Node.js 如何更改MongoDB输出JSON格式

我是MongoDb的新手，希望您能帮我解答这个问题。我编写了以下聚合管道

db.collection1.aggregate([
    { "$match": { "type" : "L" }},
    { "$facet": {
        "ON": [
            { "$match" : {"lampStatus":'ON'}},
            { "$count": "ON" }
        ],
        "OFF": [
            { "$match" : {"lampStatus": 'OFF'}},
            { "$count": "OFF" }
        ]
    } },
    { "$project": {
        "ON": { "$ifNull": [{ "$arrayElemAt": ["$ON.ON", 0] }, 0 ] },
        "OFF": { "$ifNull": [{ "$arrayElemAt": ["$OFF.OFF", 0] }, 0 ] },
    } }
])

得到了这种类型的输出

{
    "ON" : 0.0,
    "OFF": 30
} 

但是如何使用json格式获得这种类型的输出呢

/* 1 */ 
{ 
    "_id": "ON", 
    "count" : 0.0 
} 

/* 2 */ 
{ 
    "_id": "OFF", 
    "count" : 30.0 
} 

有谁能推荐我吗

实际产出

{
    "ON" : 0.0,
    "OFF" : 30
}

预期输出：

{
    "_id" : "ON",
    "COUNT" : 0.0
}

/* 2 */
{
    "_id" : "OFF",
    "COUNT" : 30.0
}

---

升级

db.collection.aggregate([
    { "$match": { "type" : "L" }},
    { "$facet": {
        "ON": [
            { "$match" : {"lampStatus":'ON'}},
            { "$count": "ON" }
        ],
        "OFF": [
            { "$match" : {"lampStatus": 'OFF'}},
            { "$count": "OFF" }
        ]
    } },
    {"$project": { 
            myArray: [ 
                {_id:"ON", count: { "$ifNull": [{ "$arrayElemAt": ["$ON.ON", 0] }, 0 ]} },
                {_id:"OFF", count: { "$ifNull": [{ "$arrayElemAt": ["$OFF.OFF", 0] }, 0 ] }}
        ] 
        }},

        {"$unwind":"$myArray"},

        {
            "$replaceRoot": { newRoot: "$myArray" }
        }

])

---

升级

db.collection.aggregate([
    { "$match": { "type" : "L" }},
    { "$facet": {
        "ON": [
            { "$match" : {"lampStatus":'ON'}},
            { "$count": "ON" }
        ],
        "OFF": [
            { "$match" : {"lampStatus": 'OFF'}},
            { "$count": "OFF" }
        ]
    } },
    {"$project": { 
            myArray: [ 
                {_id:"ON", count: { "$ifNull": [{ "$arrayElemAt": ["$ON.ON", 0] }, 0 ]} },
                {_id:"OFF", count: { "$ifNull": [{ "$arrayElemAt": ["$OFF.OFF", 0] }, 0 ] }}
        ] 
        }},

        {"$unwind":"$myArray"},

        {
            "$replaceRoot": { newRoot: "$myArray" }
        }

])

---

如果你需要更多的聚合技巧

db.collection1.aggregate([
  { "$match": { "type" : "L" }},
  { "$facet": {
    "ON": [{ "$match" : {"lampStatus": "ON" }}, { "$count": "ON" }],
    "OFF": [{ "$match" : {"lampStatus": "OFF" }}, { "$count": "OFF" }]
  }},
  { "$project": {
    "ON": { "$ifNull": [{ "$arrayElemAt": ["$ON.ON", 0] }, 0 ] },
    "OFF": { "$ifNull": [{ "$arrayElemAt": ["$OFF.OFF", 0] }, 0 ] },
  }},
  { "$project": {
    "data": {
      "$map": { "input": { "$objectToArray": "$$ROOT" }, "in": { "_id": "$$this.k", "count": "$$this.v" }}
    }
  }},
  { "$unwind": "$data" },
  { "$replaceRoot": { "newRoot": "$data" }}
])

---

如果你需要更多的聚合技巧

db.collection1.aggregate([
  { "$match": { "type" : "L" }},
  { "$facet": {
    "ON": [{ "$match" : {"lampStatus": "ON" }}, { "$count": "ON" }],
    "OFF": [{ "$match" : {"lampStatus": "OFF" }}, { "$count": "OFF" }]
  }},
  { "$project": {
    "ON": { "$ifNull": [{ "$arrayElemAt": ["$ON.ON", 0] }, 0 ] },
    "OFF": { "$ifNull": [{ "$arrayElemAt": ["$OFF.OFF", 0] }, 0 ] },
  }},
  { "$project": {
    "data": {
      "$map": { "input": { "$objectToArray": "$$ROOT" }, "in": { "_id": "$$this.k", "count": "$$this.v" }}
    }
  }},
  { "$unwind": "$data" },
  { "$replaceRoot": { "newRoot": "$data" }}
])

---

Iam尝试此代码Iam没有得到确切的结果Iam得到类似{u id:“ON”，“myArray:[0.0,0.0]}/*2*/{u id:“OFF”，“myArray:[0.0,0.0]}的类型结果@abdul@what关于现在？我正在尝试这个{$group:{“\u id”：“$lampStatus”，COUNT:{$sum:1}}}，我得到的是json的精确结果，但很小的ISSUE lampstatus:“ON”没有记录，所以没有得到{“\u id”：“ON”，“COUNT”：0.0}，value@NareshVarunGuttula更新了我的答案，这是另一种解决方案。非常感谢@abdulIam尝试此代码我没有得到确切的结果我得到了类似{“\u id”：“ON”的类型结果，myArray:[0.0,0.0]}/*2*/{“\u id:“OFF”，“myArray:[0.0,0.0]}@abdul@what关于现在？我正在尝试这个{$group:{“\u id”：“$lampStatus”，COUNT:{$sum:1}}}，我得到的是json精确的结果，但很小的是，Issue lampStatus:“ON”没有记录，所以没有得到{“\u id”：“开”，“计数”：0.0}，value@NareshVarunGuttula更新了我的答案，这是另一种解决方案。非常感谢@abdulGot it非常感谢@Anthony Winzlet非常感谢@Anthony Winzlet