使用MongoDB聚合返回数字字段上具有最大值的文档 [英] Return a document with the max value on a number field using MongoDB aggregation

问题描述

例如，如果我有一堆文件

If I have a bunch of documents for instance

{
  _id: mongoId,
  subcatId: mongoId,
  viewCount: 2
}

_id是唯一的，但subcatId不是唯一的.

_id is unique but subcatId isn't.

如果我想返回每个subcatId的viewCount最高的每个文档，我该如何使用Mongoose/MongoDB的聚合功能呢?

If I wanted to return each document that had the highest viewCount per subcatId, how would I do that using aggregation with Mongoose/MongoDB?

推荐答案

您可以这样做:

db.test.aggregate([
  // Sort the docs by viewCount descending so that the highest ones come first
  {$sort: {viewCount: -1}},
  // Group by subcatId and take the first doc from each group
  {$group: {_id: '$subcatId', doc: {$first: '$$ROOT'}}}
])

$$ROOT 系统变量已添加到2.6，表示管道中该阶段正在处理的整个文档.所有系统变量都以$$前缀引用.

对于较旧的MongoDB版本，您需要分别将所需的每个字段添加到$group:

For older versions of MongoDB, you need to individually add each field you need to the $group:

db.test.aggregate([
  {$sort: {viewCount: -1}},
  {$group: {
    _id: '$subcatId', 
    doc_id: {$first: '$_id'},
    viewCount: {$first: '$viewCount'}
  }}
])

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