Mongodb group and sort

Adding to previous answers, if you want to sort on the sum (the result of the aggregate) instead of on the actual column, you can do this:

db.your_collection.aggregate([
    {
        $group: {_id: "$your_document_name", count: {$sum: 1}}
    }, 
    {
        $sort: {"count":-1}
    }
])

This would be the equivalent of the following standard-SQL-ish syntax:

select col_a, count(col_a) as b
from table
group by col_a
order by b desc

you can use the $group and $sort in aggregate in the order $group > $sort . Using the sort before group won't work. Consider the example below:

let data = await ABCModel.aggregate([
      {
        $match: { city: { $nin: ['', null] }},
      },
      {
        $group: {
          _id: { $toLower: '$city' }
        },
      },
      { $sort: { _id: 1 } },
    ]);

I want to add the following query, as an example, it may be useful in case of two groupings.

Query:

db.getCollection('orders').aggregate([
    {$match:{
        tipo: {$regex:"[A-Z]+"}
        }
    },
    {$group:
        { 
            _id:{
                codigo:"1",
                fechaAlta:{$substr:["$fechaAlta",0,10]},
            },
            total:{$sum:1}
        }
    },
    {$sort:
        {"_id":1}
    },
    {$group:
        {
            _id:"$_id.codigo",
            fechasAltas:
            {
                $push:
                {
                    fechaAlta:"$_id.fechaAlta",
                    total:"$total"
                }
            },
            totalGeneral:{$sum:"$total"}
        }
    }
]); 

Response:

{
"_id" : "1",
"fechasAltas" : [ 
    {
        "fechaAlta" : "1940-01-01",
        "total" : 13.0
    }, 
    {
        "fechaAlta" : "2007-05-14",
        "total" : 115.0
    }, 
    {
        "fechaAlta" : "2008-09-30",
        "total" : 58.0
    }, 

    .
    .
    .
],
"totalGeneral" : 50620.0
}