CODEWITHSUNDEEP
node.jsmongodbaggregation-framework
Timathon
Timathon

Reputation: 1049

Aggregation with nodejs driver - grouping by whether a field is empty

The docs are simple as:

[
  {'id': '1', 'type': 'a', 'startedAt': '2017-06-11'},
  {'id': '2', 'type': 'b', 'startedAt': ''},
  {'id': '3', 'type': 'b', 'startedAt': '2017-06-11'}
]

And the expected aggregated result:

[
  {'type': 'a', 'started': true, 'count': 1},
  {'type': 'b', 'started': true, 'count': 1},
  {'type': 'b', 'started': false, 'count': 1}
]

How to get above result with mongodb nodejs driver?

I've tried like below, but it didn't work ('started' was always null):

db.collection('docs').group(
      {'type': '$type', 'started': {
        $cond: [{$eq: ['$startedAt': '']}, false, true ]
      }}, 
      {}, 
      {'total': 0},
      'function(curr, result) {result.total++}'
    )

Upvotes: 0

Views: 50

Answers (1)

Neil Lunn
Neil Lunn

Reputation: 151132

You use .aggregate() here and not .group(), which is a different function altogether:

db.collection('docs').aggregate([
  { "$group": {
    "_id": {
      "type": "$type",
      "started": {
        "$gt": [ "$startedAt", "" ]
      }
    },
    "count": { "$sum": 1 }
  }}
],function(err, results) {
   console.log(results);
})

The $gt operator returns true when the condition is met. In this case any content in a string is "greater than" an empty string.

If the field is actually "not present at all" then we can adapt with $ifNull. This gives a default have if the property does not actually exist, or otherwise evaluates to null.

db.collection('docs').aggregate([
  { "$group": {
    "_id": {
      "type": "$type",
      "started": {
        "$gt": [ { "$ifNull": [ "$startedAt", ""] }, "" ]
      }
    },
    "count": { "$sum": 1 }
  }}
],function(err, results) {
   console.log(results);
})

This would produce:

{ "_id" : { "type" : "b", "started" : true }, "count" : 1 }
{ "_id" : { "type" : "b", "started" : false }, "count" : 1 }
{ "_id" : { "type" : "a", "started" : true }, "count" : 1 }

You can optionally $project afterwards to change the fields from being within _id in the results, but you really should not since this means an additional pass through results, when you can just as easily access the values anyway.

So just .map() on the result:

console.log(
  results.map(function(r) { 
    return { type: r._id.type, started: r._id.started, count: r.count }
 })
);

But with $project:

db.collection('docs').aggregate([
  { "$group": {
    "_id": {
      "type": "$type",
      "started": {
        "$gt": [ { "$ifNull": [ "$startedAt", ""] }, "" ]
      }
    },
    "tcount": { "$sum": 1 }
  }},
  { "$project": {
    "_id": 0,
    "type": "$_id.type",
    "started": "$_id.started",
    "count": "$tcount"
  }}
],function(err, results) {
   console.log(results);
})

Resulting in your desired format

{ "type" : "b", "started" : true, "count" : 1 }
{ "type" : "b", "started" : false, "count" : 1 }
{ "type" : "a", "started" : true, "count" : 1 }

For reference, the correct usage with .group() would be:

  db.collection('docs').group(
    function(doc) {
      return {
        "type": doc.type,
        "started": (
          (doc.hasOwnProperty('startedAt') ? doc.startedAt : "") > ""
        )
      }
    },
    [],
    { "count": 0 },
    function(curr,result) {
      result.count += 1
    },
    function(err,results) {
      console.log(results);
    }
  );

Which returns:

[
    { "type" : "a", "started" : true, "count" : 1  },
    { "type" : "b", "started" : false, "count" : 1 },
    { "type" : "b", "started" : true,  "count" : 1 }
]

But you really should no use that since .group() relies on JavaScript evaluation that runs much slower than what you can do with .aggregate()

Upvotes: 2

Related Questions