MongoDB aggregaat vullen ontbrekende dagen

aggregeren gebruiken om deze vraag volledig te behandelen is lastig.
Maar het kan worden bereikt.
(MongoDB V2.6+ vereist)

var proj1 = {
    "$project" : {
        "created" : 1,
        "_id" : 0,
        "h" : {
            "$hour" : "$created"
        },
        "m" : {
            "$minute" : "$created"
        },
        "s" : {
            "$second" : "$created"
        },
        "ml" : {
            "$millisecond" : "$created"
        }
    }
};

var proj2 = {
    "$project" : {
        "created" : {
            "$subtract" : [ "$created", {
                "$add" : [ "$ml", {
                    "$multiply" : [ "$s", 1000 ]
                }, {
                    "$multiply" : [ "$m", 60, 1000 ]
                }, {
                    "$multiply" : [ "$h", 60, 60, 1000 ]
                } ]
            } ]
        }
    }
};

var group1 = {
        $group : {
            _id : "$created",
            count : {
                $sum : 1
            }
        }
    };

var group2 = {
        $group : {
            _id : 0,
            origin : {
                $push : "$$ROOT"
            },
            maxDate : {
                $max : "$_id"
            }
        }
};

var step = 24 * 60 * 60 * 1000; // milliseconds of one day

var project3 = {
    $project : {
        origin : 1,
        extents : {
            $map : {
                "input" : [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29],
                "as" : "e",
                "in" : {
                    _id : {
                        $subtract : [ "$maxDate", {
                            $multiply : [ step, "$$e"]
                        }]
                    },
                    count : {
                        $add : [0]
                    }
                }
            }
        }
    }
};

var project4 = {
        $project : {
            _id : 0,
            values : {
                $setUnion : [ "$origin", "$extents"]
            }
        }
};

var unwind1 = {
        $unwind : "$values"
};

var group3 = {
        $group : {
            _id : "$values._id",
            count : {
                $max : "$values.count"
            }
        }
};

db.product.aggregate([ proj1, proj2, group1, group2, project3, project4,
        unwind1, group3, {
            $sort : {
                _id : 1
            }
        } ]);

Ik zou graag het ontbrekende deel aan het einde van de aanvraag willen invullen, zoiets als dit voor uw referentie:

function sortResult(x, y) {
    var t1 = x._id.getTime();
    var t2 = y._id.getTime();
    if (t1 < t2) {
        return -1;
    } else if (t1 == t2) {
        return 0;
    } else {
        return 1;
    }
}

var result = db.product.aggregate();

var endDateMilliseconds = result[result.length - 1]._id.getTime();
var step = 24 * 60 * 60 * 1000; // milliseconds of one day

var map = {};
for (var i in result) {
    map[ result[i]._id.getTime() ] = result[i];
}

for (var ms = endDateMilliseconds, x = 1; x < 30; x++) {
    ms -= step;
    if ( ! ( ms in map ) ) {
        map[ms] = {_id : new Date(ms), count : 0};
    }
}

var finalResult = [];
for (var x in map) {
    finalResult.push(map[x]);
}
finalResult.sort(sortResult);
printjson(finalResult);