How to hide _id from Aggregation?

Question

I\'ve this query:

produits = yield motor.Op(db.users.aggregate, [{\"$unwind\":\"$pup\"},{\"$match\":{\"pup.spec.np\":nomp}}, {\"$group\":{\"_id\":\"$pup.spec         


        

Answer 1

This is not exaclty a mongoWay of doing it, but you can use this factory to generate an object that includes all but _id

/**
 * Factory that returns a $project object that excludes the _id property https://docs.mongodb.com/v3.0/reference/operator/aggregation/project/ 
 * @params {String} variable list of properties to be included  
 * @return {Object} $project object including all the properties but _id
 */
function includeFactory(/* properties */){
    var included = { "_id": 0 };
    Array.prototype.slice.call(arguments).forEach(function(include){
        included[include] = true
    })

    return { "$project": included }
}

Then use it like this:

cities.aggregate(
{ "$group": { "_id": null, "max": { "$max": "$age" }, "min": { "$min": "$age" }, "average": { "$avg": "$age" }, "total": { "$sum": "$count" } } },
        includeFactory('max','min','average','total') 
)