Get index of an element mongodb aggregation - arrays

Question

Here is my collection
{
"_id" : ObjectId("5c225f9a66d39d55c036fa66"),
"name" : "Sherlock",
"mobile" : "999999",
"adress" : [
{
"street" : "221b baker street",
"city" : "london"
},
{
"street" : "ben street",
"city" : "london"
}
],
"tags" : [
"Detective",
"Magician",
"Avenger"
]
}
Now I want to get the first or second value inside address array.
for that I'm using this command.
> db.agents.findOne({"name" : "Sherlock"},{"adress" : 1})
but instead of giving a single result it is giving the entire array like
{
"_id" : ObjectId("5c225f9a66d39d55c036fa66"),
"adress" : [
{
"street" : "221b baker street",
"city" : "london"
},
{
"street" : "ben street",
"city" : "london"
}
]
}
It can be done by comparing array value like
db.agents.find({"adress.street": "ben street"}, {_id: 0, 'adress.$': 1});
But I don't want to compare just to print the array indexes. How can I get the single result?
Any help is appreciated..

Answer 1

You can use $arrayElemAt to get the specific element from the array
db.collection.aggregate([
{ $addFields: { "$arrayElemAt": ["$adress", 0] }} //index
])
and if you want to get the sliced element then you can use $slice projection
db.collection.find({}, { adress: { $slice: [2, 1] }}) // 2 index and 1 number of element

Answer 2

You can $unwind with includeArrayIndex to get the index of address array
db.t11.aggregate([
{$match : {"adress.street" : "ben street"}},
{$unwind : {path : "$adress", includeArrayIndex : "idx"}},
{$match : {"adress.street" : "ben street"}}
]).pretty()
you can add $project to filter the fields not required
result
> db.t11.aggregate([{$match : {"adress.street" : "ben street"}},{$unwind : {path : "$adress", includeArrayIndex : "idx"}},{$match : {"adress.street" : "ben street"}}]).pretty()
{
"_id" : ObjectId("5c225f9a66d39d55c036fa66"),
"name" : "Sherlock",
"mobile" : "999999",
"adress" : {
"street" : "ben street",
"city" : "london"
},
"tags" : [
"Detective",
"Magician",
"Avenger"
],
"idx" : NumberLong(1)
}
>