使用组猫鼬时如何总结字段？

I have documents mongoDB in behind.

我后面有mongoDB文件。

{ "_id" : ObjectId("5ac06b6fb49124f4e5602817"), "status" : -1, "createdAt" : ISODate("2018-03-31T05:17:35.557Z") }

{“_ id”：ObjectId（“5ac06b6fb49124f4e5602817”），“status”： - 1，“createdAt”：ISODate（“2018-03-31T05：17：35.557Z”）}

{ "_id" : ObjectId("5ac0b51cb08b6c0014d1d0c1"), "status" : 0, "createdAt" : ISODate("2018-04-01T10:31:56.199Z") }

{“_ id”：ObjectId（“5ac0b51cb08b6c0014d1d0c1”），“status”：0，“createdAt”：ISODate（“2018-04-01T10：31：56.199Z”）}

{ "_id" : ObjectId("5ac0b538b08b6c0014d1d0c2"), "status" : 2, "createdAt" : ISODate("2018-04-01T10:32:24.542Z") }

{“_ id”：ObjectId（“5ac0b538b08b6c0014d1d0c2”），“status”：2，“createdAt”：ISODate（“2018-04-01T10：32：24.542Z”）}

i used $group, $match, $sum and i only can return: { _id: { month: 4, day: 1, year: 2018 }, count: 2 }

我使用$ group，$ match，$ sum我只能返回：{_ id：{month：4，day：1，year：2018}，count：2}

i want return: { _id: { month: 4, day: 1, year: 2018 }, count : 2, countStatus0 : 1, countStatus2 : 1 }

我想要返回：{_ id：{month：4，day：1，year：2018}，count：2，countStatus0：1，countStatus2：1}

i wish you help me... Thank you so much. Sorry my English is bad.

我希望你帮助我......非常感谢你。对不起，我的英语不好。

1 个解决方案

#1

You can try below aggregation:

您可以尝试以下聚合：

db.col.aggregate([
  {
    $group: {
      _id: {
        year: { $year: "$createdAt" },
        month: { $month: "$createdAt" },
        day: { $dayOfMonth: "$createdAt" },
        status: "$status"
      },
      count: { $sum: 1 }
    }
  },
  {
    $group: {
      _id: {
        year: "$_id.year",
        month: "$_id.month",
        day: "$_id.day"
      },
      count: { $sum: "$count" },
      statuses: { $push: { k: { $concat: ["status", {$substr:["$_id.status", 0, -1 ]}] }, v: "$count" } }  
    }
  },
  {
    $project: {
      _id: 1,
      count: 1,
      statuses: { $arrayToObject: "$statuses" }
    }
  },
  {
    $replaceRoot: {
      newRoot: {
        $mergeObjects: [ "$$ROOT", "$statuses" ]
      }
    }
  },
  {
    $project: {
      statuses: 0
    }
  }
])

To get total sum and sums for each status you have to use $group twice. Each partial status can be converted to a pair of k and v where key is a dynamically generated string which is a concatenation of status and its value like status0, status2. Having those pairs you can use $arrayToObject to generate object fields from that array. Then you can use $mergeObjects with $replaceRoot to merge root with nested statuses object.

要获得每个状态的总和和总和，您必须使用$ group两次。每个部分状态可以转换为一对k和v，其中key是动态生成的字符串，它是状态及其值的串联，如status0，status2。拥有这些对，您可以使用$ arrayToObject从该数组生成对象字段。然后，您可以使用$ mergeObjects和$ replaceRoot将root与嵌套状态对象合并。

There is one trick here. Since MongoDB does not allow to concatenate string with double we have to use $substr on double to convert it to a string.

这里有一个技巧。由于MongoDB不允许将字符串与double连接，我们必须在double上使用$ substr将其转换为字符串。

#1