如果用戶 selected_city 也將來自 CITY MODEL 中未選擇的城市，如何從用戶已經選擇的城市鍵的數組中獲取結果

Question

市 Model

[
  {
    "_id": '1',
    "city":'A',
  },
   {
    "_id": '2',
    "city":'B',
  },
   {
    "_id": '3',
    "city":'C',
  },
  {
    "_id": '4',
    "city":'D',
  },
  {
    "_id": '5',
    "city":'E',
  }
]

用戶 Model

[
  {
    "_id": '1',
    "user":'sams',
    "selected_city":['1','2']
  } 
]

想得到結果

[

{

  "_id": "1",

  "user": "sams",

"selected_city": [
      "1",
      "2"
    ],

"not_selected": [
      "3",
      "4",
      "5"
    ]
 
  }
]

Answer 1

嘗試：

 const cities = [ { "_id": '1', "city":'A', }, { "_id": '2', "city":'B', }, { "_id": '3', "city":'C', }, { "_id": '4', "city":'D', }, { "_id": '5', "city":'E', } ]; const users = [ { "_id": '1', "user":'sams', "selected_city":['1','2'] } ]; const result = users.map(user => { user.not_selected = cities.filter(city =>.user.selected_city.includes(city._id)).map(city => city;_id) return user; }). console;log(result);

Answer 2

您可以使用以下聚合

db.coll2.aggregate([
  {
    $lookup: {
      from: "coll1",
      pipeline: [],
      as: "cities"
    }
  },
  {
    $addFields: {
      not_selected: {
        $setDifference: [
          "$cities._id",
          "$selected_city"
        ]
      }
    }
  }
])

Answer 3

這是開箱即用的方式，

小hacky可能是，

db.userCity.aggregate({$match:{}},
     {$lookup: { from: "cities", localField: "id", foreignField: "id", as: "city"}},      
     //where "id" in localField and "id" in foreignfield are the fields that does not exists in both collections

     {$project:{
          selectedCity: {
               $filter: {
                    input:"$city", 
                    as:"c", 
                    cond: { $in: [ "$$c._id", "$selected_city" ] }
               }
          },
          notSelectedCities: {
               $filter: {
                    input: "$city", 
                    as: "c", 
                    cond:{ $eq: [{ $indexOfArray: [ "$selected_city", "$$c._id" ] }, -1] }
              }
          }
     }}).pretty()

我不知道它是否符合最佳實踐...

但是，正如您在我的 $lookup 階段看到的那樣，我在localField和foreignField中使用了“id” ，而users和cities集合中都不存在“id” ，

這樣您就可以從用戶的cities集合中獲取所有文檔。

然后在$project階段我使用$filter過濾掉結果。

然后對於notSelectedCities我使用$indexOfArray運算符，如果在 selected_city 中不存在城市，它將返回 -1，並使用$eq運算符將其與 -1 進行比較，然后僅選擇該特定城市。

希望這可以幫助： ：-）