MongoDB Aggregate and keep all values while setting an active key - mongodb

Question

I want all the values returned from "Items" and when there is a match I want "isActive" to be true.
exports.listUserItems = function (req, res) {
// Aggregate results
User.aggregate([{
"$match": {
"username": req.params.username
}
}, {
"$lookup": {
"from": "Items",
"localField": "itemIds",
"foreignField": "_id",
"as": "items"
}
}, {
"$unwind": {
"path": "$items"
}
}, {
"$project": {
"item": "$items.item_name",
"isActive": '1',
"_id": 0
}
}], (err, result) => res.json(result));
};
What is the best way to go about accomplishing this?
I was going to return all the items and users items seperately, then compare them and object.value them etc. etc... that seems like overkill. Can it be done on the model side?
I'm unable to post the document structure because stackoverflow doesn't let me, but you should get the idea.
Edit:
User Document
{
"username" : "anonuser",
"items" : [
ObjectId("5ba8345f1e56fe8e6caaaa07"),
ObjectId("5ba706d64e82292e72e9ae71")
]
}
Then I have an "Items" collection which has 3 documents like this.
{
"_id" : ObjectId("5ba706d64e82292e72e9ae71"),
"item_name" : "Salary"
}
My expected json api output is to be.
[{"_id":"5ba706d64e82292e72e9ae71","item_name":"Salary","isActive":true},{"_id":"5ba8345f1e56fe8e6caaaa07","item_name":"Fulltime","isActive":true},{"_id":"5ba9af6c1e56fe8e6cab521e","item_name":"Advisor","isActive":false}]

Answer 1

You can try below aggregation
Items.aggregate([
{ "$lookup": {
"from": "users",
"let": { "itemsId": "$_id" },
"pipeline": [
{ "$match": { "$expr": { "$in": ["$$itemsId", "$items"] }}},
{ "$project": { "isActive": { "$literal": true }}}
],
"as": "items"
}},
{ "$addFields": {
"isActive": {
"$ifNull": [{ "$arrayElemAt": ["$items.isActive", 0] }, { "$literal": false }]
}
}},
{ "$project": { "items": 0 }}
])