MongoDB 订单/销售聚合组每月总和 + 计数字段

谁知道更好的解决方案来按日期和总计对订单进行分组并按来源进行计数。当然，我可以按来源分组，然后我仅获得该来源的总计，之后我可以更改结果以获得所需的结果。但我想知道是否可以通过一种简单的方式$group陈述。

Eg. ordersByApp = 1, ordersByWEB = 2

订单收集

{
 _id: 'XCUZO0',
 date: "2020-02-01T00:00:03.243Z"
 total: 9.99,
 source: 'APP'
},
{
 _id: 'XCUZO1',
 date: "2020-01-05T00:00:03.243Z"
 total: 9.99,
 source: 'WEB'
},
{
 _id: 'XCUZO2',
 date: "2020-01-02T00:00:03.243Z"
 total: 9.99,
 source: 'WEB'
}

我当前的聚合

Order.aggregate([
   {
     $group: {
        _id: {
           month: { $month: "$date",
           year: { $year: "$date" } 
        },
        total: {
           $sum: "$total"
        } 
     }
   }
])

目前结果

[
  {
    _id: { month: 01, year: 2020 },
    total: 19.98
  },
  {
    _id: { month: 02, year: 2020 },
    total: 9.99
  }
]

期望的结果，我怎样才能实现以下目标？

[
  {
    _id: { month: 01, year: 2020 },
    total: 19.98,
    countByApp: 1, <---
    countByWEB: 0, <---
  },
  {
    _id: { month: 02, year: 2020 },
    total: 9.99,
    countByWEB: 2, <---
    countByAPP: 0  <---
  }
]

您可以使用$cond像下面这样：

Order.aggregate([
    {
        $group: {
            _id: {
                month: { $month: "$date" },
                year: { $year: "$date" } 
            },
            total: { $sum: "$total" },
            countByApp: { $sum: { $cond: [ {$eq: [ "$source", "APP" ]} , 1, 0] } },
            countByWeb: { $sum: { $cond: [ {$eq: [ "$source", "WEB" ]} , 1, 0] } },
        }
    }
])

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