mongo - count return no docoument found instead of 0 - mongodb

Question

In SQL query
select count(*) from table where id=1
would return 0 as result where there isn't any record with such id.
I would like to get exactly the same behavior but in mongo. Unfortunately I can only use aggregate function.
I was trying something like this
db.collection.aggregate([
{
"$match": {
"key": 1
}
},
{
$count: "s"
}
])
It works but only with records with key:1 but when this key does not exist there is "no document found"

Answer 1

You can use this aggregation query using $facet to create two possible ways: If document exists or if document does not exists.
First $facet to create the two ways
Into notFound way the result will always be {count: 0} ; into found way there is the match
Then $replaceRoot merging results to get desired value.
db.collection.aggregate([
{
"$facet": {
"notFound": [
{
"$project": {
"_id": 0,
"count": {
"$const": 0
}
}
},
{
"$limit": 1
}
],
"found": [
{
"$match": {
"key": 1
}
},
{
"$count": "count"
}
]
}
},
{
"$replaceRoot": {
"newRoot": {
"$mergeObjects": [
{
"$arrayElemAt": [
"$notFound",
0
]
},
{
"$arrayElemAt": [
"$found",
0
]
}
]
}
}
}
])
Example here where key exists and here where key doesn't exists.
Also I've tested with this using $ifNull instead of $mergeObjects and seem works ok too.

Answer 2

I think the right way to do it is with the driver code, if you get empty results you make that document {"count" : 0} you dont need i think to do anything in the database.
Another solution can be this (replace the 5 with the key value you want)
Test code here
creates 2 groups the matched(count>0) and the not matched(count=0)
sort by {"count" : -1}
take the first, if there was a match count will be the one matched,
else it will be 0
aggregate(
[ {
"$group" : {
"_id" : {
"$cond" : [ {"$eq" : [ "$key", 5 ]}, "$key", "not_match" ]
},
"count" : {
"$sum" : {"$cond" : [ {"$eq" : [ "$key", 5 ]}, 1, 0 ]}
}
}
},
{"$sort" : {"count" : -1}},
{
"$group" : {
"_id" : null,
"count" : {"$first" : "$count"}
}
},
{"$project" : {"_id" : 0}}
])

Answer 3

I did it by using $facet,$project and when there were no documents to project it was showing undefined, so I used $ifNull expression. I've kept zero value for replacement expression value (see the $ifNull docs).
db.collection.aggregate([
{
"$facet": {
"keyFound": [
{
"$match": {
"key": 1
}
},
{
"$count": "count"
}
]
}
},
{
"$project": {
"keyFoundCount": {
"$ifNull": [
{
"$arrayElemAt": [
"$keyFound.count",
0
]
},
0
]
}
}
}
])
testCodeHere