Get count of “loglevel” for each “name”

Question

For example I have following collection:

db.names.find({})

{ \"_id\" : ObjectId(\"5768d9b4bc6f464899594570\"), \"name\"          


        

Answer 1

Remove the { "$sum": 1 } and { "$sum": 0 } expressions in your if/else conditional blocks, substitute them with the values 1 and 0 (respectively for each conditional block).

The final pipeline should look like this, using the other $cond syntax which omits the if/else blocks:

db.names.aggregate([
    {
        "$group": {
            "_id": "$name",
            "error": { 
               "$sum": { 
                   "$cond": [ { "$eq": [ "$loglevel",  "ERROR" ] }, 1, 0] 
               }
           },
           "warning":{
               "$sum": { 
                   "$cond": [ { "$eq": [ "$loglevel", "WARNING" ] }, 1, 0 ]
                }
           },
           "info": { 
               "$sum": { 
                   "$cond": [ { "$eq": [ "$loglevel",  "INFO" ] }, 1, 0 ]
               }
           }
        }
    }
])

---

Or dynamically create the pipeline, given an array of possible statuses:

var statuses = ["ERROR", "WARNING", "INFO"],
    groupOperator = { "$group": { "_id": "$name" } };

statuses.forEach(function (status){ 
    groupOperator["$group"][status.toLowerCase()] = { 
       "$sum": { 
           "$cond": [ { "$eq": [ "$loglevel",  status ] }, 1, 0] 
       }
   }
});

db.names.aggregate([groupOperator]);

Output

/* 1 */
{
    "_id" : "t1",
    "error" : 2,
    "warning" : 3,
    "info" : 1
}

/* 2 */
{
    "_id" : "t2",
    "error" : 4,
    "warning" : 0,
    "info" : 1
}

/* 3 */
{
    "_id" : "t3",
    "error" : 0,
    "warning" : 0,
    "info" : 1
}

Answer 2

If I understand correctly, you want the sum of loglevels against each name. shouldn't this help?

   db.names.aggregate([{$group:{_id:{name:"$name",loglevel:"$loglevel"},sum:{$sum:1}}}])