Mongodb计数与多个组字段不同

Question

我有交易表,由员工带来的假期填充.我需要帮助在mongodb中跟踪sql场景.

select employee,month,year,count(distinct (holiday_type) from 
transactions group by employee,month,year

我需要在mongodb中使用聚合,并且像这样创建了mongo查询,这给了我错误的解决方案

db.transactions.aggregate([
    { "$group": { 
        "_id": { 
            "Month": { "$month" : "$date" }, 
            "Year": { "$year" : "$date" },
            "employee" : "$employee",
            "holiday_type" : "$holiday_type"
        },
        "Count_of_Transactions" : { "$sum" : 1 }
     }}
 ]);

我很困惑在mongodb中使用count distinct逻辑.任何建议都会有所帮助

Answer 1

部分方式,但你需要首先得到"holiday_type"的"不同"值,然后你$group再次:

db.transactions.aggregate([
    { "$group": { 
        "_id": { 
            "employee" : "$employee",
            "Month": { "$month" : "$date" }, 
            "Year": { "$year" : "$date" },
            "holiday_type" : "$holiday_type"
        },
     }},
     { "$group": {
         "_id": {
            "employee" : "$_id.employee",
            "Month": "$_id.Month",
            "Year": "$_id.Year"
         },
         "count": { "$sum": 1 }
     }}
 ], { "allowDiskUse": true }
 );

这是一般过程,因为SQL中的"distinct"本身就是一种分组操作.所以这是一个双重$group操作,以获得正确的结果.