MongoDB - MongoEngine - How do I follow a reference from "the other side"?

Here is a test case showing how to query it:

import unittestfrom mongoengine import *class StackOverFlowTest(unittest.TestCase):    def setUp(self):        conn = connect(db='mongoenginetest')    def test_one_two_many(self):        class MyClass(Document):            other_classes = ListField(ReferenceField("OtherClass"))        class OtherClass(Document):            text = StringField()        MyClass.drop_collection()        OtherClass.drop_collection()        o1 = OtherClass(text='one').save()        o2 = OtherClass(text='two').save()        m = MyClass(other_classes=[o1, o2]).save()        # Lookup MyClass that has o1 in its other_classes        self.assertEqual(m, MyClass.objects.get(other_classes=o1))        # Lookup MyClass where either o1 or o2 matches        self.assertEqual(m, MyClass.objects.get(other_classes__in=[o1, o2]))

The main question is do you need to store a list of references in the MyClass? It might be more efficient to store the relationship just on OtherClass..

Try this query:

oc = Other_Class()MyClass.objects.find( other_classes__all = [oc.id] )

While thinking about my problem I came up with a solution.

I just add the ID of my referenced class to my model.

Here's an example:

class MyClass(Document):    ...    other_classes = ListField(ReferenceField(Other_Class))class Other_Class(Document):    myclass = ReferenceField(MyClass)

I'm not quite sure if this is the Mongo-way to do it but I'm pretty sure it works :)

Optionally you can omit the other_classes attribute in MyClass to avoid redundancy but then you need a query like this to get the "child" objects:

Other_Class.objects(myclass = myclass.id)