我可以在 MongoDB 的聚合查询中申请 forEach 吗？

Question

我有一个成员集合，通过特定条件查找成员，获取成员后我需要对每个成员进行一些计算。在同一个集合上计算需要查询。

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？

谁能帮帮我:)

例如：

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[{
    "_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 中的预期结果如：

[
  { 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 }
]

Answer 1

聚合管道无法做到这一点。你应该明白 MongoDB 聚合是一系列应用于集合的特殊运算符。当您执行聚合管道时，MongoDB 将运算符相互传递，即运算符的输出成为以下运算符的输入。每个运算符的结果是一个新的文档集合。

因此，您在上面尝试实现的目标可以简单地重写为以下管道，而无需先创建文档数组：

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

跟进您的问题的更改，运行以下聚合管道将为您提供所需的结果：

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);
});