How to count records with one distinct field in mongoose?

Here's an alternative answer as I get an exception when I try Reddest's approach with Mongoose 3.1.2 (which seems like a bug in Mongoose to me as Reddest's approach should be fine).

You can call the distinct method on your collection's model, specifying the name of the user-identifying field of that collection:

Record.distinct('user_id').exec(function (err, user_ids) {    console.log('The number of unique users is: %d', user_ids.length);});

Or if you want to chain the distinct call from a find, include the callback in the distinct call (this did work for me):

Record.find().distinct('user_id', function (err, user_ids) { ... });

UPDATE

If you just want the count without getting the values, stick a count() call in the chain:

Record.distinct('user_id').count().exec(function (err, count) {    console.log('The number of unique users is: %d', count);});

NOTE: this doesn't work in the latest Mongoose code (3.5.2).

Aggregation will work for you. Something like that:

Transaction.aggregate(    { $match: { seller: user, status: 'completed'  } },     { $group: { _id: '$customer', count: {$sum: 1} } }).exec() 

If you just want get the number of queried collections, you can use this:

Record.find()      .distinct('user_id')      .count(function (err, count) {          //The number of unique users is 'count'      });