mongo query: remove object when objects has matching values (email) - mongodb

Question

I'm trying to find all the objects with type==1 (works already) and, then check duplicates from email addresses. (not working)
Output would be something like this:
[ { _id:
{ id: 59d0e3f728150e39b008013f,
firstName: 'pera',
lastName: 'pekko',
phone: '04040404040',
massageType: 'jipii massage',
time: 45,
price: 30,
startsAt: '2017-10-02T11:00:00+03:00',
endsAt: '2017-10-02T11:45:00+03:00' },
email: [ 'gg#gmail.com' ],
{ _id:
{ id: 59ce25975f4d1a096cdd2733,
firstName: 'sdad',
lastName: 'autismus',
phone: '04040404040',
massageType: 'jipii massage',
time: 60,
price: 33,
startsAt: '2017-10-02T12:45:00+03:00',
endsAt: '2017-10-02T13:45:00+03:00' },
email: [ 'YES#hotmail.com' ]
}]
So basically it would check duplicate "email"-values. If there is a duplicate value --> remove WHOLE object from result.
My code so far (result prints all type==1 objects WITH duplicates):
db3.massagerTimes.aggregate(
{$match: {_id: mongo.helper.toObjectID(data.timesData)}},
{$unwind: "$times"},
{$match: {"times.type": 1}},
{
$group: {
_id: {
id: "$times._id",
firstName: "$times.firstName",
lastName: "$times.lastName",
phone: "$times.phone",
massageType: "$times.massageType",
time: "$times.time",
price: "$times.price",
startsAt: "$times.startsAt",
endsAt: "$times.endsAt"
},
email: {$addToSet: "$times.email"},
count: {"$sum": 1}
}
},
{
$match: {
count: {"$lte": 1}
}
},
{$sort: {"times._id": -1}},
function (err, result) {
if (err)
throw err;
if (result) {
console.log(result);
res.json(result);
}
});
}
Collection:
{
"_id": {
"$oid": "59ca4eb610b3be1f8499f6d5"
},
"times": [
{
"_id": {
"$oid": "59ce274f5f4d1a096cdd2745"
},
"type": 0,
"startsAt": "2017-10-02 13:30 +03:00",
"endsAt": "2017-10-02 13:45 +03:00"
},
{
"_id": {
"$oid": "59ce274f5f4d1a096cdd2744"
},
"type": 0,
"startsAt": "2017-10-02 13:15 +03:00",
"endsAt": "2017-10-02 13:30 +03:00"
},
{
"_id": {
"$oid": "59ce28415f4d1a096cdd275b"
},
"title": "sdad autismus",
"firstName": "sdad",
"lastName": "autismus",
"type": 1,
"email": "YES#hotmail.com",
"phone": "04040404040",
"massageType": "jipii massage",
"time": 45,
"price": 30,
"additionalInfo": null,
"startsAt": "2017-10-01T08:30:00+03:00",
"endsAt": "2017-10-01T09:15:00+03:00"
},
{
"_id": {
"$oid": "59ce28665f4d1a096cdd275e"
},
"title": "sdad autismus",
"firstName": "sdad",
"lastName": "autismus",
"type": 1,
"email": "YES#hotmail.com",
"phone": "04040404040",
"massageType": "jipii massage",
"time": 45,
"price": 30,
"additionalInfo": null,
"startsAt": "2017-10-01T09:30:00+03:00",
"endsAt": "2017-10-01T10:15:00+03:00"
},
{
"_id": {
"$oid": "59d0e3f728150e39b008013f"
},
"title": "pera pekko",
"firstName": "pera",
"lastName": "pekko",
"type": 1,
"email": "gg#gmail.com",
"phone": "04040404040",
"massageType": "some massage",
"time": 45,
"price": 30,
"additionalInfo": null,
"startsAt": "2017-10-02T11:00:00+03:00",
"endsAt": "2017-10-02T11:45:00+03:00"
}
]
}
Question: How I check duplicates from value and then drop it from results?
(in this case YES#hotmail.com <-- 2x, remove duplicate object, not just the duplicate value)

Answer 1

The core of what you need would be something like that:
db3.massagerTimes.aggregate({
$unwind: "$times"
}, {
$group: {
_id: "$times.email", // group all items with the same email together
doc: {$first: "$$ROOT"} // keep only the first document for each email address
}
}, {
$replaceRoot: {
newRoot: "$doc" // move the element we remembered during the grouping stage to the top of our document
}
});