How to filter array in subdocument with MongoDB
Above solution works best if multiple matching sub documents are required. $elemMatch also comes in very use if single matching sub document is required as output
db.test.find({list: {$elemMatch: {a: 1}}}, {'list.$': 1})
Result:
{
"_id": ObjectId("..."),
"list": [{a: 1}]
}
Using aggregate
is the right approach, but you need to $unwind
the list
array before applying the $match
so that you can filter individual elements and then use $group
to put it back together:
db.test.aggregate([
{ $match: {_id: ObjectId("512e28984815cbfcb21646a7")}},
{ $unwind: '$list'},
{ $match: {'list.a': {$gt: 3}}},
{ $group: {_id: '$_id', list: {$push: '$list.a'}}}
])
outputs:
{
"result": [
{
"_id": ObjectId("512e28984815cbfcb21646a7"),
"list": [
4,
5
]
}
],
"ok": 1
}
MongoDB 3.2 Update
Starting with the 3.2 release, you can use the new $filter
aggregation operator to do this more efficiently by only including the list
elements you want during a $project
:
db.test.aggregate([
{ $match: {_id: ObjectId("512e28984815cbfcb21646a7")}},
{ $project: {
list: {$filter: {
input: '$list',
as: 'item',
cond: {$gt: ['$$item.a', 3]}
}}
}}
])
Use $filter aggregation
Selects a subset of the array to return based on the specified condition. Returns an array with only those elements that match the condition. The returned elements are in the original order.
db.test.aggregate([
{$match: {"list.a": {$gt:3}}}, // <-- match only the document which have a matching element
{$project: {
list: {$filter: {
input: "$list",
as: "list",
cond: {$gt: ["$$list.a", 3]} //<-- filter sub-array based on condition
}}
}}
]);