我想过滤report字段，其中m_name = "ABC"，并将其分组。下面是我的示例数据：

{
    "_id" : ObjectId("63f82b8619e8f81b7009d2c3"),
    "type" : "test"
    "created_at" : date
    "created_by" : "admin",
    "report" : [
        {
            "name" : "Day 3",
            "date" : date
            "m_name" : "ABC",
            "created_by" : "admin",
            "created_at" :date
        },
        {
            "name" : "Day 2",
            "date" :date,
            "created_by" : "admin",
            "m_name" : "ABC",
            "created_at" : date
        },
        {
            "name" : "test",
            "date" : date
            "m_name": "123"
            "created_by" : "admin",
            "created_at" : date,
        },
    ],
   ... 20 fields

}

这是我的管道：

.aggregate([ 
       {
          $unwind: "report",
        },
        {
          $match: {
            "report.m_name": "ABC",
          },
        },
        {
          $group: {
            _id: "$_id",
            report: {
              $push: "$report",
            },
            
          },
        },
       ])

如果这样做，输出只有_id和report，如何返回所有其余字段？我已经尝试了$$ROOT，但它嵌入在一个对象中，我希望输出如下：

{
    "_id" : ObjectId("63f82b8619e8f81b7009d2c3"),
    "type" : "test"
    "created_at" : date
    "created_by" : "admin",
    "report" : [
        {
            "name" : "Day 3",
            "date" : date
            "m_name" : "ABC",
            "created_by" : "admin",
            "created_at" :date
        },
        {
            "name" : "Day 2",
            "date" :date,
            "created_by" : "admin",
            "m_name" : "ABC",
            "created_at" : date
        },
    ],
   ... 20 fields

}

我知道project（类型：{$first：“$type”}，）就可以了，但是有20个字段，还有其他更简单的方法吗？