MongoDB聚合查询，基于参数返回作为模型参考的集合

Question

我有以下查询，

MyModel.aggregate([
    {
      $project: {
        field: 1,
      }
    },
    {
      $lookup: {
        from: "AnotherModel",
        localField: "_id",
        foreignField: "someField",
        as: "field"
      },
    },
    {
      $unwind: {
        path: "$_id"
      }
    }
])

此作品已执行，但未如预期。在AnotherModel中，有一个名为show的字段，我希望此查询返回MyModel集合，只有在show上将AnotherModel设置为true的情况下。我尝试过，

MyModel.aggregate([
    {
      $project: {
        field: 1,
      }
    },
    {
      $lookup: {
        from: "AnotherModel",
        localField: "_id",
        foreignField: "someField",
        as: "field"
      },
      $where: {
        show: true // <-- this part
      }
    },
    {
      $unwind: {
        path: "$_id"
      }
    }
])

但是那引发了一个错误。我该如何实现？

Answer 1

尝试这个：Uncorrelated Sub-queries

MyModel.aggregate([
  {
    $project: {
      field: 1
    }
  },
  {
    $lookup: {
      from: "AnotherModel",
      let: {
        id: "$_id"
      },
      pipeline: [
        {
          $match: {
            show: true,
            $expr: {
              $eq: [
                "$someField",
                "$$id"
              ]
            }
          }
        }
      ],
      as: "field"
    }
  },
  {
    $unwind: {
      path: "$_id"
    }
  }
])

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