如何在 MongoDB $lookup 中使用变量

par*_*ars 2 mongodb aggregation-framework

假设我有 3 个集合,cars, bikes, vehicles

cars 集合为:

{
  {
    "_id": "car1",
    "carBrand": "Audi",
    "color": "blue"
  },
  {
    "_id": "car2",
    "carBrand": "BMW",
    "color": "white"
  }
}
Run Code Online (Sandbox Code Playgroud)

bikes 集合为:

{
  {
    "_id": "bike1",
    "bikeBrand": "Audi",
    "color": "red"
  },
  {
    "_id": "bike2",
    "carBrand": "BMW",
    "color": "white"
  }
}
Run Code Online (Sandbox Code Playgroud)

并且该 vehicles 集合实际上只有对carsbikes集合的引用

{
  {
    "_id": "vehicle1",
    "vehicleType": "cars",
    "vehicleId": "car1"
  },
  {
    "_id": "vehicle2",
    "vehicleType": "cars",
    "vehicleId": "car2"
  },
  {
    "_id": "vehicle3",
    "vehicleType": "bikes",
    "vehicleId": "bike1"
  },
  {
    "_id": "vehicle4",
    "vehicleType": "bikes",
    "vehicleId": "bike2"
  },

}
Run Code Online (Sandbox Code Playgroud)

我想加入vehicles收藏carsbikes收藏。我试图将"$vehicleType"变量设置为$lookup'sfrom字段。然而,它并没有像我预期的那样工作。它根本不连接表。没有错误。

db.collection.aggregate([{

  $lookup: {
      from: "$vehicleType",
      localField: "vehicleId",
      foreignField: "_id",
      as: "vehicleDetails"
  }

}]);
Run Code Online (Sandbox Code Playgroud)

我期待有这样的结果

{
  {
    "_id": "vehicle1",
    "vehicleType": "cars",
    "vehicleId": "car1",
    "vehicleDetails": {    
      "_id": "car1",
      "carBrand": "Audi",
      "color": ""
    }
  },
  {
    "_id": "vehicle2",
    "vehicleType": "cars",
    "vehicleId": "car2",
    "vehicleDetails":    {
      "_id": "car2",
      "carBrand": "BMW",
      "color": "white"

  },
  {
    "_id": "vehicle3",
    "vehicleType": "bikes",
    "vehicleId": "bike1",
    "vehicleDetails":    {
      "_id": "bike1",
      "bikeBrand": "Audi",
      "color": "red"
    }
  },
  {
    "_id": "vehicle4",
    "vehicleType": "bikes",
    "vehicleId": "bike2",
    "vehicleDetails":    {
      "_id": "bike2",
      "carBrand": "BMW",
      "color": "white"
    }
  },

}    
Run Code Online (Sandbox Code Playgroud)

Ale*_*lex 5

如果汽车和自行车没有通用 ID,您可以在单独的数组中依次查找,然后将它们与$setUnion 组合

db.vehicles.aggregate([
  {$lookup: {
      from: "cars",
      localField: "vehicleId",
      foreignField: "_id",
      as: "carDetails"
  }},
  {$lookup: {
      from: "bikes",
      localField: "vehicleId",
      foreignField: "_id",
      as: "bikeDetails"
  }},
  {$project: {
     vehicleType: 1,
     vehicleId: 1,      
     vehicleDetails:{$setUnion: [ "$carDetails", "$bikeDetails" ]}
  }},
  {$project: {
      carDetails:0,
      bikeDetails:0,
  }}
]);
Run Code Online (Sandbox Code Playgroud)

否则,您将需要在查找前使用$facet按类型过滤车辆:

db.vehicles.aggregate([
   {
     $facet: {
         "cars": [
            {$match: {"vehicleType": "cars"}},
            {$lookup: {
               from: "cars",
               localField: "vehicleId",
               foreignField: "_id",
               as: "vehicleDetails"
             }},
         ],
         "bikes": [
            {$match: {"vehicleType": "bikes"}},
            {$lookup: {
               from: "bikes",
               localField: "vehicleId",
               foreignField: "_id",
               as: "vehicleDetails"
             }}
         ]
     }
   },
   {$project: {all: {$setUnion: ["$cars", "$bikes"]}}},
   {$unwind: "$all"},
   {$replaceRoot: { newRoot: "$all" }}
])
Run Code Online (Sandbox Code Playgroud)