Mongoose 相当于 JOIN ... WHERE答案

【问题标题】：Mongoose equivalent of JOIN ... WHEREMongoose 相当于 JOIN ... WHERE
【发布时间】：2020-05-11 17:35:03
【问题描述】：

我有两个模型

Opinion {
  _id: string;
  creator: string;
  teacher: ObjectId;
  text: string;
  date: Date;
}

Teacher {
  _id: string;
  name: string;
  isVerified: boolean;
}

我需要获得 3 个最新意见 WHERE { teacher.isVerified: true }

我试过了

    const newOpinions = await Opinion.find(
      null,
      "creator teacher text",
      {
        sort: {
          date: -1,
        },
      }).populate(
        {
          path: 'teacher',
          model: 'Teacher',
          select: 'name',
          match: {
            isVerified: true,
          },
        }
      ).limit(3);

但是（经过简短分析）它按设计工作 - 无论教师是否经过验证，我都会收到 3 个最新意见（在教师领域，如果未经验证，我会收到 null）

任何人都可以尝试为我指出正确的方向吗？我想可能是Model.aggregate()。

【问题讨论】：

标签： javascript mongodb express mongoose non-relational-database

【解决方案1】：

这是一种方法。有了这个，您将首先过滤教师。也请咨询$lookup documentation

Teacher.aggregate([{
  $match: { isVerified: true }
}, {
  $lookup: {
    from: 'opinions' // assume you have collection named `opinions` for model `Opinion`,
    localField: '_id', // join teacher._id ...
    foreignField: 'teacher', // ... with opionion.teacher
    as: 'opinions'
  } // from here you well get verified teachers with embedded opinions, 
// further stages are optional. We will modify the shape of output to be opinions only
}, { 
  $unwind: '$opinions' // spread out opinions array into separate documents
}, {
  $replaceRoot: '$opinions' // replace the root document with opinion only
}])

【讨论】：

【解决方案2】：

在mongodb中相当于join的是$lookup。一种猫鼬方法是使用填充，但您必须在模型中提供 ref 键

用于查找用途 https://docs.mongodb.com/manual/reference/operator/aggregation/lookup/

猫鼬参考

Opinion {
  _id: string;
  creator: string;
  teacher: {
      type: Schema.Types.ObjectId, ref: 'Teacher'
  },
  text: string;
  date: Date;
}

Teacher {
  _id: string;
  name: string;
  isVerified: boolean;
}

$lookup 方法更加灵活和可定制

【讨论】：