组数 2 项对象数组 mongodb答案

【问题标题】：Group count 2 item object array mongodb组数 2 项对象数组 mongodb
【发布时间】：2019-02-10 00:57:28
【问题描述】：

我有以下数组：

[
  {
    “counter”: 123456,
    “user”: “USER1”,
    “last”: “USER1”
  },
  {
    “counter”: 123,
    “user”: “USER1”,
    “last”: “USER2”
  },
  {
    “counter”: 111,
    “user”: “USER2”,
    “last”: “USER2”
  },
  {
    “counter”: 1122,
    “user”: “USER2”,
    “last”: “USER2”
  },
  {
    “counter”: 112233,
    “user”: “USER1”,
    “last”: “USER2”
  },
]

我在 mongodb 上做如下查询：

{$group: {
  _id: “$user”,
  total: {$sum: 1},
  last: {$sum: {$cond: [
    {$eq: ['$last’, '$user']}, 1, 0
  ]}}
}}

我想得到以下结果：

[
  {
    _id: “USER1”
    total: 3,
    last: 1
  },
{
    _id: “USER2”
    total: 2,
    last: 4
  }
]

但我明白了：

[
  {
    _id: “USER1”
    total: 3,
    last: 1
  },
{
    _id: “USER2”
    total: 2,
    last: 2
  }
]

当我分组时，我无法满意地计算最后一项

我怎样才能得到预期的结果？感谢您的帮助。

【问题讨论】：

标签： mongodb aggregation-framework

【解决方案1】：

您只有 2 个“USER2”，然后不能在该 _id 上 $sum 4 $last。如果您需要用户中的索引，请尝试使用 compose _id。

db.teste.aggregate([
  {$group:{
    _id: {user:"$user",last:"$last"},
    total: {$sum:1},
    last: {
      $sum: {
        $cond:[ {$eq: ["$last", user]}, 1, 0]
      }
    }
  }}
])

如果您需要在不同的字段中添加字符串或其他匹配项。查询是：

db.teste.aggregate([
  {$group:{
    _id: "$user"
  }},
  {$lookup:{
    from: "teste",
    let: { indice: "$_id" },
    pipeline: [
      {$group:{
        _id: null,
        user:{$sum:{$cond:[
          {$eq:["$user", "$$indice"]}, 1, 0
        ]}},
        last:{$sum:{$cond:[
          {$eq:["$last", "$$indice"]}, 1, 0
        ]}}
      }},
      {$project:{
        _id: 0
      }}
    ],
    as: "res"
  }}
])

从零到英雄：

//First we'll extract only what we need find, on this case distinct users
db.teste.aggregate([
  {$group:{
    _id: "$user"
  }}
])

//Here we will get the indexes of the search and reinsert in our filter using $let (nodejs var)
db.teste.aggregate([
  {$group:{
    _id: "$user"
  }},
  {$lookup:{
    from: "teste",
    let: { indice: "$_id" },
    pipeline: [],
    as: "res"
  }}
])

//Let's counter the total of reinserted elements
db.teste.aggregate([
  {$group:{
    _id: "$user"
  }},
  {$lookup:{
    from: "teste",
    let: { indice: "$_id" },
    pipeline: [
      {$group:{
        _id: null,
        total:{$sum:1}
      }}
    ],
    as: "res"
  }}
])

//Now let's test a true condition
db.teste.aggregate([
  {$group:{
    _id: "$user"
  }},
  {$lookup:{
    from: "teste",
    let: { indice: "$_id" },
    pipeline: [
      {$group:{
        _id: null,
        user:{$sum:{$cond:[
          {$eq:[true, true]}, 1, 0
        ]}}
      }}
    ],
    as: "res"
  }}
])

//cond tested let's extract what we want
db.teste.aggregate([
  {$group:{
    _id: "$user"
  }},
  {$lookup:{
    from: "teste",
    let: { indice: "$_id" },
    pipeline: [
      {$group:{
        _id: null,
        user:{$sum:{$cond:[
          {$eq:["$user", "$$indice"]}, 1, 0
        ]}},
        last:{$sum:{$cond:[
          {$eq:["$last", "$$indice"]}, 1, 0
        ]}}
      }}
    ],
    as: "res"
  }}
])

//Let's take the _id of the sub-colection because we do not need it
db.teste.aggregate([
  {$group:{
    _id: "$user"
  }},
  {$lookup:{
    from: "teste",
    let: { indice: "$_id" },
    pipeline: [
      {$group:{
        _id: null,
        user:{$sum:{$cond:[
          {$eq:["$user", "$$indice"]}, 1, 0
        ]}},
        last:{$sum:{$cond:[
          {$eq:["$last", "$$indice"]}, 1, 0
        ]}}
      }},
      {$project:{
        _id: 0
      }}
    ],
    as: "res"
  }}
])

【讨论】：