我可以在MongoDB中的汇总查询中申请forEach吗? [英] Can I apply forEach in aggregate query in MongoDB?

问题描述

我有一个成员集合，并根据特定条件查找成员，获得成员后，我需要为每个成员进行一些计算.要计算同一集合上的需求查询.

I have a member collection and find members by specific conditions and after get members I need to do some calculation for each member. To calculate need query on the same collection.

我的过程是

var eachMemberInfo = [];
var members = db.collection('member').find({ createdDate: currentDate, country: 'BD'}).toArray();
members.forEach(function(doc) {
  var result = db.collection('member').aggregate([
    { $match: { memberType: doc.memberType, country : doc.country } },
    {
      $group: {
        _id: {memberType:"$memberType",country:"$country"},
        memberCount: { $sum: {$cond:[{$gt: ["$numberOfInvitees",0]},1,0]} },
        lessCount: { $sum: {$cond:[{$and:[{$lt:["$numberOfInvitees", doc.numberOfInvitees]}, {$gt: ["$numberOfInvitees",0]}]},1,0]} },
        sameCount: { $sum: {$cond:[{$eq: ["$numberOfInvitees",doc.numberOfInvitees]},1,0]} }
      }
    }
  ]).toArray();

   eachMemberInfo.push({memberId:doc.memberId,memberCount: result[0].memberCount, lessCount: result[0].lessCount});

});

我的问题是如何使用单个聚合查询来做到这一点?

任何人都可以帮助我吗:)

can any one help me pls :)

例如:

成员集合，例如:

[{
    "_id" : ObjectId("57905b2ca644ec06142a8c06"),
    "memberID" : 80,
    "memberType" : "N",
    "numberOfInvitees" : 2,
    "createdDate" : ISODate("2016-07-21T00:00:00.000Z"),
    "country" : "BD"
},
{
    "_id" : ObjectId("57905b2ca644ec06142a8c09"),
    "memberID" : 81,
    "memberType" : "N",
    "numberOfInvitees" : 3,
    "createdDate" : ISODate("2016-07-21T00:00:00.000Z"),
    "country" : "BD"
}
{
    "_id" : ObjectId("57905b2ca644ec06142a8fgh"),
    "memberID" : 82,
    "memberType" : "N",
    "numberOfInvitees" : 4,
    "createdDate" : ISODate("2016-07-21T00:00:00.000Z"),
    "country" : "BD"
}
{
    "_id" : ObjectId("57905b2ca644ec06142a8cfgd"),
    "memberID" : 83,
    "memberType" : "N",
    "numberOfInvitees" : 1,
    "createdDate" : ISODate("2016-07-21T00:00:00.000Z"),
    "country" : "BD"
}
{
    "_id" : ObjectId("57905b2ca644ec06142a8cfgd"),
    "memberID" : 84,
    "memberType" : "N",
    "numberOfInvitees" : 2,
    "createdDate" : ISODate("2016-07-21T00:00:00.000Z"),
    "country" : "BD"
}
..............
]

eachMemberInfo 中的预期结果，例如:

Expected result in eachMemberInfo like:

[
  { memberID : 80, memberCount:5,lessCount: 1,sameCount:2 },
  { memberID : 81, memberCount:5,lessCount: 3,sameCount:1 },
  { memberID : 82, memberCount:5,lessCount: 4,sameCount:1 },
  { memberID : 83, memberCount:5,lessCount: 0,sameCount:1 },
  { memberID : 84, memberCount:5,lessCount: 1,sameCount:2 }
]

推荐答案

使用聚合管道无法做到这一点.您应该了解MongoDB聚合是应用于集合的一系列特殊运算符.当您执行聚合管道时，MongoDB将运算符彼此管道传递，即运算符的输出成为后续运算符的输入.每个运算符的结果都是新的文档集合.

You can't do that with the aggregation pipeline. You should understand that MongoDB aggregation is a series of special operators applied to a collection. When you execute an aggregation pipeline, MongoDB pipes operators into each other i.e. the output of an operator becomes the input of the following operator. The result of each operator is a new collection of documents.

因此，您在上面想要实现的目标可以简单地重写为以下管道，而无需首先创建文档数组:

Hence what you are trying to achieve in the above can be simply rewritten as the following pipeline without the need to create an array of documents first:

var collection = db.collection('member'), 
    pipeline = [
        { "$match": { createdDate: currentDate, country: 'BD' } },
        {
            "$group": {
                "_id": { "memberType": "$memberType", "country": "$country" },
                "memberCount": { 
                    "$sum": { "$cond":[ { "$gt": ["$numberOfInvitees", 0] }, 1, 0 ] } 
                },
                "sameCount": { "$sum": 1 } 
            }
        }
    ];

collection.aggregate(pipeline, function(err, result){
    if (err) throw err;
    console.log(result);
});

---

更新

在对问题进行更改之后，运行以下聚合管道将为您提供所需的结果:

Follow-up to the changes to your question, running the following aggregation pipeline will give you the desired result:

var collection = db.collection('member'), 
    pipeline = [   
        { "$match": { createdDate: currentDate, country: 'BD' } },
        {
            "$group": {
                "_id": { 
                    "memberType": "$memberType", 
                    "country": "$country" 
                },            
                "invitees":{ 
                    "$push":  {
                        "memberID": "$memberID",
                        "count": "$numberOfInvitees"
                    }
                },
                "inviteesList": { "$push": "$numberOfInvitees" },
                "memberCount": { "$sum": 1 } 
            }
        },
        { "$unwind": "$invitees" },
        { "$unwind": "$inviteesList" },
        { 
            "$group": {
                "_id": "$invitees.memberID",
                "sameInviteesCount": { 
                     "$sum": { 
                        "$cond": [ 
                            { "$eq": ["$inviteesList", "$invitees.count"] }, 
                            1, 0 
                        ] 
                    }
                },
                "lessInviteesCount": { 
                    "$sum": { 
                        "$cond":[ 
                            { "$lt": ["$inviteesList", "$invitees.count"] }, 
                            1, 0 
                        ] 
                    }
                },
                "memberCount": { "$first": "$memberCount" }
            }
        }
    ];

collection.aggregate(pipeline, function(err, result){
    if (err) throw err;
    console.log(result);
});

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