有没有一种方法可以使用聚合管道方法进行此查询 [英] Is there a way to use the aggregate pipeline method for this query

问题描述

我在一个在线商店中有一个针对商品的2000多个评论的数据库，我正在寻找所有至少发表5条评论的用户，并且我想列出他们的姓名，评论数量和评论者ID

I have a database of over 2000 reviews for products on an online store, and I am trying to find all of the users who have written minimum 5 reviews, and I would like to list their names, number of reviews, and reviewer ID

我知道我必须使用聚合方法来执行此操作，但是在尝试时我没有运气，因为我还是mongodb的新手

I understand that I have to use aggregate method to do this but when trying I have no luck as I am still new to mongodb

我添加了一张图片，以显示数据库中某些数据的示例

I have added a picture to show an example of some of the data in the database

一个进入数据库的示例如下；

An example of one entry into the database is as below;

{
    "_id": ObjectId("5d0b70f2d7367de7f5fa0ee1"),
    "SH_reviewerID": "A2G0LNLN79Q6HR",
    "SH_asin": "0000031887",
    "SH_reviewerName": "aj_18 \"Aj_18\"",
    "SH_helpful": [
      1,
      1
    ],
    "SH_reviewText": "This was a really good",
    "SH_overall": 4.0,
    "SH_summary": "Really Cute but rather short.",
    "SH_unixReviewTime": 1337990400,
    "SH_reviewTime": "05 26, 2012"
}

我已经尝试了好几天，但是没有运气，我认为我们需要使用$ match，$ group，$ project

I have tried to work it out for over several days but have no luck, I think we need to use $match, $group, $project

推荐答案

我想您需要运行此聚合:

I guess you need to run this aggregation:

db.review.aggregate([
  {
    $group: {
      _id: "$SH_reviewerID",
      name: {
        $first: "$SH_reviewerName"
      },
      number: {
        $sum: 1
      }
    }
  },
  {
    $match: {
      number: {
        $gte: 5
      }
    }
  },
  //Optional - Order views descendant
  {
    $sort: {
      number: -1
    }
  },
  //Optional - Change field names
  {
    $project: {
      _id: 0,
      SH_reviewerID: "$_id",
      SH_reviewerName: "$name",
      SH_reviewerViews: "$number"
    }
  }
])

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