在Mongodb / Nodejs中使用2個單獨的集合進行復雜的查詢

Question

我需要在Mongo DB / Node js中為我要構建的Web應用程序執行以下公式：

Average = (SUM (Collection_2.Amount where Collection2.Date is for previous month (31 days) and Collection 1.external_id is 20)) / (Total Number of documents in Collection 1 where Collection1.Status = "Active" and Collection 1.external_id = 20).

這是集合的示例。

Collection_1
{_id, common_field_1, external_id, status}

Collection_2
{_id, common_field_2, Date, Amount}

下面是兩個單獨查詢的示例。 本質上，我需要划分Total_2 / Total_1 。

    try {
Collection_1.aggregate([
{"$match" : {external_id : 20, Status : "Active"},
{"$group" : {_id: null, "Total_1": {$sum: 1}}} 

    try {
Collection_2.aggregate([
{"$lookup" : {"from" : "Collection_1", "localField" : "common_field_2", "foreignField" : "common_field_1", "as" : "JOIN"}},
{"$match" : {
"JOIN.external_id" : 20, 
"$expr": { "$gte": [ "$Date", new Date("2019-07-23T00:00:00.000+00:00")]}}},
{"$group" : {_id: null, "Total_2": {$sum: "$Amount"}}} 

有誰知道是否可以在單個查詢中執行此操作？

Answer 1

這不是很漂亮，但是我設法使用$ facet找到了一個解決方案：

Collection_1.aggregate([
{ 
    "$lookup" : {
        "from" : "Collection_2", 
        "localField" : "common_field_1", 
        "foreignField" : "common_field_2", 
        "as" : "JOIN"
    }
},
{
    "$match" : {external_id : 20}
},
{
    "$facet": {
        "Total_1": [
            { "$match" : {Status : "Active"}},
            { "$group": { _id: null, "Total" : { $sum: 1 } } }
        ],
        "Total_2": [
            { "$unwind" : "$JOIN"},
            { "$match" : {"JOIN.Date" : { $gte : new Date("2019-07-23T00:00:00.000+00:00")}}},
            { "$group" : {_id: null, "Total": {$sum: "$JOIN.Amount"}}}
        ]
    }            
},
{
    "$unwind" : "$Total_1"
},
{
    "$unwind" : "$Total_2"
},
{
    "$project" : { "Result" : {"$divide" : ["$Total_2.Total", "$Total_1.Total"]}}
}
]