Mongo_count for sub-documents in mongodb C driver

Question

Here's the structure part of my collection :

{
   ...
   list: [
      { id:'00A', name:'None 1' },
      { id:'00B', name:'None 2' },
   ],
   ...
}

After having used the function "distinct" (see the link : sub-object in sub-array in mongodb-C)

What would be the best solution for me to count the number of sub-documents please ?

Because the use of mongo_count being not compatible with distinct and mongo_run_command() function.

Answer 1

It would be helpful if you included more information in your question, like what version you are using and what your overall goal is. Mongo is extremely use case specific. I understand that you are looking to get the count of distinct elements in an array. When arrays are involved the aggregation framework [AF] is often the easiest way to get your task accomplished (regrettably at this time you may also need to use map reduce for more complex queries). You should be to modify the example below for your specific case. This example works with C driver 0.8 and mongodb 2.4.6. There is one massive caveat on Mongo differing from SQL beyond the use of distinct. Subdocuments are compared in the binary representation (an example is below). So the meaning of distinct in Mongo is slightly different then the SQL term for subdocuments. Therefore {..., subdoc:[{a:1,b:1}]} {..., subdoc:[{b:1,a:1}]} represent two distinct values of subdoc. This is only the case for subdocuments; top-level fields can be in any order. Additionally, Mongo does not guarantee field order in a document except for arrays. So while aggregation on subdoc.a will guarantee the results you get(assuming that a itself isn’t also a subdocument), aggregation on subdoc will not.

If you want to compare a subdoc field by field map reduce can be used. The complexity map reduce is several orders of magnitude higher then the AF. In the event you want to read about MapReduce: http://docs.mongodb.org/manual/core/map-reduce/

Now that we’ve gotten through all the disclamers and warnings: The AF shell command below will count everything properly with the same id: db.collection.aggregate({$unwind:"$list"}, {$group: {_id:"$list.id", count:{$sum:1}}}) This command sums up the number of distinct documents in a collection by list. Unfortunately, the C driver doesn’t have an aggregation helper command. The mongo_run_command function needs to be used to run an aggregation. Be advised that this can only return a bson document and not a cursor; therefore results are limited to the document size limit (Mongo 2.5.3, 16Megs). The two C examples sum on the subdoc, as a whole for equality purposes, so for these ordering matters. You need to add . to the subdoc to get a specific field (which again cannot be a subdoc or you will have the same constraints already stated).

Here is the example using BCON (http://api.mongodb.org/c/current/bcon.html), the recommended method for easy of use(it is ~10% slower then example two):

#include <mongo.h>
#include <bcon.h>
#include <stdio.h>

int main() {
    /*
     * We assume objects in the form of {_id:<any_id>, list:[{a:<int>,b:<int>}, ...]}
     */
    char table[] = "agg";
    mongo conn[1];
    mongo_init(conn);
    if(MONGO_OK != mongo_client(conn, "127.0.0.1", 27017))
        return 1;
    bson b[1], b_result[1];

    /*create the aggregation command in bson*/
    bcon cmd_aggregate[] = { "aggregate", BRS(table),
            "pipeline",
            "[",
                "{",
                    "$unwind", "$list",
                "}",
                "{",
                    "$group",
                    "{",
                        "_id", "$list",
                        "distinct_count",
                        "{",
                            "$sum", BI(1),
                        "}",
                    "}",
                "}",
            "]",
            BEND
    };
    bson_from_bcon(b, cmd_aggregate);

    /*So you can see your command*/
    bson_print(b);

    /*run the command*/
    mongo_run_command(conn, "test", b, b_result);

    /*command results*/
    bson_print(b_result);

    bson_destroy(b_result);
    bson_destroy(b);
    mongo_destroy(conn);
    return 0;
}

Here is the slightly faster style:

#include <mongo.h>
#include <stdio.h>

int main() {
    /*
     * We assume objects in the form of {_id:<any_id>, list:[{a:<int>,b:<int>}, ...]}
     */
    char table[] = "agg";
    mongo conn[1];
    mongo_init(conn);
    if(MONGO_OK != mongo_client(conn, "127.0.0.1", 27017))
        return 1;
    bson b[1], b_result[1];
    /*create the aggregation command in bson*/
    bson_init(b);
        bson_append_string(b, "aggregate", "agg");
        bson_append_start_array(b, "pipeline");
            bson_append_start_object(b,"0");
                bson_append_string(b, "$unwind", "$list");
            bson_append_finish_object(b);
            bson_append_start_object(b,"1");
                bson_append_start_object(b,"$group");
                    bson_append_string(b,"_id", "$list");
                    bson_append_start_object(b, "_count");
                        bson_append_int(b, "$sum", 1);
                    bson_append_finish_object(b);
                bson_append_finish_object(b);
            bson_append_finish_object(b);
        bson_append_finish_array(b);
        bson_finish(b);

    /*So you can see your command*/
    bson_print(b);

    /*run the command*/
    mongo_run_command(conn, "test", b, b_result);

    /*command results*/
    bson_print(b_result);

    bson_destroy(b_result);
    bson_destroy(b);
    mongo_destroy(conn);
    return 0;
}

Last but no least, connecting the database in the shell and then running the following commands will allow you to see the queries (if you started the database directly from the command line and didn't fork it you can see them in that terminal):

use admin
db.runCommand( { setParameter: 1, logLevel: 1 } )

This is useful for debugging the command you think you are sending to the database. It should in that terminal look the same as on you've created in a mongo shell.

Hopefully that covers all aspects of your question. Let me know if you need anything else.

Best, Charlie