MongoDB：获取具有嵌套数组元素匹配的内部数组对象的计数

Question

我有 mongo 集合，其中包含每个用户提交的调查答案。我想获得选择作为选项的用户数。只有一位用户选择了选项 O12。输出应该是 1。

{
    "_id" : ObjectId("5ea179eb39ff117948f19266"),
    "_class" : "model.survey.Answer",
    "survey_id" : "5ea178c239ff117948f19265",
    "survey_user" : [ 
        {
            "user_id" : 1072,
            "user_option" : [ 
                {
                    "question_id" : "Q1",
                    "option_id" : "O11"
                }, 
                {
                    "question_id" : "Q2",
                    "option_id" : "O21"
                }, 
                {
                    "question_id" : "Q3",
                    "option_id" : "O31"
                }, 
                {
                    "question_id" : "Q4",
                    "option_id" : "O41"
                }
            ]
        }, 
        {
            "user_id" : 1073,
            "user_option" : [ 
                {
                    "question_id" : "Q1",
                    "option_id" : "O12"
                }, 
                {
                    "question_id" : "Q2",
                    "option_id" : "O21"
                }, 
                {
                    "question_id" : "Q3",
                    "option_id" : "O31"
                }, 
                {
                    "question_id" : "Q4",
                    "option_id" : "O41"
                }
            ]
        }
    ]
}

Answer 1

您可以使用 MongoDB 的 aggregation-pipeline 来做到这一点：

不同的方法，一种方法是使用$unwind：

类型 1 - 查询 1：

db.collection.aggregate([
    /** Optional but will be good on huge collections to lessen data for further stages */
    {
      $match: { "survey_user.user_option.option_id": "O12" }
    },
    {
      $unwind: "$survey_user"
    },
    /** When you unwind a each object/element in array gets it's own document after `unwind` stage */
    {
      $match: { "survey_user.user_option.option_id": "O12" }
    },
    /** After match you'll only have objects which met the criteria in `survey_user` array */
    /** group on `_id` & push entire original doc to data field */
    {
      $group: { _id: "$_id", survey_user: { $push: "$survey_user" }, data: {  $first: "$$ROOT" } }
    },
    /** Add `survey_user` array to `data.survey_user` & it's size to `data.optedCount` field */
    {
      $addFields: { "data.survey_user": "$survey_user", "data.optedCount": {  $size: "$survey_user" } }
    },
    /** Make `data` as new root to doc */
    {
      $replaceRoot: { newRoot: "$data" }
    }
  ])

测试： mongoplayground

以防万一您只需要计数但不需要返回整个文档，则上述查询将发生细微变化：

类型 1 - 查询 2：

db.collection.aggregate([
    {
      $match: { "survey_user.user_option.option_id": "O12" }
    },
    {
      $unwind: "$survey_user"
    },
    {
      $match: { "survey_user.user_option.option_id": "O12" }
    },
    /** Just group on `_id` & count no.of docs, maintain `survey_id` */
    {
      $group: { _id: "$_id", optedCount: { $sum: 1 }, survey_id: { $first: "$survey_id" } }
    }
  ])

测试： mongoplayground

使用数组迭代器$reduce，如果您的集合数据如此庞大，这可能会有所帮助，因为展开会爆炸您的文档。

类型 2 - 查询：

db.collection.aggregate([
  {
    $match: {
      "survey_user.user_option.option_id": "O12",
    },
  },
  /** Instead of `$addFields`, you can use `$project` to project fewer needed fields (which can be help improve query with performance benefits ) */
  {
    $addFields: {
      optedCount: {
        $reduce: {
          input: "$survey_user",
          initialValue: 0,
          in: {
            $cond: [
              { $in: ["O12", "$$this.user_option.option_id"] },
              { $add: ["$$value", 1] },
              "$$value",
            ]
          }
        }
      }
    }
  }
]);

测试： mongoplayground