mongodb - how to find and then aggregate

Question

I have collection that contains documents with below schema. I want to filter/find all documents that contain the gender female and aggregate the sum of brainscore. I tried the

Answer 1

You have to use $match:

db['!all'].aggregate([
  {$match:
    {'GENDER': 'F',
     'DOB':
      { $gte: 19400801,
        $lte: 20131231 } } },
  {$group:
     {_id: "$GENDER",
     totalscore:{ $sum: "$BRAINSCORE"}}}
])

Outputs:

{ "_id" : "F", "totalscore" : 109 }

Answer 2

Sample working query :

db.getCollection('NOTIF_EVENT_RESULT').aggregate([
{$match:
    {'userId': {'$in' : ['user-900', 'user-1546']},
    'criteria.operator': 'greater than', 'criteria.thresold' : '90', 'category' : 'capacity'}
},
{"$group" :  {_id : {userId:"$userId"}, "count" : { "$sum" : 1} } }
])

Answer 3

Here is an answer if the DOB numbers needs to be converted to Date then compared. If not, a number or Date such as 1970 will be incorrectly $gte to 19400801 (you can try):

db['!all'].aggregate([
    {
        $addFields: {
            "_temp_DOB": {
                $dateFromString: {
                    dateString: {$toString: {$toLong: "$DOB"}},
                    format: "%Y%m%d"
                }
            }
        }   
    },
    {
        $match: {
            'GENDER': 'F', 
            '_temp_DOB': { $gte: new Date("1940-08-01"),  
                           $lte: new Date("2013-12-31") }
        }
    },
    {
        $group: {
            _id: "$GENDER", 
            totalscore: { $sum: "$BRAINSCORE" }
        }
    }
])

Outputs:

{ "_id" : "F", "totalscore" : 109 }