如何在 MongoDB $lookup 中使用变量答案

【问题标题】：How to use variables with MongoDB $lookup如何在 MongoDB $lookup 中使用变量
【发布时间】：2017-03-29 20:24:56
【问题描述】：

假设我有 3 个集合，cars、bikes、vehicles。

cars收藏为：

{
  {
    "_id": "car1",
    "carBrand": "Audi",
    "color": "blue"
  },
  {
    "_id": "car2",
    "carBrand": "BMW",
    "color": "white"
  }
}

bikes收藏为：

{
  {
    "_id": "bike1",
    "bikeBrand": "Audi",
    "color": "red"
  },
  {
    "_id": "bike2",
    "carBrand": "BMW",
    "color": "white"
  }
}

和 vehicles 集合实际上只是引用了cars 和bikes 集合作为

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

}

我想与cars 和bikes 集合一起加入vehicles 集合。我试图将"$vehicleType" 设置为$lookup 的from 字段的变量。但是它并没有像我预期的那样工作。它根本不加入表格。没有错误。

db.collection.aggregate([{

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

}]);

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"
    }
  },

}

【问题讨论】：

标签： mongodb aggregation-framework

【解决方案1】：

如果汽车和自行车没有共同的 ID，您可以在单独的数组中顺序查找，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,
  }}
]);

否则，您需要在查找前使用$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" }}
])

【讨论】：

谢谢。性能怎么样？我想，这将加入cars 和bikes 集合，而不管vehicleType。有没有办法在$lookup之前根据vehicleType进行过滤？
是的，正如我所说，在更一般的情况下，您需要使用构面。我已经用代码 sn-p 更新了答案。