Mongoose - finding subdocuments by criteria

Question

I've just got stuck with this problem. I've got two Mongoose schemas:

var childrenSchema = mongoose.Schema({
    name: {
        type: String
    },
    age: {
        type: Number,
        min: 0
    }
});

var parentSchema = mongoose.Schema({
    name : {
        type: String
    },
    children: [childrenSchema]
});

Question is, how to fetch all subdocuments (in this case, childrenSchema objects) from every parent document? Let's suppose I have some data:

var parents = [
    { name: "John Smith",
    children: [
        { name: "Peter", age: 2 }, { name: "Margaret", age: 20 }
    ]},
    { name: "Another Smith",
    children: [
        { name: "Martha", age: 10 }, { name: "John", age: 22 }
    ]}
];

I would like to retrieve - in a single query - all children older than 18. Is it possible? Every answer will be appreciated, thanks!

Do you want it to only return the parent if the child is over 18 or do you want it to only populate the children that are over 18 on each parent? — mr.freeze, May 30 '13 at 20:48

score 61 · Accepted Answer · answered May 30 '13 at 21:12

You can use $elemMatch as a query-projection operator in the most recent MongoDB versions. From the mongo shell:

db.parents.find(
    {'children.age': {$gte: 18}},
    {children:{$elemMatch:{age: {$gte: 18}}}})

This filters younger children's documents out of the children array:

{ "_id" : ..., "children" : [ { "name" : "Margaret", "age" : 20 } ] }
{ "_id" : ..., "children" : [ { "name" : "John", "age" : 22 } ] }

As you can see, children are still grouped inside their parent documents. MongoDB queries return documents from collections. You can use the aggregation framework's $unwind method to split them into separate documents:

> db.parents.aggregate({
    $match: {'children.age': {$gte: 18}}
}, {
    $unwind: '$children'
}, {
    $match: {'children.age': {$gte: 18}}
}, {
    $project: {
        name: '$children.name',
        age:'$children.age'
    }
})
{
    "result" : [
        {
            "_id" : ObjectId("51a7bf04dacca8ba98434eb5"),
            "name" : "Margaret",
            "age" : 20
        },
        {
            "_id" : ObjectId("51a7bf04dacca8ba98434eb6"),
            "name" : "John",
            "age" : 22
        }
    ],
    "ok" : 1
}

I repeat the $match clause for performance: the first time through it eliminates parents with no children at least 18 years old, so the $unwind only considers useful documents. The second $match removes $unwind output that doesn't match, and the $project hoists children's info from subdocuments to the top level.

It works! :) Thanks for help! :) Where can I get some docs about advanced MongoDB operations? Its docs are enough? I wish to learn as much as possible about MongoDB :) — Kuba T, Jun 02 '13 at 18:21
Take 10gen's online MongoDB For Developers class, read Rick Copeland's MongoDB Design Patterns book. — A. Jesse Jiryu Davis, Jun 03 '13 at 16:00
I know, i'm quite late, but what if a parent have more than one children over 18? I tried the first query on mongo shell, but it only retrieves one match for each parent. Am I right? — andrea.spot., Sep 25 '14 at 15:50
Thanks!@A.JesseJiryuDavis I have one more question about it, I saw the results only contain "_id" of parent object, what could I do if I want keep all parent's data and only with children over 18, like: { "_id" : ..., "name": "John Smith", "children" : [ { "name" : "Margaret", "age" : 20 } ] } — Jiang, Apr 14 '15 at 20:32
For those who end up here and are frustrated with this answer like I was... It can be done just using find. Here's an example: https://mongoplayground.net/p/qfAskjZcNGz I got there from this question, though it's specifically mentioning nesting. https://stackoverflow.com/questions/36229123/return-only-matched-sub-document-elements-within-a-nested-array — Ethan, Jan 05 '22 at 02:53

score 23 · Answer 2 · answered Feb 08 '15 at 14:45

23

In Mongoose, you can also use the elegant .populate() function like this:

parents
.find({})
.populate({
  path: 'children',
  match: { age: { $gte: 18 }},
  select: 'name age -_id'
})
.exec()

answered Feb 08 '15 at 14:45

AbdelHady

9,334
8
56
83

12

`populate` works when referencing documents in other collections. The OP is using sub docs. – Lulylulu Aug 02 '17 at 15:48
1

It is a long time since I made this answer :) & I can't try it now, but I think it should work also for embedded docs! – AbdelHady Aug 02 '17 at 15:59
If it does not match returns `null` – nicogaldo Feb 25 '19 at 14:49
4

@AbdelHady it wont work for embedded documents, `populate` only works for referenced documents – Ravi Shankar Bharti Mar 27 '19 at 05:26
Not relevant for the question, and could lead to confusion and bugs been introduced to peoples code - as this examines multiple collections. – omeanwell Jun 12 '23 at 14:44

score 1 · Answer 3 · answered May 29 '19 at 10:53

A. Jesse Jiryu Davis's response works like a charm, however for later versions of Mongoose (Mongoose 5.x) we get the error:

Mongoose 5.x disallows passing a spread of operators to Model.aggregate(). Instead of Model.aggregate({ $match }, { $skip }), do Model.aggregate([{ $match }, { $skip }])

So the code would simply now be:

> db.parents.aggregate([{
    $match: {'children.age': {$gte: 18}}
}, {
    $unwind: '$children'
}, {
    $match: {'children.age': {$gte: 18}}
}, {
    $project: {
        name: '$children.name',
        age:'$children.age'
    }
}])
{
    "result" : [
        {
            "_id" : ObjectId("51a7bf04dacca8ba98434eb5"),
            "name" : "Margaret",
            "age" : 20
        },
        {
            "_id" : ObjectId("51a7bf04dacca8ba98434eb6"),
            "name" : "John",
            "age" : 22
        }
    ],
    "ok" : 1
}

(note the array brackets around the queries)

Hope this helps someone!

Mongoose - finding subdocuments by criteria

3 Answers3

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