$ group by之后的动态键

Question

我有以下收藏

{
    "_id" : ObjectId("5b18d14cbc83fd271b6a157c"),
    "status" : "pending",
    "description" : "You have to complete the challenge...",
}
{
    "_id" : ObjectId("5b18d31a27a37696ec8b5773"),
    "status" : "completed",
    "description" : "completed...",
}
{
    "_id" : ObjectId("5b18d31a27a37696ec8b5775"),
    "status" : "pending",
    "description" : "pending...",
}
{
    "_id" : ObjectId("5b18d31a27a37696ec8b5776"),
    "status" : "inProgress",
    "description" : "inProgress...",
}

我需要分组status并动态获取其中的所有密钥status

[
  {
    "completed": [
      {
        "_id": "5b18d31a27a37696ec8b5773",
        "status": "completed",
        "description": "completed..."
      }
    ]
  },
  {
    "pending": [
      {
        "_id": "5b18d14cbc83fd271b6a157c",
        "status": "pending",
        "description": "You have to complete the challenge..."
      },
      {
        "_id": "5b18d31a27a37696ec8b5775",
        "status": "pending",
        "description": "pending..."
      }
    ]
  },
  {
    "inProgress": [
      {
        "_id": "5b18d31a27a37696ec8b5776",
        "status": "inProgress",
        "description": "inProgress..."
      }
    ]
  }
]

Answer 1

并不是说我认为这是个好主意，主要是因为我根本看不到任何“聚合”，是在将“分组”添加到数组后，您可以$push通过"status"分组键将所有类似的内容添加到数组中，然后转换为键在文档的$replaceRoot使用$arrayToObject：

db.collection.aggregate([
  { "$group": {
    "_id": "$status",
    "data": { "$push": "$$ROOT" }
  }},
  { "$group": {
    "_id": null,
    "data": {
      "$push": {
        "k": "$_id",
        "v": "$data"
      }
    }
  }},
  { "$replaceRoot": {
    "newRoot": { "$arrayToObject": "$data" }
  }}
])

返回值：

{
        "inProgress" : [
                {
                        "_id" : ObjectId("5b18d31a27a37696ec8b5776"),
                        "status" : "inProgress",
                        "description" : "inProgress..."
                }
        ],
        "completed" : [
                {
                        "_id" : ObjectId("5b18d31a27a37696ec8b5773"),
                        "status" : "completed",
                        "description" : "completed..."
                }
        ],
        "pending" : [
                {
                        "_id" : ObjectId("5b18d14cbc83fd271b6a157c"),
                        "status" : "pending",
                        "description" : "You have to complete the challenge..."
                },
                {
                        "_id" : ObjectId("5b18d31a27a37696ec8b5775"),
                        "status" : "pending",
                        "description" : "pending..."
                }
        ]
}

如果
您实际上事先进行了“汇总”，那可能就可以了，但是在任何实际大小的集合上，所有要做的就是将整个集合强行放入一个文档中，这很可能会超出BSON限制（16MB），所以我只是不建议在此步骤之前，甚至尝试不进行“分组”操作。

坦白地说，以下相同的代码可以完成相同的操作，并且没有聚合技巧，也没有BSON限制问题：

var obj = {};

// Using forEach as a premise for representing "any" cursor iteration form
db.collection.find().forEach(d => {
  if (!obj.hasOwnProperty(d.status))
    obj[d.status] = [];
  obj[d.status].push(d);
})

printjson(obj);

或更短：

var obj = {};

// Using forEach as a premise for representing "any" cursor iteration form
db.collection.find().forEach(d => 
  obj[d.status] = [ 
    ...(obj.hasOwnProperty(d.status)) ? obj[d.status] : [],
    d
  ]
)

printjson(obj);

聚合用于“数据精简”，而任何简单地“重塑结果”却不实际减少从服务器返回的数据的方法通常都可以通过客户端代码更好地处理。无论您做什么，仍将返回所有数据，并且游标的客户端处理的开销要小得多。而且没有限制。