MongoDB get SubDocument

Question

I would like to retrieve a sub document from a document in MongoDB. I have the following document:

{
    "_id" : "10000",
    "password" : "password1",
    "name" : "customer1",
    "enabled" : true,
    "channels" : [ 
        {
            "id" : "10000-1",
            "name" : "cust1chan1",
            "enabled" : true
        }, 
        {
            "id" : "10000-2",
            "name" : "cust1chan2",
            "enabled" : true
        }
    ]
}

The result I would like is:

{
    "id" : "10000-1",
    "name" : "cust1chan1",
    "enabled" : true
}

However, the best I can do so far is using the following query:

db.customer.find({"channels.id" : "10000-1"}, {"channels.$" : 1, "_id" : 0})

But this gives me the following result:

{
    "channels" : [ 
        {
            "id" : "10000-1",
            "name" : "cust1chan1",
            "enabled" : true
        }
    ]
}

Does anyone know if it is possible to write a query that will give me my desired result? Any help would be much appreciated.

If you're doing it from the mongoshell, something like db.customer.find({"channels.id" : "10000-1"}, {"channels.$" : 1, "_id" : 0}).channels[0] should do it — Mzzl
– Mzzl, Commented Jan 14, 2014 at 12:36
Thanks for the suggestion. I'm using the spring mongoTemplate (java driver) so I don't think this will work for me. I tried this in mongo shell and I get the error: TypeError: Cannot read property '0' of undefined. — Stuart
– Stuart, Commented Jan 14, 2014 at 12:47

pleerock · Accepted Answer · 2015-11-23 17:27:46Z

14

You can do it with Aggregation Framework. Query will be something like :

db.customer.aggregate([
    {$unwind : "$channels"},
    {$match : {"channels.id" : "10000-1"}},
    {$project : {_id : 0, 
                 id : "$channels.id", 
                 name : "$channels.name", 
                 enabled : "$channels.enabled"}}
])

edited Nov 23, 2015 at 17:27

pleerock

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answered Jan 14, 2014 at 12:32

Parvin Gasimzade

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4 Comments

Stuart Over a year ago

Thanks for replying so quickly. This is close but it's not exactly what I need. This gives me: { "result" : [ { "name" : "cust1chan1", "enabled" : true, "id" : "10000-1" } ], "ok" : 1 } rather than: { "id" : "10000-1", "name" : "cust1chan1", "enabled" : true }

Parvin Gasimzade Over a year ago

I don't think this is possible to get what you want. Why don't you want to extract fields from the result on the application layer?

Stuart Over a year ago

I think you're right that it's not possible. Yes, I can extract the fields but it's just nicer and cleaner to not have to do that. I posted this question because I looked over the Mongo documentation and I don't see any reference to doing this so I was wondering if anyone else had done it. An alternative could be to put a reference from one document to another subdocument(s) and store them in separate collections but then you have the issue of Mongo not supporting transactions.

Jonathan Muller Over a year ago

I think you can use $cond in $project to get it done, not sure though

chridam · Accepted Answer · 2018-09-10 15:08:17Z

Using MongoDB 3.4.4 and newer, the aggregation framework offers a number of operators that you can use to return the desired subdocument.

Consider running an aggregate pipeline that uses a single $replaceRoot stage to promote the filtered subdocument to the top-level and replace all other fields.

Filtering the subdocument requires the $filter operator which selects a subset of an array to return based on the specified condition i.e. returns an array with only those elements that match the condition. You can then convert the single array element to a document by using the $arrayElemAt operator

Overall running this aggregate operation will yield the desired result:

db.customer.aggregate([
    { "$replaceRoot": { 
        "newRoot": {
            "$arrayElemAt": [
                { "$filter": {
                   "input": "$channels",
                   "as": "channel",
                   "cond": { /* resolve to a boolean value and determine if an element should be included in the output array. */
                       "$eq": ["$$channel.id", "10000-1"]
                    } 
                } },
                0 /* the element at the specified array index */
            ]
        }
    } }
])

Output

{
    "id" : "10000-1",
    "name" : "cust1chan1",
    "enabled" : true
}

MeVimalkumar · Accepted Answer · 2023-03-14 04:23:43Z

0

I know it may be a bit late, but I was in a similar situation and I figure out the solutions for this. Inside the aggregation pipeline, you can first filter out the data by matching sub-documents and then just unwind the nested array field and then replace the root.

db.customer.aggregate([
    [
        { $match: { "channels.id": "10000-1" }},
        { $unwind: "$channels" },
        { $replaceRoot: { newRoot: "$channels" } }
    ]
])

answered Mar 14, 2023 at 4:23

MeVimalkumar

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