mongodb aggregation get all fields sum group by date - mongodb

Question

Collection
{
"_id" : ObjectId("5a143a79ca78479b1dc90161"),
"createdAt" : ISODate("2017-11-21T14:38:49.375Z"),
"amount" : 227.93359186,
"pair" : "ant_eth"
}
Expected output
{
"12-12-2012": [
{
"pair": "ant_eth",
"sum": "sum of amounts in 12-12-2012"
},
{
"pair": "new_pair",
"sum": "sum of amounts in 12-12-2012"
},
],
"13-12-2012": [{
"pair": "ant_eth",
"sum": "sum of amounts in 13-12-2012"
}]
}
What I achieved so far from my knowledge is;
const criteria = [
{ $group: {
_id: '$pair',
totalAmount: { $sum: '$amount' } } }
]
Any help to achieve the expected output is much appreciated.

Answer 1

OK, so you want to sum up amount by just the date portion of a datetime and pair, and then "organize" all the pair+sum by date. You can do this by "regrouping" as follows. The first $group creates the sums but leaves you with repeating dates. The second $group fixes up the output to almost what you wish except that the dates remain as rvals to the _id instead of becoming lvals (field names) themselves.
db.foo.aggregate([
{
$group: {
_id: {d: {$dateToString: { format: "%Y-%m-%d", date: "$createdAt"}}, pair: "$pair"},
n: {$sum: "$amount"}
}
},
{
$group: {
_id: "$_id.d",
items: {$push: {pair: "$_id.pair", sum: "$n"}}
}
}
]);
If you REALLY want to have field names, then add these two stages after the second $group:
,{$project: {x: [["$_id","$items"]] }}
,{$replaceRoot: { newRoot: {$arrayToObject: "$x"} }}

Answer 2

This is what I could get to:
db.collection.aggregate([{
$group: {
_id: {
year: {
"$year": "$createdAt"
},
month: {
"$month": "$createdAt"
},
day: {
"$dayOfMonth": "$createdAt"
},
pair: "$pair"
},
sum: {
$sum: "$amount"
}
}
}])
For rest of the thing, you probably need to do app side parsing to generate output you want