Return first element if no match found in array - mongodb

Question

I have the following document:
{
_id: 123,
state: "AZ",
products: [
{
product_id: 1,
desc: "P1"
},
{
product_id: 2,
desc: "P2"
}
]
}
I need to write a query to return a single element from the products array where state is "AZ" and product_id is 2. If the matching product_id is not found, then return the first (or any) element from the products array.
For example: If product_id is 2 (match found), then the result should be:
products: [
{
product_id: 2,
desc: "P2"
}
]
If the product_id is 3 (not found), then the result should be:
products: [
{
product_id: 1,
desc: "P1"
}
]
I was able to meet one condition when the match is found but not sure how to satisfy the second condition in the same query:
db.getCollection('test').find({"state": "AZ"}, {_id: 0, state: 0, products: { "$elemMatch": {"product_id": "2"}}})
I tried using the aggregation pipeline as well but could not find a working solution.
Note: This is different from the following question as I need to return a default element if the match is not found:
Retrieve only the queried element in an object array in MongoDB collection

Answer 1

You can try below aggregation
Basically you need to $filter the products array and check for the $condition if it doesn't contain any element or equal to [] then you have to $slice with the first element of the products array.
db.collection.aggregate([
{ "$addFields": {
"products": {
"$cond": [
{
"$eq": [
{ "$filter": {
"input": "$products",
"cond": { "$eq": ["$$this.product_id", 2] }
}},
[]
]
},
{ "$slice": ["$products", 1] },
{ "$filter": {
"input": "$products",
"cond": { "$eq": ["$$this.product_id", 2] }
}}
]
}
}}
])
or even using $let aggregation
db.collection.aggregate([
{ "$addFields": {
"products": {
"$let": {
"vars": {
"filt": {
"$filter": {
"input": "$products",
"cond": { "$eq": ["$$this.product_id", 2] }
}
}
},
"in": {
"$cond": [
{ "$eq": ["$$filt", []] },
{ "$slice": ["$products", 1] },
"$$filt"
]
}
}
}
}}
])

Answer 2

If you don't care which element you get back then this is the way to go (you'll get the last element in the array in case of no match since $indexOfArray will return -1):
db.getCollection('test').aggregate([{
$addFields: {
"products": {
$arrayElemAt: [ "$products", { $indexOfArray: [ "$products.product_id", 2 ] } ]
},
}
}])
If you want the first then do this instead ($max will take care of transforming -1 into index 0 which is the first element):
db.getCollection('test').aggregate([{
$addFields: {
"products": {
$arrayElemAt: [ "$products", { $max: [ 0, { $indexOfArray: [ "$products.product_id", 2 ] } ] } ]
},
}
}])
Here is a version that should work on v3.2 as well:
db.getCollection('test').aggregate([{
"$project": {
"products": {
$slice: [{
$concatArrays: [{
$filter: {
"input": "$products",
"cond": { "$eq": ["$$this.product_id", 2] }
}},
"$products" // simply append the "products" array
// alternatively, you could append only the first or a specific item like this [ { $arrayElemAt: [ "$products", 0 ] } ]
]
},
1 ] // take first element only
}
}
}])