猫鼬排序，每个字段只得到一个答案

【问题标题】：Mongoose sort and get only one per field猫鼬排序，每个字段只得到一个
【发布时间】：2018-06-15 15:33:05
【问题描述】：

我需要使用 mongoose 获取项目列表，但是，我只需要选择任何类型的一个项目（我有 5 个可能的值），并且始终是该类型的最后插入的项目。

例如，到下面的列表中，没有排序（我知道，这里是排序的）。

[
  {
    "_id": "5b226abab64b2309de01bfd5",
    "documentType": "type3",
    "createdAt": "2018-06-14T13:16:42.321Z"
  },
  {
    "_id": "5b226a5da93b0e09b8aee447",
    "documentType": "type3",
    "createdAt": "2018-06-14T13:15:09.458Z"
  },
  {
    "_id": "5b226a21f454d6097ffa461c",
    "documentType": "type2",
    "createdAt": "2018-06-14T13:14:09.159Z"
  },
  {
    "_id": "5b2267fb445d7709590e1695",
    "documentType": "type1",
    "createdAt": "2018-06-14T13:04:59.742Z"
  },
  {
    "_id": "5b2267de76708b094696b0fe",
    "documentType": "type3",
    "createdAt": "2018-06-14T13:04:30.349Z"
  },
  {
    "_id": "5b2267ce2f3724092b1a14a3",
    "documentType": "type2",
    "createdAt": "2018-06-14T13:04:14.410Z"
  },
  {
    "_id": "5b2267a92f3724092b1a14a2",
    "documentType": "type4",
    "createdAt": "2018-06-14T13:03:37.079Z"
  },
  {
    "_id": "5b22647b63017e08a69a3043",
    "documentType": "type5",
    "createdAt": "2018-06-14T12:50:03.999Z"
  },
  {
    "_id": "5b2264471778a20880d4ba64",
    "documentType": "type5",
    "createdAt": "2018-06-14T12:49:11.773Z"
  }
]

预期的结果是这样的：

[
  {
    "_id": "5b226abab64b2309de01bfd5",
    "documentType": "type3",
    "createdAt": "2018-06-14T13:16:42.321Z"
  },
  {
    "_id": "5b226a21f454d6097ffa461c",
    "documentType": "type2",
    "createdAt": "2018-06-14T13:14:09.159Z"
  },
  {
    "_id": "5b2267fb445d7709590e1695",
    "documentType": "type1",
    "createdAt": "2018-06-14T13:04:59.742Z"
  },
  {
    "_id": "5b2267a92f3724092b1a14a2",
    "documentType": "type4",
    "createdAt": "2018-06-14T13:03:37.079Z"
  },
  {
    "_id": "5b22647b63017e08a69a3043",
    "documentType": "type5",
    "createdAt": "2018-06-14T12:50:03.999Z"
  }
]

基本上每种类型只有一个，按id排序

我该怎么做？

【问题讨论】：

标签： node.js mongodb mongoose

【解决方案1】：

您可以在mongodb中使用$group来区分documentType

db.collection.aggregate([
  { "$group": {
    "_id": "$documentType",
    "createdAt": { "$first": "$createdAt" },
    "id": { "$first": "$_id" }
  }},
  { "$project": {
    "_id": "$id",
    "createdAt": 1,
    "documentType": "$_id"
  }}
])

输出将是

[
  {
    "_id": "5b226abab64b2309de01bfd5",
    "createdAt": "2018-06-14T13:16:42.321Z",
    "documentType": "type3"
  },
  {
    "_id": "5b2267fb445d7709590e1695",
    "createdAt": "2018-06-14T13:04:59.742Z",
    "documentType": "type1"
  },
  {
    "_id": "5b22647b63017e08a69a3043",
    "createdAt": "2018-06-14T12:50:03.999Z",
    "documentType": "type5"
  },
  {
    "_id": "5b2267a92f3724092b1a14a2",
    "createdAt": "2018-06-14T13:03:37.079Z",
    "documentType": "type4"
  },
  {
    "_id": "5b226a21f454d6097ffa461c",
    "createdAt": "2018-06-14T13:14:09.159Z",
    "documentType": "type2"
  }
]

【讨论】：

我把我的问题描述错了，我需要的是这个，但是，我想要一个按 createdAt 的 desc 排序，而不是按 id 排序
我已经解决了 { $sort: { createdAt: -1, }, },