Question

这里我有一个示例嵌套数组。我在这个深度嵌套的集合上编写正确的查询时遇到了问题。

{
"productUUID" : "craft001",
"providers": [
{
  "providerUUID": "prov001",
  "orgs": [
    {
      "orgUUID": "org001",
      "location": {
        "buildings": [
          {
            "buildingUUID": "sit001",
            "floors": [
              {
                "floorUUID": "GrndFlr",
                "assets": [ ],
                "agents": [ ],
                "users": [ ]
              },
              {
                "floorUUID": "1stFlr",
                "assets": [ ],
                "agents": [ ],
                "users": [ ]
              }
            ]
          },
          {
            "buildingUUID": "ist001",
            "floors": [ ]
          }
        ]
      }
    },
    {
      "orgUUID": "org002",
      "location": {
        "buildings": [ ]
      }
    }
  ]
},
{
  "providerUUID": "prov002",
  "orgs": [ ]
}
]
}

用简单的词语提问，“1。获取orgUUID以下的所有providerUUID: "prov001"”。同样，“2。将所有floorUUID设为"buildingUUID": "sit001"”。

如果有人可以帮我解决第一个问题，我希望自己可以解决第二个问题。

Answer 1

Mongo aggregation用于查找嵌套文档。首先unwind所有providers数组然后使用match匹配providerUUID给定prov001，然后使用project获取所有orgUUID和聚合查询为：

db.collectionName.aggregate({"$unwind":"$providers"},
                            {"$match":{"providers.providerUUID":"prov001"}},
                            {"$project":{"orgUUID":"$providers.orgs.orgUUID"}},
                            {"$unwind":"$orgUUID"},
                            {"$project":{"_id":0,"orgUUID":1}}
                           ).pretty()

这将返回数组中的所有orgUUID。

如果您使用$elemMacth，则此运营商将其自身限制为

$ elemMatch运算符匹配包含数组字段的文档，其中至少有一个元素符合所有指定的查询条件。

elemMatch查询为：

 db.collectionName.find({"providers":{"$elemMatch":{"providerUUID":"prov001"}}},
                        {"providers.$.providerUUID.orgs.orgUUID":1}).pretty()

它返回整个匹配的providers数组。

我希望你能自己找出＆＃34; 2＆＃34; 问题查询，如果你在找到＆＃34; 2＆＃34; 查询我也会发布＆＃34; 2＆＃34; 查询。尝试自己找出第二个查询答案：）

Answer 2

出于某种原因，我必须更改集合中的数据，如下所示。

{
"productUUID": "prod001",
"providers": [
{
  "providerUUID": "prov001",
  "orgs": [
    {
      "orgUUID": "org001",
      "floors": [
        { "floorUUID": "SIT_GrndFlr" },
        { "floorUUID": "SIT_1stFlr" }
      ],
      "assets": [{},{}],
      "agents": [{},{}],
      "users": [{},{}]
    },
    {
      "orgUUID": "org002",
      "floors": [
        { "floorUUID": "IST_1stFlr" },
        { "floorUUID": "IST_2ndFlr" }
      ],
      "assets": [{},{}],
      "agents": [{},{}],
      "users": [{},{}]
    }
  ]
},
{
  "providerUUID": "prov002",
  "orgs": [
    {
      "orgUUID": "org001",
      "floors": [{},{}],
      "assets": [{},{}],
      "agents": [{},{}],
      "users": [{},{}]
    },
    {
      "orgUUID": "org002",
      "floors": [{},{}],
      "assets": [{},{}],
      "agents": [{},{}],
      "users": [{},{}]
    }
  ]
}
]
}

所以，现在在@yogesh的帮助下，我被介绍到aggregate并且能够为我的问题编写查询。

1. Get all `orgUUID`s under `providerUUID: "prov001"`.
db.collectionName.aggregate({"$unwind":"$providers"},
                        {"$match":{"providers.providerUUID":"prov001"}},
                        {"$project":{"orgUUID":"$providers.orgs.orgUUID"}},
                        {"$unwind":"$orgUUID"},
                        {"$project":{"_id":0,"orgUUID":1}}
                       )

2. Get all `floorUUID`s under `orgUUID : "org001"`.
db.collectionName.aggregate(
    { "$unwind" : "$providers" },    
    { "$match" : { "providers.providerUUID" : "prov001" } },    
    { "$unwind" : "$providers.orgs" },    
    { "$match" : { "providers.orgs.orgUUID" : "org001" } },    
    { "$project" : { "floorUUID" : "$providers.orgs.floors.floorUUID" } },    
    { "$unwind" : "$floorUUID" },    
    { "$project" : { "_id":0 , "floorUUID" : 1 } }
)

从MongoDB中的嵌套数组中查找单个字段

2 个答案: