如何在 mongodb 的数组中合并具有相同键的对象？答案

【问题标题】：How can I merge objects with the same key in an array in mongodb?如何在 mongodb 的数组中合并具有相同键的对象？
【发布时间】：2021-12-18 11:06:36
【问题描述】：

我使用 mongodb 的聚合导出了以下数据。

{
    "first_count": 2,
    "second_count": 1,
    "third_count": 2,
    "test_count": 2,
    "sido": "대구",
    "guguns": [
        {
            "gugun": "남시",
            "first_count": 1,
            "second_count": 1,
            "third_count": 1
        },
        {
            "gugun": "부천군",
            "first_count": 1,
            "second_count": 0,
            "third_count": 1
        },
        {
            "gugun": "남시",
            "test_count": 1
        },
        {
            "gugun": "부천군",
            "test_count": 1
        }
    ]
}

这是合并两个方面数据的结果。但我想要的结果是：

{
    "first_count": 2,
    "second_count": 1,
    "third_count": 2,
    "test_count": 2,
    "sido": "대구",
    "guguns": [
        {
            "gugun": "남시",
            "first_count": 1,
            "second_count": 1,
            "third_count": 1,
            "test_count": 1
        },
        {
            "gugun": "부천군",
            "first_count": 1,
            "second_count": 0,
            "third_count": 1,
            "test_count": 1
        }
    ]
}

guguns.gugun如何将相同的值合二为一？如果可能的话，我想使用 mongodb 聚合来处理它。

【问题讨论】：

标签： mongodb mongoose aggregation-framework

【解决方案1】：

$unwind 解构 guguns 数组
$mergeObjects 将guguns 的当前对象与其他对象合并
$group by _id 和 guguns.gugun 属性并获取必填字段第一个值和 guguns 合并对象
$group 仅由_id 获取所需字段的第一个值并构造guguns 对象的数组

db.collection.aggregate([
  { $unwind: "$guguns" },
  {
    $group: {
      _id: {
        _id: "$_id",
        gugun: "$guguns.gugun"
      },
      first_count: { $first: "$first_count" },
      second_count: { $first: "$second_count" },
      third_count: { $first: "$third_count" },
      test_count: { $first: "$test_count" },
      sido: { $first: "$sido" },
      guguns: { $mergeObjects: "$guguns" }
    }
  },
  {
    $group: {
      _id: "$_id._id",
      first_count: { $first: "$first_count" },
      second_count: { $first: "$second_count" },
      third_count: { $first: "$third_count" },
      test_count: { $first: "$test_count" },
      sido: { $first: "$sido" },
      guguns: { $push: "$guguns" }
    }
  }
])

Playground

【讨论】：