合并和删除嵌套数组中的元素

Question

我有这个数组，我想合并嵌套数组中对象内的所有元素并删除重复项..该数组是 mongo db populate 的输出，所以从那里得到的答案或只是 js 会很棒:)

 "visitors": [
        [
            {
                "name": "matan",
                "id": "61793e6a0e08cdcaf213c0b1"
            },
            {
                "name": "shani",
                "id": "61793e910e08cdcaf213c0b5"
            }
        ],
        [
            {
                "name": "david",
                "id": "6179869cb4944c6b19b05a23"
            },
            {
                "name": "orit",
                "id": "617986e535fdf4942ef659bd"
            }
        ],
        [
            {
                "name": "david",
                "id": "6179869cb4944c6b19b05a23"
            },
            {
                "name": "orit",
                "id": "617986e535fdf4942ef659bd"
            }
        ]
    ]

想要这个输出 -

"visitors": [
        {
            "name": "matan",
            "id": "61793e6a0e08cdcaf213c0b1"
        },
        {
            "name": "shani",
            "id": "61793e910e08cdcaf213c0b5"
        },
        {
            "name": "david",
            "id": "6179869cb4944c6b19b05a23"
        },
        {
            "name": "orit",
            "id": "617986e535fdf4942ef659bd"
        },
]

这些是我的收藏，我需要让所有访客都在一个太阳系上，所以 > 太阳能 > 行星 > 访客

const solarsModel = new Schema({
    planets: [ { type: Schema.Types.ObjectId ,ref:'planet'} ],
    starName: { type: String, required: true, default: "" }
})

const planetModel = new Schema({
    planetName: { type: String, required: true, default: "" },
    system:{type: Schema.Types.ObjectId, ref: 'solar'},
    visitors: [{ type: Schema.Types.ObjectId , ref: 'visitor'}]
})

const visitorModel = new Schema({
    visitorName:{ type: String, required: true, default: "" },
    homePlanet: {type: Schema.Types.ObjectId, ref:"planet" },
    visitedPlanets: [{ type: Schema.Types.ObjectId, ref:"planet" }]
})

这就是我为实现结果所做的工作，我很想使用 Aggregate ..

  const response = await solarModel
    .findById({ _id: data.id })
    .select({ starName: 1, _id: 0 })
    .populate({
      path: "planets",
      select: { visitors: 1, _id: 0 },
      populate: {
        path: "visitors",
        select: "visitorName",
      },
    })
    .exec();

Answer 1

(1)展平数组数组

visitors = visitors.flat();

这给了我们这个：

[
  { name: 'matan', id: '61793e6a0e08cdcaf213c0b1' },
  { name: 'shani', id: '61793e910e08cdcaf213c0b5' },
  { name: 'david', id: '6179869cb4944c6b19b05a23' },
  { name: 'orit', id: '617986e535fdf4942ef659bd' },
  { name: 'david', id: '6179869cb4944c6b19b05a23' },
  { name: 'orit', id: '617986e535fdf4942ef659bd' }
]

(2) 获取唯一id

let uniqueIds= [...new Set(visitors.map(v => v.id)]

这给了我们这个：

[
   '61793e6a0e08cdcaf213c0b1',
   '61793e910e08cdcaf213c0b5',
   '6179869cb4944c6b19b05a23',
   '617986e535fdf4942ef659bd'
]

(3) 仅根据 uniqueIds 获取新的访问者列表

visitors = uniqueIds.map(id => {
   let name = visitors.find(v => v.id === id).name;

   return {
      id,
      name
   }
});

这给了我们这个：

[
  { name: 'matan', id: '61793e6a0e08cdcaf213c0b1' },
  { name: 'shani', id: '61793e910e08cdcaf213c0b5' },
  { name: 'david', id: '6179869cb4944c6b19b05a23' },
  { name: 'orit', id: '617986e535fdf4942ef659bd' },
]

Answer 2

您可以像这样使用aggregate() ：

• 由于嵌套数组， $unwind两次
• $group使用$addToSet来避免重复。

db.collection.aggregate([
  {
    "$unwind": "$visitors"
  },
  {
    "$unwind": "$visitors"
  },
  {
    "$group": {
      "_id": null,
      "visitors": {
        "$addToSet": {
          "id": "$visitors.id",
          "name": "$visitors.name"
        }
      }
    }
  }
])

示例在这里

Answer 3

询问

• 用 concat 减少以展平
• union 与一个空数组，只是为了删除重复项
• 如果您有除访客以外的其他字段，则它们不受影响

*在您的驱动程序上试一试，操场有现场订单问题，有时会丢失，在这里我们比较文档以删除重复项。

aggregate(
[{"$set": 
   {"visitors": 
     {"$setUnion": 
       [{"$reduce": 
         {"input": "$visitors",
          "initialValue": [],
          "in": {"$concatArrays": ["$$value", "$$this"]}}},
        []]}}}])

结果

[{
  "visitors": [
    {
      "name": "david",
      "id": "6179869cb4944c6b19b05a23"
    },
    {
      "name": "matan",
      "id": "61793e6a0e08cdcaf213c0b1"
    },
    {
      "name": "orit",
      "id": "617986e535fdf4942ef659bd"
    },
    {
      "name": "shani",
      "id": "61793e910e08cdcaf213c0b5"
    }
  ]
}]