Taking sum of “column” in MongoDB

Question

In MySQL, I would simply go \"SELECT sum(pts) FROM table\" to get the sum of the pts column on the table. However, I am converting this application to MongoDB and cannot fin

Answer 1

$mongo = new Mongo();

$map = new MongoCode("function() { emit('total', this.type); }");
$reduce = new MongoCode("function(k, vals) {
    var sum = 0;
    for(i in vals) {
        sum += vals[i];
    }

    return sum;
}");

$sumtotal = $mongo->db->command(array(
    'mapreduce' => 'tableName',
    'map' => $map,
    'reduce' => $reduce,
    // 'query' => array('col' => 'value'),
    'out' => array('merge' => 'tableName_sum')
));

$result = $mongo->db->selectCollection($sumtotal['result'])->findOne();
$sum = $result['value'];

Answer 2

Another solution is to create a script in the database, that will sum the require data and return your result something like:

The Script

function sum() {
   var sum = 0;
   for (var query = db.collection.find({}, {type: 1, _id:0}); query.hasNext();) {
      sum += query.next();
   }
   return sum;
}

PHP

In php after setting the connection with the db, it will be something like:

$toexec = 'function(x, y) { return sum() }';
$response = $database->execute($toexec, array());

Look this tutorial to learn more about scripting in mongo with php.

http://pointbeing.net/weblog/2010/08/getting-started-with-stored-procedures-in-mongodb.html

Answer 3

This can be done with the following map reduce functions:

var map = function () {
    emit("total", {sum:this.type}) 
    }

var reduce = function (key, values) {
    var result = {sum:0};
    values.forEach(function (value) {result.sum += value.type;});
    return result;
    }

db.table.mapReduce(map, reduce, {out : "type_totals"})

or in php:

$db->table->mapReduce( array( 
   'map' =>
   'function () {
      emit("total", {sum:this.type});
 }' 
,'reduce' =>
   'function (key, values) {
      var result = {sum:0};
      values.forEach(function (value) {result.sum += value.type;});
      return result;
     }'));

Answer 4

Try:
$m = new MongoClient();
$con = $m->selectDB("dbName")->selectCollection("collectionName");
$cond = array(
    array(
        '$group' => array(
            '_id' => null,
           'total' => array('$sum' => '$type'),
        ),
    )
);

$out = $con->aggregate($cond);
print_r($out);

More detail click here

Answer 5

Utilizing MongoDB's new Aggregation Framework:

$result = $db->command(array(

    'aggregate' => 'COLLECTION_NAME',

    'pipeline'  => array(
        // Match:
        // We can skip this in case that we don't need to filter 
        // documents in order to perform the aggregation
        array('$match' => array(
            // Criteria to match against
        )),

        // Group:
        array('$group'  => array(
            // Groupby fields:
            '_id'       => "$fieldname1, $fieldname2, $or_empty, $etc",
            // Count:
            'count'     => array('$sum' => 1),
            // Sum:
            'type_sum'  => array('$sum' => '$type'),
        ))

        // Other aggregations may go here as it's really a PIPE :p

    ),
));

Answer 6

Other answers have already given lot many sources. Just giving answer for asked question:

db.YOUR_COLLECTTION.aggregate([ { 
    $group: { 
        _id: null, 
        total: { 
            $sum: "$type" 
        } 
    } 
} ] )