Question

我从一个集合中提取了大量结果。我想先显示过去30天内所有条目，然后按下载次数对其余条目进行排序。一次只加载10个，滚动时会加载更多。

Data.aggregate([
      {'$match': match},
      {"$sort": {downloads : -1, createdAt: -1}},
      {"$skip": page * entriesPerPage},
      {"$limit": entriesPerPage},
      {"$lookup": {
        "from": "examples",
        "localField": "_id",
        "foreignField": "docId",
        "as": "examples"
      }},
    ]).exec(function(err, docs) {});

现在我尝试通过获取所有结果然后将它们分成两个数组来手动完成，但现在我意识到这意味着我还必须手动执行$ skip部分。同样不是很重要，我必须加载并查看所有结果。

有没有办法在聚合查询中执行此操作？

编辑：这是一份文件：

{ "_id" : ObjectId("58ddc3f76853d8286b22bc7b"), 
"updatedAt" : ISODate("2017-11-12T09:39:22.031Z"), 
"createdAt" : ISODate("2017-03-31T02:50:31.631Z"), 
"name" : "Dragon", 
"slug" : "dragon", 
"description" : "A dragon is a legendary creature, typically scaled or fire-spewing and with serpentine, reptilian or avian traits, that features in the myths of many cultures around world.", 
"downloads" : 18, 
"type" : "monster", 
"numberOfColors" : 16, "__v" : 1, 
"tags" : [ "monster" ] }

目前大约有100个。我想再次显示创建过去30天内的所有内容，然后按下载次数显示其余内容。

Answer 1

您将需要一个临时帮助字段，您可以使用$addFields管道阶段（或$project处理MongoDB版本＆lt; 3.4）来获取该字段：

var thirtyDaysInMilliSeconds = 30 * 24 * 60 * 60 * 1000;

Data.aggregate([
      {"$match": match},
      {'$addFields': { // let's add a field that contains either 0 or 1 depending on whether the "createdAt" value is greater than "today - 30 days"
          "sortHelperField": { $cond: [ { $gt: [ "$createdAt", { $subtract: [ new Date(), thirtyDaysInMilliSeconds ] } ] }, "$createdAt", 0] }
      }},
      {"$sort": {"sortHelperField": -1, downloads: -1}}, // then we can sort by that field, too, and make it the field with the highest priority
      /* all the rest of your stages can stay the way the are */
    ]);

Mongo DB使用aggregate将更新的结果放在查询的顶部

1 个答案: