仅返回在文档中搜索的嵌入式属性的最后一级 [英] return only the last level of the embedded property that is searched in a document

问题描述

我有这些文件.

db.test.find({"house.floor":1})



    db.test.insertMany([{
        "name":"homer",
        "house": {
            "floor": 1,
            "room": 
            {
                "bed": "bed_pink",
                "chair":"chair_pink"
            }
        }

    },
    {
        "name":"marge",
        "house": {
            "floor": 1,
            "room": 
            {
                "bed": "bed_blue",
                "chair":"chair_red"
            }
        }

    }]
    )    

     db.test.find({"house.room.bed":"bed_blue"})

我只想返回搜索的最后一级的值.在这种情况下:

I want to return only the value of the last level of my search. in this case:

            {
                "bed": "bed_blue",
                "chair":"chair_red"
            }

我得到的答案是整个文档，我想前进到查询的特定级别.我该怎么办?

the answer I get, is the whole document, I want to advance to a certain level of the query. how can I do it?

推荐答案

您可以将aggregate与$match和$project一起使用，如下所示:

You can use aggregate together with $match and $project for this as follows:

db.test.aggregate([
  {
    $match: {
      "house.room.bed": "bed_blue"
    }
  },
  {
    $project: {
      "_id": 0,
      "bed": "$house.room.bed",
      "chair": "$house.room.chair"
    }
  }
])

此外，您可以根据需要修改$project部分.

Further, you can modify the $project part as needed.

这是一个演示: https://mongoplayground.net/p/Fvll0LFIvQy

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