mongodb - 聚合 $group 以了解集合中的上一个和下一个文档

Question

从这个 mongodb 操场示例： https ://mongoplayground.net/p/2yip2tRycYd

数据：

[
  {
    text: "bond",
    createdAt: ISODate("2021-07-20T07:02:39.299+00:00"),
    category: "films"
  },
  {
    text: "monty python",
    createdAt: ISODate("2021-07-21T07:06:39.299+00:00"),
    category: "films"
  },
  {
    text: "blow",
    createdAt: ISODate("2021-07-21T07:12:39.299+00:00"),
    category: "films"
  },
  {
    text: "cake",
    createdAt: ISODate("2021-07-24T10:01:39.299+00:00"),
    category: "food"
  },
  {
    text: "shoes",
    createdAt: ISODate("2021-07-24T11:02:39.299+00:00"),
    category: "clothes"
  },
  {
    text: "bread",
    createdAt: ISODate("2021-07-24T12:02:39.299+00:00"),
    category: "food"
  },
  {
    text: "bmw",
    createdAt: ISODate("2021-07-24T12:05:39.299+00:00"),
    category: "cars"
  },
  {
    text: "tree",
    createdAt: ISODate("2021-07-24T12:05:39.299+00:00"),
    category: "world"
  },
  {
    text: "fork",
    createdAt: ISODate("2021-07-24T12:05:39.299+00:00"),
    category: "kitchen"
  },
  
]

查询

db.collection.aggregate([
  {
    $group: {
      _id: {
        day: {
          $dayOfMonth: "$createdAt"
        },
        month: {
          $month: "$createdAt"
        },
        year: {
          $year: "$createdAt"
        },
        category: "$category"
      },
      count: {
        $sum: 1
      },
      createdAt: {
        "$addToSet": "$createdAt"
      },
      text: {
        "$addToSet": "$text"
      },
      category: {
        "$addToSet": "$category"
      },
      
    },
    
  },
  {
    $sort: {
      "createdAt": 1
    }
  },
  {
    $skip: 0
  },
  {
    $limit: 5
  },
  
])

结果

[
  {
    "_id": {
      "category": "films",
      "day": 20,
      "month": 7,
      "year": 2021
    },
    "category": [
      "films"
    ],
    "count": 1,
    "createdAt": [
      ISODate("2021-07-20T07:02:39.299Z")
    ],
    "text": [
      "bond"
    ]
  },
  {
    "_id": {
      "category": "films",
      "day": 21,
      "month": 7,
      "year": 2021
    },
    "category": [
      "films"
    ],
    "count": 2,
    "createdAt": [
      ISODate("2021-07-21T07:06:39.299Z"),
      ISODate("2021-07-21T07:12:39.299Z")
    ],
    "text": [
      "blow",
      "monty python"
    ]
  },
  {
    "_id": {
      "category": "food",
      "day": 24,
      "month": 7,
      "year": 2021
    },
    "category": [
      "food"
    ],
    "count": 2,
    "createdAt": [
      ISODate("2021-07-24T10:01:39.299Z"),
      ISODate("2021-07-24T12:02:39.299Z")
    ],
    "text": [
      "bread",
      "cake"
    ]
  },
  {
    "_id": {
      "category": "clothes",
      "day": 24,
      "month": 7,
      "year": 2021
    },
    "category": [
      "clothes"
    ],
    "count": 1,
    "createdAt": [
      ISODate("2021-07-24T11:02:39.299Z")
    ],
    "text": [
      "shoes"
    ]
  },
  {
    "_id": {
      "category": "world",
      "day": 24,
      "month": 7,
      "year": 2021
    },
    "category": [
      "world"
    ],
    "count": 1,
    "createdAt": [
      ISODate("2021-07-24T12:05:39.299Z")
    ],
    "text": [
      "tree"
    ]
  }
]

结果如预期 - 我们将 2 部电影和 2 种食物组合在一起。

但是，是否有更多的配置可以注入到查询中，其中的项目介于或不在要分组的项目之间。

在上面的示例中，是否可以更改聚合以将所有第一部电影组合在一起（因为在集合中它们都彼此相邻），而不是将食物组合在一起，因为它们之间有衣服？

标签： mongodbaggregation-framework