Accessing an array nested in a dictionary in an array object

Question

I created an aggregate query that uses the 'parts' key as the '_id' and returns all the 'subparts' numbers associated in an array. The output from the aggregate query is undesirable as I get the array 'result' which contains 74755 dictionary objects. Each object contains an int 32 for the _id and an array for '#associated_subparts'. I can access sub-elements via:

db.agg_result.find({}, {_id: 0, result: { $slice: [0,2]}})

but this is just a positional method. I tried variants of nested $elemMatch and $all queries with no success. I want to be able to return the object with id 468183339 for example. I might be projecting my original aggregate query incorrectly, but I'd like to return a collection without the nest so each object in the collection is one of these parts > subparts objects.

Example Query:

var agg_out = db.collection.aggregate( { $group :{ '_id' : "$parts#",
    'associated_subparts#' : { $addToSet : "$sub_part" }}});
db.agg_result.insert(agg_out);
db.agg_result.find();

Example Output:

{
    "_id" : ObjectId("52b4c4c6e984c01d69ff176f"),
    "result" : [ 
        {
            "_id" : 468183339,
            "associated_subparts#" : [ 
                -1408237536
            ]
        }, 
        {
            "_id" : 782155933,
            "associated_subparts#" : [ 
                -1408237536
            ]
        }, 
        {
            "_id" : 1583659973,
            "associated_subparts#" : [ 
                -1408237535, 
                -1408237535, 
                -1408237535, 
                -1408237535, 
                -1408237535, 
                -1408237535, 
                -1408237535, 
                -1408237535, 
                -1408237535, 
                -1408237535, 
                -1408237535, 
                -1408237535
            ]
        }
]
}

Calimero · Accepted Answer · 2013-12-21 21:44:15Z

1

If I understand your problem correctly this time, the proper way to populate your agg_result collection would be to iterate over the result field of your aggregation query result document before inserting each subitem of the 'result' index in your agg_result collection, something like :

agg_out.result.forEach(function(x){db.agg_result.insert(x);});

edited Dec 21, 2013 at 21:44

answered Dec 21, 2013 at 8:40

Calimero

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7 Comments

KLD Over a year ago

Hi Calimero, maybe I wasn't clear, I want to maintain the one to many relationship. So that I can see all the subparts related to unique parts. For example Car A would have Wheels X, Tires Y etc. Unwinding would get me Car A - Wheels X then another record Car A - Tires Y.

Calimero Over a year ago

Thanks for the clarification, and sorry for the misunderstanding - just realized you did the exact opposite with your current aggregation query, and you posted the resulting collection. Hopefully my next update will be more helpful !

KLD Over a year ago

Ah thanks for making me re-read I just changed "example data" to "example result" that was confusing.

Calimero Over a year ago

updated the answer. Haven't tested it though, but it should work as you expect.

KLD Over a year ago

Looks like you did it my friend :) I'm guessing there is no way to get the aggregate output like this in the first place, so I'll have to add this extra proc time. The _id gets converted from an int to a double with '.000000' appended to the original _id number (parts#), is there a way to specify that the _id remain the same? I know how to convert the variables after the fact, but I don't want to add extra computational overhead: db.collection.find({_id : {$exists : true}}).forEach(function(obj) { obj._id = new NumberInt(obj._id);

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