如何使用聚合检查 MongoDB 数据库中的所有文档以获取特定字段的最新值？答案

【问题标题】：How to check all the documents in a MongoDB database for the latest value of a particular field using aggregation?如何使用聚合检查 MongoDB 数据库中的所有文档以获取特定字段的最新值？
【发布时间】：2020-03-07 07:00:10
【问题描述】：

我是 MongoDB 新手，正在使用它处理 NodeJS 代码。

我需要使用聚合函数来拉取每本书的最新book_version。

这是我的数据库的外观：

[
  {
    "book_id": "ab12nld”,
    "book_version": "0”,
    "author": “Sam”,
    “name”: “Sample Book”,
    “comments”: “Done”
  },
  {
    "book_id": "ab12nld”,
    "book_version": "1",
    "author": "Martin",
    "name": "Sample Book",
    “comments”: “In Progress”
  },
  {
    "book_id": "ab12nld”,
    "book_version": "2",
    "author": "Roy",
    "name": "Sample Book",
    “comments”: “To-Do”
  }
]
[
  {
    "book_id": "bcj123n”,
    "book_version": "0”,
    "author": “Don”,
    “name”: “Another Book”,
    “comments”: “Done”
  },
  {
    "book_id": "bcj123n”,
    "book_version": "1",
    "author": "Ray",
    "name": "Another Book",
    “comments”: “In Progress”
  },
  {
    "book_id": "bcj123n”,
    "book_version": "2",
    "author": "Max",
    "name": "Another Book",
    “comments”: “To-Do”
  }
]
[
  {
    "book_id": "k23d1d3”,
    "book_version": "0”,
    "author": “Sunny”,
    “name”: “New Book”,
    “comments”: “Done”
  },
  {
    "book_id": "k23d1d3”,
    "book_version": "1",
    "author": "Archer",
    "name": "New Book",
    “comments”: “In Progress”
  }
]
[
  {
    "book_id": "o902f3s”,
    "book_version": "0”,
    "author": “Joy”,
    “name”: “Nice Book”,
    “comments”: “Done”
  },
  {
    "book_id": "o902f3s”,
    "book_version": "1",
    "author": "James",
    "name": "Nice Book",
    “comments”: “In Progress”
  },
{
    "book_id": "o902f3s”,
    "book_version": "2”,
    "author": “Pam”,
    “name”: “Nice Book”,
    “comments”: “Done”
  },
  {
    "book_id": "o902f3s”,
    "book_version": "3",
    "author": "Jonah",
    "name": "Nice Book",
    “comments”: “Completed”
  }
]

因此，我应该为每个 book_id 以及整个对象获得 book_version 的最大值：

{
    "book_id": "ab12nld”,
    "book_version": "2",
    "author": "Roy",
    "name": "Sample Book",
    “comments”: “To-Do”
  },
{
    "book_id": "bcj123n”,
    "book_version": "2",
    "author": "Max",
    "name": "Another Book",
    “comments”: “To-Do”
  },
{
    "book_id": "k23d1d3”,
    "book_version": "1",
    "author": "Archer",
    "name": "New Book",
    “comments”: “In Progress”
  },
  {
    "book_id": "o902f3s”,
    "book_version": "3",
    "author": "Jonah",
    "name": "Nice Book",
    “comments”: “Completed”
  }

【问题讨论】：

标签： mongodb mongoose aggregation-framework aggregation

【解决方案1】：

您可以使用$sort 和$group 的组合来达到预期的效果：

db.collection.aggregate([
  {
    $sort: {
      // Sort by the book_version from highest to lowest 
      book_version: -1
    }
  },
  {
    $group: {
      // Group the documents by their book_id
      _id: "$book_id",
       book_version: {
        // The first book_version is the highest book version
        // The same thing applies to every other first-of other book fields 
        $first: "$book_version"
      },
      book_id: {
        $first: "$book_id"
      },
      author: {
        "$first": "$author"
      },
      name: {
        "$first": "$name"
      },
      comments: {
        "$first": "$comments"
      }
    }
  }
])

【讨论】：

【解决方案2】：

您可以通过book_id $group 并使用$max 运算符捕获lastVersion。然后你需要$filter属于特定组的所有书籍，并使用$replaceRoot将单个找到的最新版本提升到根级别。

db.collection.aggregate([
    {
        $group: {
            _id: "$book_id",
            lastVersion: { $max: "$book_version" },
            versions: { $push: "$$ROOT" }
        }
    },
    {
        $replaceRoot: {
            newRoot: {
                $let: {
                    vars: {
                        lastVer: {
                            $filter: { 
                                input: "$versions", 
                                cond: { $eq: [ "$$this.book_version", "$lastVersion" ] } 
                            }
                        }
                    },
                    in: { $arrayElemAt: [ "$$lastVer", 0 ] }
                }
            }
        }
    }
])

Mongo Playground

【讨论】：