Question

我在一个集合中有大量的记录，如下所示：

{
    "_id" : ObjectId("5a95cf7790bd8fbf1c6a39da"),
    "dmb_reviewerID" : "AB9S9279OZ3QO",
    "dmb_asin" : "0078764343",
    "dmb_reviewerName" : "Alan",
    "dmb_helpful" : [ 
        1, 
        1
    ],
    "dmb_reviewText" : "I haven't gotten around to playing the campaign but the multiplayer is solid and pretty fun. Includes Zero Dark Thirty pack, an Online Pass, and the all powerful Battlefield 4 Beta access.",
    "dmb_overall" : 5.0,
    "dmb_summary" : "Good game and Beta access!!",
    "dmb_unixReviewTime" : 1373155200,
    "dmb_reviewTime" : "07 7, 2013"
}

我需要找到所有200条评论或以上的产品ID（dmb_asin）。

到目前为止，我已经设法计算它们并使用聚合返回一个总和，但我无法弄清楚如何仅显示大于200的那些。

我的代码：

aggregate({
  $group: {
    _id: "$dmb_asin",
    reviews: {
      $addToSet: "$dmb_asin"
    },
    count: {
      $sum: 1,},


}

});

Answer 1

尝试此查询：

db.collection.aggregate([
      {$group: {
             _id: "$dmb_asin",
              reviews: {
                       $addToSet: "$dmb_asin"
               },
             count: {
                     $sum: 1,}
       }},
       {$match:{"reviews":{$gte:200}}}

])

Answer 2

试试这段代码（如果我能正确理解你）

aggregate([
  {
    $group: {
      _id: '$dmb_asin',
      count: {
          $sum: 1
      }
    }
  },
  {
    $match: {
      count: {
          $gte: 200
      }
    }
  }
])

MongoDB基于大于表达式的分组结果

2 个答案: