How to retrieve a sub document array in MongoDB

Question

I have a collection in mongodb like this:

 db.country_list.find().pretty()
{
"_id" : ObjectId("53917321ccbc96175d7a808b"),
"countries" : [
    {
        "countryName" : "Afghanistan",
        "iso3" : "AFG",
        "callingCode" : "93"
    },
    {
        "countryName" : "Aland Islands",
        "iso3" : "ALA",
        "callingCode" : "358"
    },
    {
        "countryName" : "Albania",
        "iso3" : "ALB",
        "callingCode" : "355"
    }
              ]
}

like that i have 100 country details

i want to retrieve a country name where the calling code is 355. I have tried like this

db.country_list.find({countries: {$elemMatch :{ 'callingCode':'355'} } } )

and like this

 db.country_list.find({'countries.callingCode':'355'}).pretty()

but i am getting all records.How to get a specific record .Thanks in advance

Neil Lunn · Accepted Answer · 2014-06-06 09:37:03Z

1

What you want is the positional $ operator:

db.country_list.find(
   { "countries": { "$elemMatch" :{ "callingCode":"355"} } }
   { "countries.$": 1 }
)

Or even with the other syntax you tried:

db.country_list.find(
   { "countries.callingCode": "355"}
   { "countries.$": 1 }
)

This is because a "query" matches documents and is not a filter for the array contained in those documents. So the second argument there projects the field with the "position" that was matched on the query side.

If you need to match more than one array element, then you use the aggregation framework which has more flexibility:

db.country_list.aggregate([

    // Matches the documents that "contain" the match
    { "$match": {
        "countries.callingCode": "355"
    }},

    // Unwind the array to de-normalize as documents
    { "$unwind": "$countries" },

    // Match to "filter" the array content
    { "$match": {
        "countries.callingCode": "355"
    }},

    // Group back if you want an array
    { "$group": {
        "_id": "$_id",
        "countries": { "$push": "$countries" }
    }}
])

Or with MongoDB 2.6 or greater you can do this without the $unwind and $group:

db.country_list.aggregate([

    // Matches the documents that "contain" the match
    { "$match": {
        "countries.callingCode": "355"
    }},

    // Project with "$map" to filter
    { "$project": {
        "countries": {
            "$setDifference": [
                { "$map": {
                    "input": "$countries",
                    "as": "el",
                    "in": { 
                        "$cond": [
                            { "$eq": [ "$$el.callingCode", "355" ] }
                            "$$el",
                            false
                        ]
                    }
                }},
                [false]
            ]
        }
    }}

])

edited Jun 6, 2014 at 9:37

answered Jun 6, 2014 at 8:43

Neil Lunn

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3 Comments

Mulagala Over a year ago

It is really very useful answer for mogodb learners.Thanks once again

Mulagala Over a year ago

What if the condition matches multiple sub documents, If the first one only comes in that situation how to get all that mathced...?

Neil Lunn Over a year ago

@Mulagala This does match multiple sub-documents, which is the point and where this differs from the positional $ operator and $elemMatch in projection where you can match only one. This is kind of an old question (even if your own). So if you really need further clarification then as a new one.

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