MongoDB - Fetch an object from deep subdocuments

Question

How to query an object from an Array inside an Array, and get it as a top-level object? For example, consider the following record.

{
  "subjects": [
         {
             "name": "English",
             "teachers": [
                  {
                    "name": "Mark"   /* Trying to get this object*/
                  },
                  {
                    "name": "John"
                  }
              ]
         }
   ]
}

I am trying to get the following object out as the top-level object.

{
   "name": "Mark"
}

dyouberg · Accepted Answer · 2016-09-06 16:45:23Z

1

You need to use the aggregation framework to do exactly what you're asking for.

Here I entered the document you gave into collection: foo.

> db.foo.find().pretty()

{
    "_id" : ObjectId("57ceed3d31484d5b491eaae9"),
    "subjects" : [
        {
            "name" : "English",
            "teachers" : [
                {
                    "name" : "Mark"
                },
                {
                    "name" : "John"
                }
            ]
        }
    ]
}

Using $unwind to unravel our array we then enter our first stage of the aggregation pipeline:

> db.foo.aggregate([
...     {$unwind: "$subjects"}
... ]).pretty()

{
    "_id" : ObjectId("57ceed3d31484d5b491eaae9"),
    "subjects" : {
        "name" : "English",
        "teachers" : [
            {
                "name" : "Mark"
            },
            {
                "name" : "John"
            }
        ]
    }
}

Subjects was an array of length 1 so the only difference here is one less set of [] array brackets.

We need to unwind again.

> db.foo.aggregate([
...     {$unwind: "$subjects"},
...     {$unwind: "$subjects.teachers"}
... ]).pretty()

{
    "_id" : ObjectId("57ceed3d31484d5b491eaae9"),
    "subjects" : {
        "name" : "English",
        "teachers" : {
            "name" : "Mark"
        }
    }
}
{
    "_id" : ObjectId("57ceed3d31484d5b491eaae9"),
    "subjects" : {
        "name" : "English",
        "teachers" : {
            "name" : "John"
        }
    }
}

Now we turned our array of length '2' into two separate documents. The first one with subjects.teachers.name = Mark and the second with subjects.teachers.name = John.

We only want to return the case where name = Mark so we need to add a $match stage to our pipeline.

> db.foo.aggregate([
...     {$unwind: "$subjects"},
...     {$unwind: "$subjects.teachers"},
...     {$match: {"subjects.teachers.name": "Mark"}}
... ]).pretty()

{
    "_id" : ObjectId("57ceed3d31484d5b491eaae9"),
    "subjects" : {
        "name" : "English",
        "teachers" : {
            "name" : "Mark"
        }
    }
}

Ok! Now we are only matching on the case where name: Mark.

Let's add a $project case to shape our input how we want.

> db.foo.aggregate([
...     {$unwind: "$subjects"},
...     {$unwind: "$subjects.teachers"},
...     {$match: {"subjects.teachers.name": "Mark"}},
...     {$project: {"name": "$subjects.teachers.name", "_id": 0}}
... ]).pretty()

{ "name" : "Mark" }

answered Sep 6, 2016 at 16:45

dyouberg

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4 Comments

Muhammad Kashif Nazar Over a year ago

Thank you very much for the detailed answered.

dyouberg Over a year ago

You have to use docs.mongodb.com/manual/reference/operator/aggregation/literal/… For instance: {$project: {"name": "$subjects.teachers.name", "_id": 0, "address": {$literal: "123 ABC"}}}

Muhammad Kashif Nazar Over a year ago

No. I mean to update the information into the collection. So that the collection becomes: { "subjects": [ { "name": "English", "teachers": [ { "name": "Mark" , "address": "123 ABC XYZ" }, { "name": "John" } ] } ] }

dyouberg Over a year ago

Aggregation isn't really a write operation in the Mongo Shell. You'd have to execute this inside your server code and pass the results into a callback function where you can then do your manipulation and resave the documents.

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