汇总多个文档和计数总字段

发布于 2025-02-08 17:23:47 字数 268 浏览 1 评论 0 原文

文档：

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

汇总结果：

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

有可能吗？我知道如何计数完成，条件但是如何汇总文档数组？

原文

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?

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岁月流歌 2025-02-15 17:23:47

解决方案1 

$ set - set finish_count 和 non_finish_count 通过检查 finish> finish 字段代码>，如果匹配，则1，else 0。
。 p> $ group - 组成 null ，总结 finish_count 和 non_finish_count fields。将每个文档添加到文档数组中。
$ unset - 删除 documents.finish_count 和 documents.non_finish_count fields。

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"
    ]
  }
])

示例mongo playground（解决方案1

）代码> $ group - null 组组，然后将每个文档添加到文档 array。
$ set - 通过使用 finish （ $ size ）来设置字段>完成：true ）和 non_finish （完成：false ）来自文档 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
              ]
            }
          }
        }
      }
    }
  }
])

示例mongo playground（解决方案2）

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)

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