Mongodb查询返回计数？答案

【问题标题】：Mongo DB query to return count?Mongodb查询返回计数？
【发布时间】：2021-12-06 07:42:26
【问题描述】：

我有两个集合 Course 和 File。这是我来自Course 集合的数据

[{
    _id: 1,
    name: "Course1",
    price: "100"
},
{
    _id: 2,
    name: "Course2",
    price: "200"
}]

这是我收集的文件中的数据

[{
    _id: 1,
    name: "File1",
    isFree: true,
    courseId: 1
}
{
    _id: 2,
    name: "File2",
    isFree: false,
    courseId: 1
}
{
    _id: 3,
    name: "File3",
    isFree: false,
    courseId: 1
}]

我想用Course name,Count of files which are free,Count of files which are not free 检索课程 1。

我的尝试：

Course.aggregate([
    {$lookup:{
        from:"files",
        localField: "_id",
        foreignField: "courseId",
        as: "files"
    }},
])

如果我执行上述查询，这是我的回复：

[
    {
        "_id": "6161bdcc0aada09d81598840",
        "name": "Course1",
        "files": [
            {
                "_id": "6166a0814ad3f56b71232b9c",
                "name": "File1.jpeg",
                "isFree":true
            },
            {
                "_id": "6166a0814ad3f56b71232b9d",
                "name": "File2.jpeg",
                "isFree":false
            },
            {
                "_id": "6166a0814ad3f56b71232b9v",
                "name": "File3.jpeg",
                "isFree":false
            }
        ]
    }
]

我希望我的输出为：

[
    {
        "_id": "6161bdcc0aada09d81598840",
        "name": "Course1",
        "files": {"free": 1,"paid": 2}
    }
]

【问题讨论】：

标签： javascript database mongodb mongoose

【解决方案1】：

您可以尝试使用管道查找，

$lookup 与管道，通过 courseId 在 let
$matchcourseId条件
$group by isFeee 并获得总数
$project 以 k(key) 和 v(value) 格式显示必填字段
$cond 检查 isFree 是否为真，然后返回“免费”，否则返回“付费”
$addFields 更新files 数组
$arrayToObject 将files 键值数组转换为对象
$ifNull 检查属性是否不存在然后设置 0

await Course.aggregate([
  {
    $lookup: {
      from: "files",
      let: { courseId: "$_id" },
      pipeline: [
        { $match: { $expr: { $eq: ["$$courseId", "$courseId"] } } },
        {
          $group: {
            _id: "$isFree",
            count: { $sum: 1 }
          }
        },
        {
          $project: {
            _id: 0,
            k: { $cond: ["$_id", "free", "paid"] },
            v: "$count"
          }
        }
      ],
      as: "files"
    }
  },
  {
    $addFields: {
      files: { $arrayToObject: "$files" }
    }
  },
  {
    $addFields: {
      "files.free": { $ifNull: ["$files.free", 0] },
      "files.paid": { $ifNull: ["$files.paid", 0] }
    }
  }
])

Playground

【讨论】：

【解决方案2】：

尝试以下操作：

db.course.aggregate([
  {
    $lookup: {
      from: "files",
      localField: "_id",
      foreignField: "courseId",
      as: "files",
    },
  },
  {
    $unwind: {
      path: "$files",
      preserveNullAndEmptyArrays: true,
    },
  },
  {
    $group: {
      _id: "$_id",
      name: {
        $first: "$name",
      },
      price: {
        $first: "$price",
      },
      free: {
        $sum: {
          $cond: [
            {
              $eq: ["$files.isFree", true],
            },
            1,
            0,
          ],
        },
      },
      paid: {
        $sum: {
          $cond: [
            {
              $eq: ["$files.isFree", false],
            },
            1,
            0,
          ],
        },
      },
    },
  },
  {
    $addFields: {
      "files.free": "$free",
      "files.paid": "$paid",
    },
  },
  {
    $project: {
      _id: 1,
      name: 1,
      files: 1,
      price: 1,
    },
  },
]);

【讨论】：