如何获取特定记录？

Question

我想获取今年的记录，其中不包含停止状态这里是我的数据。是否有可能通过聚合函数来实现？

这是我的实际数据

    { [
    {
      shortMovie":"name"
       "categrory": "comedy"
      "times": [
        {
          "state": "start",
          "date": "2017-1-09T18:30:00.000Z"
        },
{
          "state": "stop",
          "date": "2017-1-09T19:30:00.000Z"
        }
      ],
      "_id": "58b017bac8966e17dc59bd95"
    }
  ],[
    {
      shortMovie":"name"
       "categrory": "comedy"
      "times": [
        {
          "state": "start",
          "date": "2017-5-09T18:30:00.000Z"
        }
      ],
      "_id": "58b017bac8966e17dc59bd95"
    }
  ],
 [
    {
      shortMovie":"name"
       "categrory": "comedy"
      "times": [
        {
          "state": "start",
          "date": "2017-6-09T18:30:00.000Z"
        },
{
          "state": "pause",
          "date": "2017-6-09T19:30:00.000Z"
        },
{
          "state": "resume",
          "date": "2017-6-09T22:30:00.000Z"
        },
{
          "state": "stop",
          "date": "2017-6-09T23:30:00.000Z"
        }
      ],
      "_id": "58b017bac8966e17dc59bd95"
    }
  ]
}

我试过

const startDate = new Date(data.year, 0, 1),
                endDate = new Date(data.year, 11, 31, 23, 59, 59, 999),
db.collection.find({$and: [{
                    $or: [{
                        "times.0.date": {
                            $gt: startDate,
                            $lt: endDate
                        }
                    }, {
                        "times.date": {
                            $gt: startDate,
                            $lt: endDate
                        }
                    }]
                }]}).toarray(function(result){

                                            result.times.map(function(get_time){
})
}