Aggregate and flatten an array field in MongoDB - node.js

Question

I have a Schema:
var ProjectSchema = new Schema({
name: {
type: String,
default: ''
},
topics: [{
type: Schema.ObjectId,
ref: 'Topic'
}],
user: {
type: Schema.ObjectId,
ref: 'User'
}
});
What I want to do is get an array with all topics from all projects. I cannot query Topic directly and get a full list because some topics are unassigned and they do not hold a reference back to a Project (for reasons of avoiding two way references). So I need to query Project and aggregate some how. I am doing something like:
Project.aggregate([{$project:{topics:1}}]);
But this is giving me an array of Project objects with the topics field. What I want is an array with topic objects.
How can I do this?

Answer 1

When dealing with arrays you typically want to use $unwind on the array members first and then $group to find the distinct entries:
Project.aggregate(
[
{ "$unwind": "$topics" },
{ "$group": { "_id": "$topics._id" } }
],
function(err,docs) {
}
)
But for this case, it is probably simplier to just use .distinct() which will do the same as above, but with just an array of results rather than documents:
Project.distinct("topics._id",function(err,topics) {
});
But wait there a minute because I know what you are really asking here. It's not the _id values you want but your Topic data has a property on it like "name".
Since your items are "referenced" and in another collection, you cannot do an aggregation pipeline or .distinct() operation on the property of a document in another collection. Put basically "MongoDB does not perform Joins" and mongoose .populate() is not a join, just something that "emulates" that with additional query(ies).
But you can of course just find the "distinct" values from "Project" and then fetch the information from "Topic". As in:
Project.distinct("topics._id",function(err,topics) {
Topic.find({ "_id": { "$in": topics } },function(err,topics) {
});
});
Which is handy because the .distinct() function already returned an array suitable for use with $in.