如何使用聚合函数mongo db-query [英] How to use aggregation function mongo db-query

问题描述

我是MongoDB的新手，我想在我要检查type == topic并获取以下输出的地方使用聚合函数

I am new in MongoDB and I would like to use the aggregation function where I want to check type == topic and get the following output

预期输出

[
    {
        conceptName : 59d98cfd1c5edc24e4024d00
        totalCount : 2
    },
    {
        conceptName : 59d98cfd1c5edc24e4024d03
        totalCount : 1
    }
]

示例输入db.GroupContents

Sample input db.GroupContents

{
    "_id" : "5a0948bb1c5edc7a5000521a",
    "type" : "topic",
    "groupID" : "5a0948bb1c5edc7a5000521a",
    "pedagogyID" : "59d98cfa1c5edc24e40249a3",
   }

示例输入db.PedagogyNodes

Sample input db.PedagogyNodes

{
    "_id" : "59d98cfa1c5edc24e40249a3",
    "latestVersion" : "59d98cfa1c5edc24e402497f_1",
    "createdAt" : "2017-10-08 04:27:06",
    "updatedAt" : "2017-10-08 04:27:06"
}

示例输入db.PedagogyVersions

Sample input db.PedagogyVersions

    {
    "_id" : "59d98cfa1c5edc24e402497f_1",
    "type" : "topic",
    "contentNodes" : {
        "LearningNodes" : [
            "59d98cfd1c5edc24e4024d00",
            "59d98cfd1c5edc24e4024d03",
            "59d98cfd1c5edc24e4024d00",
        ]
    },
    "createdAt" : "2017-10-08 04:27:06",
    "updatedAt" : "2017-10-08 04:27:06"
}

到目前为止我已经尝试过的

What I have tried so far

var groupID = "5a0948bb1c5edc7a5000521a"; // Step 1
var records;
var pnDoc;
var pvDoc;
db.GroupContents.find({groupID : groupID}).forEach(function (doc){ // Step 2
   var pedagogyID = doc.pedagogyID;
   var records = db.getSiblingDB('PedagogyService');
       records.PedagogyNodes.find({_id : pedagogyID}).forEach(function (pnDoc) { // Step 3
          var latestVersion = pnDoc.latestVersion;
          // addded aggregate function here
          records.PedagogyVersions.aggregate([
            {
                $match:{_id:latestVersion} // Step 4
            },
            {
               $unwind:"$contentNodes.LearningNodes"
            },
            {
                $group:
                {
                    _id:"$contentNodes.LearningNodes",
                    count:{$sum:1}
                }
            }
        ])
      })
});

我无法根据预期答案编写数据库查询，请帮忙.

I am unable to write db query based on my expected answer, please help.

了解我的要求

Understand my requirement

Step : 1 => I am passing `groupID = 5a0948bb1c5edc7a5000521a`
Step : 2 => we have to check from GroupContents where groupID = groupID then we have to take `pedagogyID`
Step : 3 => we have to check from PedagogyNodes where _id = pedagogyID then we have to take `latestVersion`
Step : 4 => we have to check from PedagogyVersions where _id = latestVersion then we have to take `contentNodes->LearningNodes`
Step : 5 => Finally we have to do the aggregation then we have display the result

推荐答案

要查看聚合结果，您必须传递回调以作为参数执行.

To see the result of your aggregation you have to pass the callback to be executed as parameter.

records.PedagogyVersions.aggregate([
        {
            $match:{_id:latestVersion} // Step 4
        },
        {
           $unwind:"$contentNodes.LearningNodes"
        },
        {
            $group:
            {
                _id:"$contentNodes.LearningNodes",
                count:{$sum:1}
            }
        }
    ], function(err, results) {
         console.log(results);
    });

这篇关于如何使用聚合函数mongo db-query的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！