MongoDB组最小与条款

发布于 2025-02-09 05:16:27 字数 455 浏览 0 评论 0 原文

希望将结果与最低日期汇总在一起，但是最小应该仅适用于状态为真的字段？

鉴于以下数据，$ dateModified的$ min将返回日期为2，但是我希望返回3，因为它是最低的数据，并具有真实状态，

{name: "one", dateModified: "4", status: true},
{name: "one", dateModified: "2", status: false},
{name: "one", dateModified: "3", status: true}

.aggregate([
   $group: {
   _id: "$name",
   date: {$min: "$dateModified" // where status is true},
   total: {$sum: 1}
  }
])

我还需要真实和错误条目的整体计数，因此请应用一个匹配状态：在小组之前是正确的

原文

Looking to aggregate a result with the minimum date however the min should apply only to fields where the status is true?

Given the following data, $min on $dateModified will return date as 2 however I'm looking to return 3 because it is the lowest dateModified with the status of true

{name: "one", dateModified: "4", status: true},
{name: "one", dateModified: "2", status: false},
{name: "one", dateModified: "3", status: true}

.aggregate([
   $group: {
   _id: "$name",
   date: {$min: "$dateModified" // where status is true},
   total: {$sum: 1}
  }
])

I'll also need the overall count of true and false entries so applying a match on status: true before the group may prevent that

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星 2025-02-16 05:16:27

为什么不只是添加 $ match 阶段作为第一个使用状态过滤文档的阶段：true ？

db.collection.aggregate([
  {
    $match: {
      status: true
    }
  },
  {
    $group: {
      _id: "$name",
      date: {
        $min: "$dateModified"
      }
    }
  }
])

示例mongo playground

当您热衷于保持文档的状态：false 状态：false ，然后您可以与 $ cond> $ cond 和。

db.collection.aggregate([
  {
    $group: {
      _id: "$name",
      date: {
        $min: {
          $cond: {
            if: {
              $eq: [
                "$status",
                true
              ]
            },
            then: "$dateModified",
            else: "$REMOVE"
          }
        }
      }
    }
  }
])

示例mongo Playground 2

Why not just add the $match stage as the first stage to filter the documents with status: true only?

db.collection.aggregate([
  {
    $match: {
      status: true
    }
  },
  {
    $group: {
      _id: "$name",
      date: {
        $min: "$dateModified"
      }
    }
  }
])

Sample Mongo Playground

As you are keen to remain the document with status: false, then you can work with $cond and $$REMOVE.

db.collection.aggregate([
  {
    $group: {
      _id: "$name",
      date: {
        $min: {
          $cond: {
            if: {
              $eq: [
                "$status",
                true
              ]
            },
            then: "$dateModified",
            else: "$REMOVE"
          }
        }
      }
    }
  }
])

Sample Mongo Playground 2

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