Aggregate multiple documents and count sum field答案

【问题标题】：Aggregate multiple documents and count sum fieldAggregate multiple documents and count sum field
【发布时间】：2022-12-19 09:39:43
【问题描述】：

Document:

{
  _id: "___"
  finish: false or true
  ... some field..
}

Aggregate result:

{
  finish: 10,
  non_finish: 3,
  results: [
   docuemnt,
   docuemnt,
   ...
  ]
}

Is it possible? I know how can I count finish, with condition but how to aggregate document array?

【问题讨论】：

标签： mongodb aggregation-framework

【解决方案1】：

Solution 1

$set - Set finish_count and non_finish_count fields by checking the finish, if match then 1, else 0.
$group - Group by null, sum the finish_count and non_finish_count fields. Add each document into documents array.

$unset - Remove documents.finish_count and documents.non_finish_countfields.

db.collection.aggregate([
  {
    $set: {
      finish_count: {
        $cond: {
          if: {
            $eq: [
              "$finish",
              true
            ]
          },
          then: 1,
          else: 0
        }
      },
      non_finish_count: {
        $cond: {
          if: {
            $eq: [
              "$finish",
              false
            ]
          },
          then: 1,
          else: 0
        }
      }
    }
  },
  {
    $group: {
      _id: null,
      finish: {
        $sum: "$finish_count"
      },
      non_finish: {
        $sum: "$non_finish_count"
      },
      documents: {
        $push: "$$ROOT"
      }
    }
  },
  {
    $unset: [
      "documents.finish_count",
      "documents.non_finish_count"
    ]
  }
])

Sample Mongo Playground (Solution 1)

Solution 2

$group - Group by null, and add each document into the documents array.

$set - Set the fields by counting the size ($size) of the array by filtering the document with finish (finish: true) and non_finish (finish: false) from the documents array.

db.collection.aggregate([
  {
    $group: {
      _id: null,
      documents: {
        $push: "$$ROOT"
      }
    }
  },
  {
    $set: {
      finish: {
        $size: {
          $filter: {
            input: "$documents",
            cond: {
              $eq: [
                "$$this.finish",
                true
              ]
            }
          }
        }
      },
      non_finish: {
        $size: {
          $filter: {
            input: "$documents",
            cond: {
              $eq: [
                "$$this.finish",
                false
              ]
            }
          }
        }
      }
    }
  }
])

Sample Mongo Playground (Solution 2)

【讨论】：