Question

我有一个以下格式的收藏集。

{
    "_id": "ffffc446-f33d",
    "className": "com.ezdx.vist.model.Visit",
    "organizationId": "0a0beff7-fe1e-4ab7",
    "centerId": "9aef68fe-dffd-4a7d-b0ee-f8dd3fc03303",
    "tests": [{
            "result": 157,
            "type": "PHYSICAL",
            **"name": "HEIGHT",**
            "consumableQuantity": 0,
            "testCost": 0,
            "testExpenseConsumable": 0

        },
        {
            "result": 8,
            "type": "RDT",
            **"name": "URIC",**
            "consumableQuantity": 0,
            "testCost": 0,
            "testExpenseConsumable": 0
        }

    ],
    "repeatVisit": true
}

我想要集合test.name =“ Uric”并具有特定列。

  {
    "result": 8,
    "type": "RDT",
    **"name": "Uric",**
    "consumableQuantity": 0,
    "testCost": 0,
    "testExpenseConsumable": 0
}

我设法以所需的方式进行收藏，但无法获得所需的格式。以下是我的查询

db.visits.aggregate( [ { $unwind : "$tests" }, 
{ $match: { $and: [{"tests.name":"URIC"}]
    } } ] )

Answer 1

尝试以下操作：$replaceWith（= v4.2）或$replaceRoot（> = v3.4）

db.visits.aggregate([
  {
    $unwind: "$tests"
  },
  {
    $match: {
      "tests.name": "URIC"
    }
  },
  {
    $replaceWith: "$tests"
  }
])

MongoPlayground

Answer 2

作为Valijon答案的替代方法，可以使用$filter聚合，这可能会更快，因为我们没有应用$ unwind。

db.collection.aggregate([
  {
    $project: {
      items: {
        $filter: {
          input: "$tests",
          as: "item",
          cond: {
            $eq: [
              "$$item.name",
              "URIC"
            ]
          }
        }
      }
    }
  },
  {
    $replaceRoot: {
      newRoot: {
        $arrayElemAt: [
          "$items",
          0
        ]
      }
    }
  }
])

Playground

从MongoDB获取嵌套数据

2 个答案: