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...",
}

我需要按status进行分组，并动态获取status

中的所有密钥

[
  {
    "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

并非我认为这是一个好主意，主要是因为我没有看到任何＆＃34;聚合＆＃34;在这里，＆＃34;分组＆＃34;要通过"status"分组键将所有内容添加到数组中，$push将所有内容添加到数组中，然后使用$replaceRoot转换为$arrayToObject中文档的键：

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

返回：

{
        "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..."
                }
        ]
}

那可能没事 IF 你实际上是＆＃34;聚合＆＃34;事先，但在任何实际大小的集合上，所有正在做的就是试图将整个集合强制转换为单个文档，并且这可能会破坏BSON Limit of 16MB，所以我不建议甚至在没有＆＃34;分组＆＃34;在此步骤之前的其他事情。

坦率地说，下面的代码完全相同，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);

或者更短一些：

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＆＃34;数据缩减＆＃34;而任何简单的＆＃34;重塑结果＆＃34;在没有实际减少从服务器返回的数据的情况下，通常可以在客户端代码中更好地处理。无论你做什么，你仍然会返回所有数据，并且游标的客户端处理开销要小得多。没有限制。

$ group by后的动态键

1 个答案: