Getting the number of unique values of a query - node.js

Question

I have some documents with the following structure:
{
"_id": "53ad76d70ddd13e015c0aed1",
"action": "login",
"actor": {
"name": "John",
"id": 21337037
}
}
How can I make a query in Node.js that will return the number of the unique actors that have done a specific action. For example if I have a activity stream log, that shows all the actions done by the actors, and a actorscan make a specific action multiple times, how can I get the number of all the unique actors that have done the "login" action. The actors are identified by actor.id

Answer 1

db.collection.distinct()
db.collection.distinct("actor.id", { action: "login"})
will return all unique occiriences and then you can get count of a result set.
PS
do not forget about db.collection.ensureIndex({action: 1})

Answer 2

You can use aggregation framework for this:
db.coll.aggregate([
/* Filter only actions you're looking for */
{ $match : { action : "login" }},
/* finally group the documents by actors to calculate the num. of actions */
{ $group : { _id : "$actor", numActions: { $sum : 1 }}}
]);
This query will group the documents by the entire actor sub-document and calculate the number of actions by using $sum. The $match operator will filter only documents with specific action.
However, that query will work only if your actor sub-documents are the same. You said that you're identifying your actors by id field. So if, for some reason, actor sub-documents are not exactly the same, you will have problems with your results.
Consider these these three documents:
{
...
"actor": {
"name": "John",
"id": 21337037
}
},
{
...
"actor": {
"name": "john",
"id": 21337037
}
},
{
...
"actor": {
"surname" : "Nash",
"name": "John",
"id": 21337037
}
}
They will be grouped in three different groups, even though the id field is the same.
To overcome this problem, you will need to group by actor.id.
db.coll.aggregate([
/* Filter only actions you're looking for */
{ $match : { action : "login" }},
/* finally group the documents to calculate the num. of actions */
{ $group : { _id : "$actor.id", numActions: { $sum : 1 }}}
]);
This query will correctly group your documents by looking only at the actor.id field.
Edit
You didn't specify what driver you were using so I wrote the examples for MongoDB shell.
Aggregation with Node.js driver is very similar but with one difference: Node.js is async The results of the aggregation are returned in the callback. You can check the Node.js aggregation documentation for more examples:
So the aggregation command in Node.js will look like this:
var MongoClient = require('mongodb').MongoClient;
MongoClient.connect('mongodb://127.0.0.1:27017/test', function(err, db) {
if(err) throw err;
var collection = db.collection('auditlogs');
collection.aggregate([
{ $match : { action : "login" }},
{ $group : { _id : "$actor.id", numActions: { $sum : 1 }}} ],
function(err, docs) {
if (err) console.error(err);
console.log(docs);
// do something with results
}
);
});
For these test documents:
{
"_id" : ObjectId("53b162ea698171cc1677fab8"),
"action" : "login",
"actor" : {
"name" : "John",
"id" : 21337037
}
},
{
"_id" : ObjectId("53b162ee698171cc1677fab9"),
"action" : "login",
"actor" : {
"name" : "john",
"id" : 21337037
}
},
{
"_id" : ObjectId("53b162f7698171cc1677faba"),
"action" : "login",
"actor" : {
"name" : "john",
"surname" : "nash",
"id" : 21337037
}
},
{
"_id" : ObjectId("53b16319698171cc1677fabb"),
"action" : "login",
"actor" : {
"name" : "foo",
"id" : 10000
}
}
It will return the following result:
[ { _id: 10000, numActions: 1 },
{ _id: 21337037, numActions: 3 } ]

Answer 3

The aggregation framework is your answer:
db.actors.aggregate([
// If you really need to filter
{ "$match": { "action": "login" } },
// Then group
{ "$group": {
"_id": {
"action": "$action",
"actor": "$actor"
},
"count": { "$sum": 1 }
}}
])
Your "actor" combination is "unique", so all you need to do it have the common "grouping keys" under the _id value for the $group pipeline stage and count those "distinct" combinations with $sum.