返回特定的 mongodb 嵌入文档

Question

我想查询 mongodb 文档以返回特定文档

假设我有以下文档，我只想搜索给定房间的具体细节。比如说，我想要查询返回 room_num = 210 的注释。我该怎么做？

我可以通过 db.locations.findOne({"facility.room.room_num" : room_num}); 进行查询 我不知道如何投影房间。 概括问题，我如何仅投影嵌入文档的特定条目。

我的文档是这样的--->

{
  _id: Object("xxx"),
  name: "Campus",
  facilities: [
     { name : "Science",
     rooms : [
          {
             room_num: 210,
             notes: "Chemistry Lab 1",
             Dimension: "x*x"
          },{
             room_num: 120,
             notes: "Chemistry Lab 2",
             Dimension: "x*x"
          }
      ]

    },
    { name : "Arts",
     rooms: [
          {
             room_num: 90,
             notes: "Drawing 1",
             Dimension: "x*x"
          },{
             room_num: 100,
             notes: "Drawing 2",
             Dimension: "x*x"
          }
      ]

    }
  ]

Answer 1

由于您有一个嵌入式数组（一个在另一个中），那么您确实可以使用聚合来做到这一点。你只需要做一个“双重”展开操作：

db.collection.aggregate([
    { "$unwind": "$facilities" },
    { "$unwind": "$facilities.rooms" },
    { "$match": { "facilities.rooms.room_num": 210 } }
])

如果你真的希望它看起来像原始文档，那就再进一步：

db.collection.aggregate([
    { "$unwind": "$facilities" },
    { "$unwind": "$facilities.rooms" },
    { "$match": { "facilities.rooms.room_num": 210 } },
    { "$group": {
        "_id": "$_id",
        "name": { "$first": "$name" },
        "facName": { "$first": "$facilities.name" },
        "rooms": { "$push": "$facilities.rooms" }
    }},
    { "$group": {
        "_id": "$_id",
        "name": { "$first": "$name" },
        "facities": { "$push": {
            "name": "$facName" ,
            "rooms": "$rooms"
        }}
    }}
])

所以根据您的数据，这将是结果：

{
    "_id" : {
            "0" : "x",
            "1" : "x",
            "2" : "x"
    },
    "name" : "Campus",
    "facities" : [
       {
           "name" : "Science",
           "rooms" : [
               {
                   "room_num" : 210,
                   "notes" : "Chemistry Lab 1",
                   "Dimension" : "x*x"
               }
           ]
       }
    ]
}

或者最后，如果您只需要数组的内部房间部分，那么只需在最后使用 $project 语句，如第一个示例所示：

db.campus.aggregate([
    { "$unwind": "$facilities" },
    { "$unwind": "$facilities.rooms" },
    { "$match": { "facilities.rooms.room_num": 210 } },
    { "$project": {
        "_id": 0,
        "room_num": "$facilities.rooms.room_num",
        "notes": "$facilities.rooms.notes",
        "Dimension": "$facilities.rooms.Dimension"
    }}
])

结果如下：

{ "room_num" : 210, "notes" : "Chemistry Lab 1", "Dimension" : "x*x" }

这应该可以清除它。

Answer 2

试试下面的查询

db.locations.aggregate( 
 { $unwind: "facilities.rooms" },
 { $match: { "facilities.rooms.room_num" : room_num}},
 { $project: { "facilities.rooms.notes" : 1 }}
)

Answer 3

您可以使用aggregation framework。我相信 $project 和 $unwind 运算符是您正在寻找的。

db.rooms.aggregate([{
  $project: {
    _id: 0,
    facilities: "$facilities.rooms.room_num"
  }
}, {
  $unwind: "$facilities"
}, {
  $unwind: "$facilities"
}])

你会得到类似的东西：

{
  "result" : [
    {
      "facilities" : 210
    },
    {
      "facilities" : 120
    },
    {
      "facilities" : 90
    },
    {
      "facilities" : 100
    }
  ],
  "ok" : 1
}