Mongodb aggregate group by day (php) - php

Question

I have a collection with documents that look like this:
{
_id: ObjectId("516eb5d2ef4501a804000000"),
accountCreated: "2013-04-17 16:46",
accountLevel: 0,
responderCount: 0
}
I want to group and count these documents based on the accountCreated date (count per day), but I am stuck with the handling of dates since the date includes time as well.
This is what I have, but it returns the count including the time, witch means lots of entries always with 1 as accounts.
$g = $form->mCollectionUsers->aggregate(array(
array( '$group' => array( '_id' => '$accountCreated', 'accounts' => array( '$sum' => 1 ) ) )
));
Is there a way to rewrite the date to only take day in account and skip the time?
I have found this example but I can´t really get figure out how to adapt it to this example.

Answer 1

If accountCreated is a date you can do it like this (I'll use the mongo shell syntax since I'm not familiar with the php driver):
db.mCollectionUsers.aggregate([
{$project :{
day : {"$dayOfMonth" : "$accountCreated"},
month : {"$month" : "$accountCreated"},
year : {"$year" : "$accountCreated"}
}},
{$group: {
_id : {year : "$year", month : "$month", day : "$day"},
accounts : { "$sum" : 1}
}}
]);

Answer 2

If you want to display the date properly:
db.mCollectionUsers.aggregate([
{
$group: {
_id: { $dateToString: { format: '%Y-%m-%d', date: '$accountCreated' } },
count: { $sum: 1 }
}
},
{
$project: {
_id: 0,
date: '$_id',
count: 1
}
}
])
The result would look like:
[
{
"date": "2020-11-11",
"count": 8
},
{
"date": "2020-11-13",
"count": 3
},
{
"date": "2020-11-16",
"count": 3
},
]