如何在单个MongoDB查询结果中获取TotalCount和第一个文档？

发布于 2025-02-07 01:15:14 字数 588 浏览 0 评论 0原文

有一个过滤器myFilter用于获得有关XYZ集合的一些见解。现在，我需要运行三个查询，该查询具有相同的步骤以过滤文档

db.xyz.find(myFilter).sort(mySort1).limit(1) // get first item by sort #1
db.xyz.find(myFilter).sort(mySort2).limit(1) // get first item by sort #2
db.xyz.find(myFilter).count()                // get total count

如何在单个查询或聚合中表达相同意图？

预期结果是

{
  "item_1": {/* first doc by (myFilter & mySort1) *//},

  "item_2": {/* first doc by (myFilter & mySort2) *//},

  "totalCount": /* count of all by myFilter */
}

原文

There is a filter myFilter used to get some insights about the xyz collection.
Now I need to run three queries having the same step for filtering the documents

db.xyz.find(myFilter).sort(mySort1).limit(1) // get first item by sort #1
db.xyz.find(myFilter).sort(mySort2).limit(1) // get first item by sort #2
db.xyz.find(myFilter).count()                // get total count

How to express the same intention in a single query or aggregation?

Expected result is

{
  "item_1": {/* first doc by (myFilter & mySort1) *//},

  "item_2": {/* first doc by (myFilter & mySort2) *//},

  "totalCount": /* count of all by myFilter */
}

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纸短情长 2025-02-14 01:15:14

您可以使用$ facet在一个查询中运行相同数据的几个管道：

db.collection.aggregate([
  {$match: myFilter}
  {$facet: {
    item_1: [{sort1}],
    item_2: [{sort2}],
    totalCount: [{$group: {_id: null, count: {$sum:1}}}]
  }}
])

You can use $facet to run several pipelines on the same data at one query:

db.collection.aggregate([
  {$match: myFilter}
  {$facet: {
    item_1: [{sort1}],
    item_2: [{sort2}],
    totalCount: [{$group: {_id: null, count: {$sum:1}}}]
  }}
])

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