MongoDB

Question

我需要按 featured 和 trending 对以下文档进行分组。

[
    {
        "_id" : ObjectId("5f22546cd49ffe0d6f087a2e"),
        "title" : "....",
        "image" : "...",
        "featured" : true,
        "trending" : false,
        "creationDate" : ISODate("2020-07-30T05:02:36.592Z")
    },
    {
        "_id" : ObjectId("5f22546cd49ffe0d6f087a2f"),
        "title" : "Cras non dolor",
        "image" : "...",
        "featured" : false,
        "trending" : true,
        "creationDate" : ISODate("2020-07-30T05:02:36.592Z")
    }
]

因此，分组后它们应该采用以下格式 -

[
    {
        _id: null,
        featured: [
            {
                "_id" : ObjectId("5f22546cd49ffe0d6f087a2e"),
                "title" : "....",
                "image" : "...",
                "featured" : true,
                "trending" : false,
                "creationDate" : ISODate("2020-07-30T05:02:36.592Z")
            }
        ],
        trending: [
            {
                "_id" : ObjectId("5f22546cd49ffe0d6f087a2f"),
                "title" : "Cras non dolor",
                "image" : "...",
                "featured" : false,
                "trending" : true,
                "creationDate" : ISODate("2020-07-30T05:02:36.592Z")
            }
        ]
    }
]

如何通过 aggregation 或任何其他方式获得此结果？

我一直在尝试 aggregate $group。但我无法弄清楚如何通过 featured/trending 值 $group 等于 true.

因此，在对它们进行分组时，我不需要相应的 false 值。此外，可能还有 highlighted 等其他字段以及 featured trending.

Answer 1

这很容易实现，最简单的方法是使用 $facet

db.collection.aggregate([
  {
    $facet: {
      featured: [
        {
          $match: {
            "featured": true
          }
        }
      ],
      trending: [
        {
          $match: {
            "trending": true
          }
        }
      ]
    }
  }
])

MongoPlayground

MongoDB - 按字段值对文档进行分组

MongoDB - group documents by field value

grouping

mongoose

mongoose-schema