我的模型：

const scheduleTaskSchema = new Schema({
  activity: { type: Object, required: true },
  date: { type: Date, required: true },
  crew: Object,
  vehicle: Object,
  pickups: Array,
  details: String,
});
const ScheduleTaskModel =  mongoose.model("schedule_task", scheduleTaskSchema),

以及该聚合管道：

let aggregation = [
    {
      $sort: {
        "pickups.0.time": 1,
      },
    },
    {
      $group: {
        _id: "$date",
        tasks: { $push: "$$ROOT" },
      },
    },
    { $sort: { _id: -1 } },
  ];


  if (hasDateQuery) {
    aggregation.unshift({
      $match: {
        date: { $gte: new Date(start_date), $lte: new Date(end_date) },
      },
    });
  } else {
    aggregation.push({ $limit: 2 });
  }

  const scheduledTasksGroups = await ScheduleTaskModel.aggregate(aggregation);

crew对象可以具有任意数量键，其结构如下：

crew : {
  drivers: [
    {
      _id: "656b1e9cf5b894a4f2v643bc",
      name: "john"
    },
    {
      _id: "567b1e9cf5b954a4f2c643bhh",
      name: "bill"
    }
  ],
  officers: [
    {
      _id: "655b1e9cf5b6632a4f2c643jk",
      name: "mark"
    },
    {
      _id: "876b1e9af5b664a4f2c234bb",
      name: "jane"
    }
  ],
  //...any number of keys that contain an array of objects that all have an _id
}

我正在寻找一种方法，在不知道要搜索哪个键的情况下，返回包含crew对象中任何位置的给定_id的所有文档（在排序/分组之前），它可以是许多不同的键，这些键都包含一个对象数组，这些对象都具有_id
有什么想法吗？