返回特定的mongodb嵌入文档

return specific mongodb embedded document

本文关键字：文档 mongodb 返回      更新时间：2023-09-26

我想查询mongodb文档以返回特定的文档

假设我有下面的文档，并且我只想搜索给定房间的具体细节。比方说，我希望查询返回room_num=210的注释。我该怎么做？

我能够通过db.locations.findOne（｛"facility.room.room_num"：room_num｝）进行查询；我不知道如何只投影房间。概括这个问题，我如何只投影嵌入的文档特定条目。

我的文档看起来是这样--->

{
  _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"
          }
      ]
    }
  ]

由于您有一个嵌入的数组（一个在另一个中），因此您确实可以使用聚合来实现这一点。你只需要做一个"双重"放卷操作：

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

如果你真的想让它看起来像原始文档，那就更进一步：

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"
        }}
    }}
])

因此，从你的数据来看，这将是一个结果：

{
    "_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，那么只需在末尾使用$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" }

这应该会澄清问题。

尝试以下查询

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

您可以使用聚合框架。我相信$project和$unflow操作符正是您想要的。

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

你会得到这样的东西：

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