How to group MongoDB aggregation [duplicate] - mongodb

Question

Who knows a better solution to group Orders by date and sum total and count by source. Of course I can group by Source and then I get only totals for this source only, I can alter the result thereafter to get the desired result. But I would like to know if it is possible in one simple $group statement.
Eg. ordersByApp = 1, ordersByWEB = 2
Orders collection
{
_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'
}
My current aggregation
Order.aggregate([
{
$group: {
_id: {
month: { $month: "$date",
year: { $year: "$date" }
},
total: {
$sum: "$total"
}
}
}
])
Current result
[
{
_id: { month: 01, year: 2020 },
total: 19.98
},
{
_id: { month: 02, year: 2020 },
total: 9.99
}
]
Desired result, How can I achieve the below?
[
{
_id: { month: 01, year: 2020 },
total: 19.98,
countByApp: 1, <---
countByWEB: 0, <---
},
{
_id: { month: 02, year: 2020 },
total: 9.99,
countByWEB: 2, <---
countByAPP: 0 <---
}
]

Answer 1

You can use $cond like below:
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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