MongoDB query to group by

Question

I've data in this format in test collection

{car_id: "1", owner_id: "a"}
{car_id: "1", owner_id: "b"}
{car_id: "1", owner_id: "c"}
{car_id: "1", owner_id: "d"}
{car_id: "1", owner_id: "a"}
{car_id: "1", owner_id: "b"}
{car_id: "1", owner_id: "c"}
{car_id: "1", owner_id: "d"}
{car_id: "2", owner_id: "a"}
{car_id: "2", owner_id: "b"}
{car_id: "2", owner_id: "c"}
{car_id: "2", owner_id: "d"}
{car_id: "2", owner_id: "a"}
{car_id: "2", owner_id: "b"}
{car_id: "2", owner_id: "c"}
{car_id: "2", owner_id: "d"}
{car_id: "3", owner_id: "a"}
{car_id: "3", owner_id: "b"}
{car_id: "3", owner_id: "c"}
{car_id: "3", owner_id: "a"}
{car_id: "3", owner_id: "b"}
{car_id: "3", owner_id: "c"}
{car_id: "3", owner_id: "d"}
{car_id: "3", owner_id: "e"}
    ...

we need output after query in this manner

{
    "result" : [
        {
            "for_cars_id" : 1,
            "owner_id" : "a",
            "owner_id_count: 2
        },
        {
            "for_cars_id" : 1,
            "owner_id" : "b",
            "owner_id_count: 2
        },
        {
            "for_cars_id" : 1,
            "owner_id" : "c",
            "owner_id_count: 2
        },
        {
            "for_cars_id" : 1,
            "owner_id" : "d",
            "owner_id_count: 2
        },
        {
            "for_cars_id" : 2,
            "owner_id" : "a",
            "owner_id_count: 2
        },
        {
            "for_cars_id" : 2,
            "owner_id" : "b",
            "owner_id_count: 2
        }
                       ...
                       ...

    ],
    "ok" : 1
}

I try to do this with the help of link (Group count with MongoDB using aggregation framework) but not get appropriate result please help me how I find this output

thanks

Tal · Accepted Answer · 2013-06-12 19:57:15Z

1

This is pretty straightforward:

db.test.aggregate({
  $group: {
    _id: {
      car_id: "$car_id",
      owner_id: "$owner_id"
    },
    owner_id_count: { $sum: 1 }
  }
}, {
  $project: {
    _id: 0,
    for_cars_id: "$_id.car_id",
    owner_id: "$_id.owner_id",
    owner_id_count: 1
  }
})

answered Jun 12, 2013 at 19:57

Tal

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