Get “data from collection b not in collection a” in a MongoDB shell query

Question

I have two MongoDB collections that share a common _id. Using the mongo shell, I want to find all documents in one collection that do not have a matching _id in the other co

Answer 1

In mongo 3.2 the following code seems to work

db.collectionb.aggregate([
    {
      $lookup:
        {
          from: "collectiona",
          localField: "collectionb_fk",
          foreignField: "collectiona_fk",
          as: "matched_docs"
        }
   },
   {
      $match: { "matched_docs": { $eq: [] } }
   }
]);

based on this https://docs.mongodb.com/manual/reference/operator/aggregation/lookup/#use-lookup-with-an-array example

Answer 2

Answering your follow-up. I'd use map().

Given this:

> b1 = {i: 1}
> db.b.save(b1)
> db.b.save({i: 2})
> db.a.save({_id: b1._id})

All you need is:

> vals = db.a.find({}, {id: 1}).map(function(a){return a._id;})
> db.b.find({_id: {$nin: vals}})

which returns

{ "_id" : ObjectId("4f08c60d6b5e49fa3f6b46c1"), "i" : 2 }

Answer 3

You will have to save the _ids from collection A to not pull them again from collection B, but you can do it using $nin. See Advanced Queries for all of the MongoDB operators.

Your end query, using the example you gave would look something like:

db.Test.find({"_id": {"$nin": [ObjectId("4f08a75f306b428fb9d8bb2e"), 
 ObjectId("4f08a766306b428fb9d8bb2f")]}})`

Note that this approach won't scale. If you need a solution that scales, you should be setting a flag in collections A and B indicating if the _id is in the other collection and then query off of that instead.

Updated for second part:

The second part is impossible. MongoDB does not support joins or any sort of cross querying between collections in a single query. Querying from one collection, saving the results and then querying from the second is your only choice unless you embed the data in the rows themselves as I mention earlier.

Answer 4

I've made a script, marking all documents on the second collection that appears in first collection. Then processed the second collection documents.

var first = db.firstCollection.aggregate([ {'$unwind':'$secondCollectionField'} ])

while (first.hasNext()){ var doc = first.next(); db.secondCollection.update( {_id:doc.secondCollectionField} ,{$set:{firstCollectionField:doc._id}} ); }

...process the second collection that has no mark

db.secondCollection.find({"firstCollectionField":{$exists:false}})