MongoDB如果其他集合中存在字段/引用，则从汇总中排除文档

Question

const sellerSchema = Schema(
  {
    name: String,
    url:String
  }

const productSchema = Schema(
  {
    title: String,
    sellerUrl:String
  }

下面的查询将从所有产品返回唯一的sellerUrl：

context.Product.aggregate([
        {
          $group: {
            _id: "$sellerUrl",
          }
        }
      ]);

但是我也想从汇总中排除已经保存的卖方。因此，如果url == sellerUrl汇总必须排除该卖方。请帮助我

Answer 1

您可以尝试以下查询：

db.product.aggregate([
    {
      $group: {
        _id: "", /** group on no condition & push all unique `sellerUrl` to sellerUrls array */
        sellerUrls: { $addToSet: "$sellerUrl" }
      }
    },
    {
      $lookup: {
        from: "seller",
        let: { sellerUrls: "$sellerUrls" }, // creating local variable
        pipeline: [
          { $group: { _id: "", urls: { $addToSet: "$url" } } }, /** group on no condition & push all unique `url` to urls array */
          { $project: { _id: 0, uniqueAndNotInSellerColl: { $setDifference: [ "$$sellerUrls", "$urls" ] } } } // get difference between two arrays
        ],
        as: "data" // As we're grouping will always be one doc/element in an array
      }
    },
    /** Create a new root doc from getting first element(though it will have only one) from `data` array */
    {
      $replaceRoot: { newRoot: { $arrayElemAt: [ "$data", 0 ] } }
    }
  ])

Test： mongoplayground