Node.js MongoDb aggregation Sum all and Sum all with specific field

Question

Here's my collection [test].

{"_id" : "Test1", "enabled": "on", "value" : 10},
{"_id" : "Test2", "enabled": "on", "value" : 50},
{"_id" : "Test3", "enabled": "on", "value" : 10},
{"_id" : "Test4", "value" : 5},
{"_id" : "Test5", "value" : 2}

I would like to get all the total of the value and total value of the field with "enabled":"on" like these:

Desired result:

[ { _id: null, totalValue: 77, totalEnabled: 70 } ]

Here's what i have so far but no luck.

db.collection('test').aggregate({
    $group: {
          _id: null,
          totalValue : {
              $sum: "$value"
          },
          totalEnabled: $sum : {"enabled":{$exists:true}}
      }
  }, function(err, result) {
    if (err) return console.dir(err)

    console.log(result);
});

Neil Lunn · Accepted Answer · 2014-02-21 10:16:22Z

3

You were close but $exists doesn't function is aggregation and has a different function. What you were looking for is $cond

db.items.aggregate([
    {$group: { 
        _id: null,
         totalValue: {$sum: "$value"}, 
         enabledValue: {$sum: {
             $cond: [
                 // Condition to test 
                 {$eq: ["$enabled", "on"] },
                 // True
                 "$value",
                 // False
                 0
            ] 
         }}
    }}
])

The usage is to provide a different value depending on whether the condition is evaluated to true or false.

edited Feb 21, 2014 at 10:16

answered Feb 21, 2014 at 9:58

Neil Lunn

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2 Comments

per.eight Over a year ago

That's more like it! I was trying to figure what your code does until you commented out the condition structure. Thanks!

Neil Lunn Over a year ago

@per.eight Glad that helped and that coming back for the edit was worthwhile. That is how I usually do it in these types of questions.

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