获取mongo嵌套属性作为列

Question

下面是我的mongo结构

{ 
    "_id" : ObjectId("5bfbc18e28198e2f74e1d2d0"), 
    "address" : {
        "building" : "1007", 
        "coord" : [
            -73.856077, 
            40.848447
        ], 
        "street" : "Morris Park Ave", 
        "zipcode" : "10462"
    }, 
    "borough" : "Bronx", 
    "cuisine" : "Bakery"
}

启动db.Restaurant.find（{“ address.building”：“ 1007”}，{“ address.building”：1，“ address.street”：1，“ address.zipcode”：1}）  查询结果为

{ 
    "_id" : ObjectId("5bfbc18e28198e2f74e1d2d0"), 
    "address" : {
        "building" : "1007", 
        "street" : "Morris Park Ave", 
        "zipcode" : "10462"
    }
}

因此它仍以嵌套文档的形式返回结果，但我想将这些嵌套属性作为列名使用，并且不希望使用嵌套结构。我该如何修改此查询？

Answer 1

find查询只能限制字段。如果要重塑数据，则需要使用aggregation

db.collection.aggregate([
  { "$match": { "address.building": "1007" }},
  { "$replaceRoot": { "newRoot": "$address" }},
  { "$project": { "coord": 0 }}
])

Answer 2

您可以尝试以下汇总。它首先匹配您的任何条件，然后投影必填字段，而不会嵌套。

db.test.aggregate([
    {
        $match : {
            "address.building":"1007"
        }
    },
    {
        $project : {
            building : "$address.building",
            street : "$address.street",
            zipcode : "$address.zipcode"
        }
    }
])

输出：

{
    "_id" : ObjectId("5bfbc18e28198e2f74e1d2d0"),
    "building" : "1007",
    "street" : "Morris Park Ave",
    "zipcode" : "10462"
}

让我知道这是否有帮助！