Meteor: how to search for only distinct field values aka a collection.distinct("fieldname") similar to Mongo's

Question

I'm using Meteor, and I'm trying to find only distinct (unique) values of a field. Mongodb has the command

Collection.distinct("fieldname");

but it's not implemented in the Meteor driver for Mongo. I've tried using the meteor-mongo-extensions package but still got an "undefined" on the client console for a client-side Collection.distinct("fieldname");

what, essentially, is a distinct looking for?

I feel like someone this is such a basic query- like for example a list of students with their teacher as a field, and then making a list of students by teachers out of that, for example...

if I can just wrap my lazy errant brain around the general concept I can probably bash something workable out.

Answer 1

You can just use underscore.js, which comes with Meteor - should be fine for the suggested use case. You might run into performance problems if you try this on a vast data set or run it repeatedly, but it should be adequate for common-or-garden usage:

var distinctEntries = _.uniq(Collection.find({}, {
    sort: {myField: 1}, fields: {myField: true}
}).fetch().map(function(x) {
    return x.myField;
}), true);

That will return the distinct entries in myField for all documents in Collection. Apologies if the one-liner looks a bit unwieldy; the sort and fields options and the true flag are just to make it more efficient.

Answer 2

If you'd prefer to do this on the server and thus save client memory and bandwidth try this aggregator package. It wraps mongodb's "distinct" function. https://github.com/zvictor/meteor-mongo-server/

Answer 3

A Non-Underscore, pure mongo solution would be to access the raw collection directly. This involves a Promise (though you can use await to await the resolved value).

Note: it will run on the MongoDB server and return distinct values to the client.

Collection._collection.rawCollection().distinct('fieldname').then(distinctValues => console.log(distinctValues)) 

Answer 4

Improving richsilv's answer:

function distinct(collection, field) {
  return _.uniq(collection.find({}, {
    sort: {[field]: 1}, fields: {[field]: 1}
  }).fetch().map(x => x[field]), true);
}

How to use:

var arrResults = distinct(MyCollection, 'myfield');