如何在MongoDB$lookup中使用变量

假设我有3个系列，

汽车

，

自行车

，

汽车

cars

collection as:

bikes

collection as:

而

vehicles

集合实际上只是引用了

cars

和

bikes

集合

我想加入

vehicles

系列，加入

cars

和

bikes

系列。我试图将

“$vehicleType”

作为变量设置为

$lookup

的

from

字段。然而，它并没有像我预期的那样工作。它只是不连接表。没有错误

db.collection.aggregate([{

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

}]);

我本以为会有这样的结果

{
  {
    "_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"
    }
  },

}    

---

如果汽车和自行车没有通用ID，您可以在单独的数组中顺序查找，然后将它们与：

否则，在查找之前，您需要使用按类型筛选车辆：

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" }}
])

---

非常感谢。表演怎么样？我想，这将加入

汽车

和

自行车

集合，而不考虑

车辆类型

。在

$lookup

之前是否有一种基于

vehicleType

的过滤方法？是的，正如我在更一般的情况下所说的，您需要使用facet。我已经用代码片段更新了答案。

db.collection.aggregate([{

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

}]);

{
  {
    "_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"
    }
  },

}    

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,
  }}
]);

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" }}
])