需要在 Mongoose 中获取按外键分组的记录列表

Question

我有 2 个表interestTypes 和interests 表，兴趣记录存储了interestTypeId 引用。

现在我需要兴趣类型列表及其兴趣。

const interestSchema = new mongoose.Schema(
  {
    name: {
      type: String,
      trim: true,
      minlength: 3,
      maxlength: 50,
      required: true
    },
    interestTypes: [
      {
        interestTypeId: {
          type: mongoose.SchemaTypes.ObjectId,
          ref: 'interestType'
        }
      }
    ]
  },
  {
    timestamps: true
  }
)

const interestTypeSchema = new mongoose.Schema(
  {
    name: {
      type: String,
      trim: true,
      minlength: 3,
      maxlength: 50,
      required: true
    }
  },
  {
    timestamps: true
  }
)

我知道这可以通过查找聚合来完成，但我搞砸了。这段代码是用兴趣模型写的

interestSchema.statics.getTypes = function() {
    return new Promise((resolve, reject) => {

    this.aggregate([
      {
        "$lookup":
          {
            "from": "interestTypes",
            "localField": "interestType",
            "as": "interestTypes",
            "foreignField": "_id",
          }
      },
      {
        "$group": 
          {
            "_id": "$_id",
            "interestTypes": { "$push": "$interestTypes" }
          }
        }
    ],
      function(err, result) {
        if (err) return reject(err)        
        resolve(result)
      }
    )
}

无法得到想要的结果

我希望这个方法返回的是这样的

[
{
    id: interestTypeIdRef1,
    name: interestType1
    interests: [
    {
        id: interestIdRef1,
        name: interestType1
    },
    {
        id: interestIdRef2,
        name: interestType2
    },
    {
        id: interestIdRef3,
        name: interestType3
    }
    ]
},
{
    id: interestTypeIdRef2,
    name: interestType1
    interests: [
    {
        id: interestIdRef2,
        name: interestType1
    },
    {
        id: interestIdRef3,
        name: interestType2
    },
    {
        id: interestIdRef4,
        name: interestType3
    }
    ]
}
]

Answer 1

从interestTypeSchema 开始聚合以获得所需的输出。

InterestSchema.statics.getTypes = function () {
return new Promise((resolve, reject) => {

    interestType.aggregate([
        {
            $lookup: {
                from: 'interestSchema',
                localField: '_id',
                foreignField: 'interestTypeId',
                as: 'interests'
            }
        },
        {
            $project: {
                name:1,
                interests:1,
            }
        }
    ],
        function (err, result) {
            if (err) return reject(err)
            resolve(result)
        }
    )
}

Answer 2

您可以使用下面的链接来检查查询。 这是在聚合中进行查找的查询

已创建示例数据库

db={
  "interestTypes": [
    {
      "typeId": 1,
      "type": "type1"
    },
    {
      "typeId": 2,
      "type": "type2"
    }
  ],
  "interests": [
    {
      "_id": 1,
      "interestName": "interest1",
      "interestTypeId": 1
    },
    {
      "_id": 2,
      "interestName": "interest2",
      "interestTypeId": 1
    },
    {
      "_id": 3,
      "interestName": "interest3",
      "interestTypeId": 2
    },
    {
      "_id": 4,
      "interestName": "interest4",
      "interestTypeId": 2
    },
    {
      "_id": 5,
      "interestName": "interest5",
      "interestTypeId": 1
    }
  ]
}

示例答案

db.interestTypes.aggregate([
  {
    $lookup: {
      from: "interests",
      localField: "typeId",
      foreignField: "interestTypeId",
      as: "interesetTypesWithInterest"
    }
  },
  {
    $project: {
      type: 1,
      interestTypeId: "$typeId",
      interests: "$interesetTypesWithInterest"
    }
  }
])

使用此链接进行验证 https://mongoplayground.net/p/3f5AzKQMamw 既然您熟悉将 mongodb 代码转换为 Mongoose，它将对您有所帮助。