Mongodb aggregate return data conditionally - database

Question

I am trying to return ONLY data records which email not exist in other collection with mongodb aggregation. Example:
LetterCollection:
[
{
"email": "some#email.com",
"data": "someData"
}, {
"email": "some2#email.com",
"data": "someData2"
},
]
UserCollection:
[
{
"id": "1"
"email": "some#email.com",
}
]
Get all letters which user email not exist in users collection.
Expected result:
[{
"email": "some2#email.com",
"data": "someData2"
}]
How can I achieve that?

Answer 1

Run an aggregate operation on the letters collection in which the pipeline first does a $lookup to the users collection, joined on the email field.
Documents returned from this pipeline stage contain an array field with the joined documents, so you would need a further $match pipeline step to filter documents where this new field does not have any documents. This can be done by checking the length of the array field with $size and filtering using the $expr expression with the $eq comparison operator.
The result can be further $projected to remove the new field in those documents that match to give the final desired output.
The above pipeline can be expressed as:
db.letters.aggregate([
{ '$lookup': {
'from': 'users',
'localField': 'email',
'foreignField': 'email',
'as': 'users'
} },
{ '$match': { '$expr': { '$eq': [ { '$size': '$users' }, 0 ] } } },
{ '$project': { 'users': 0 } }
])