have to make two mongo calls.

Sammaye · Accepted Answer · 2014-02-19 12:02:54Z

0

Sounds like db.collection.count() would give you your result actually. This is because you are actually just summing up ALL documents in the collection there.

answered Feb 19, 2014 at 12:02

Sammaye

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1 Comment

Andrii Furmanets Over a year ago

I need count of grouped fields and total of all fields. Please take a look at my expected result. And one more thing, I should use aggregation framework

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