How to write the following MongoDB query - mongodb

Question

Suppose I have the following foobar collection:
{
"foobar": [
{
"_id": "abc",
"history": [
{
"type": "foo",
"timestamp": 123456789.0
},
{
"type": "bar",
"timestamp": 123456789.0
}
]
},
{
"_id": "dfg",
"history": [
{
"type": "baz",
"timestamp": 123456789.0
},
{
"type": "bar",
"timestamp": 123456789.0
}
]
},
{
"_id": "hij",
"history": [
{
"type": "foo",
"timestamp": 123456789.0
},
{
"type": "bar",
"timestamp": 123456789.0
},
{
"type": "foo",
"timestamp": 123456789.0
}
]
}
]
}
How can I query ($gte/$lte) the foobar items based on the timestamp prop inside their history items, but using the timestamp from the item where type: "foo"?
If there is no subdocuments with type equals to foo, the entire document is filtered out, if there is more than one subdocument with type equals to foo then it could match anyone.

Answer 1

You can try below aggregation:
var threshold = 123456788;
db.foobar.aggregate([
{
$addFields: {
foobar: {
$filter: {
input: "$foobar",
as: "doc",
cond: {
$let: {
vars: {
foo: {
$arrayElemAt: [
{
$filter: {
input: "$$doc.history",
as: "history",
cond: {
$eq: [ "$$history.type", "foo" ]
}
}
},
0]
}
},
in: {
$gt: [ "$$foo.timestamp", threshold ]
}
}
}
}
}
}
}
])
$addFields can be used to overwrite existing field (foobar). To filter out all subdocuments not matching your condition you can use $filter (the outer one). For each foobar document you can use $let to define temporary variable foo. You need inner $filter to get all history elements where type is foo and then $arrayElemAt to get first one. Then you just need $gt or $lt to apply your comparison.
For those subdocuments where there's no foo there will be undefined returned which will be then filtered out on $gt stage.