Match documents with fields not containing values in array

Question

After some "match" steps in db.collection.aggregate I got

{_id: 1, field1: "a", field2: ["1.1.1", "1.1.2", "1.1.3", "1.1.4"]}
{_id: 2, field1: "b", field2: [         "1.1.2",          "1.1.4"]}
{_id: 3, field1: "c", field2: ["1.1.1", "1.1.2"                  ]}
{_id: 4, field1: "d", field2: [                           "1.1.4"]}
{_id: 5, field1: "e", field2: [                  "1.1.3"         ]}

// I made preformatting

I want to apply yet another rule (I suppose it would be "group" rule) in aggregate to get all documents where "field2" does not contain "1.1.1" OR "1.1.4". So I need documents without "1.1.1" OR "1.1.4" OR ("1.1.1" AND "1.1.4").

I want to get something like this:

{_id: 5, field1: "e", field2: [                  "1.1.3"         ]}

So, I have field2, which is array in each document, and I want to get all the documents which have no single element in a given array (e.g. ["1.1.1", "1.1.4"]).

P.S. I need this for "single statement aggregation command". No additional JS accepted.

UPD

@blakes-seven , tnx

My mistake is that I used

{ "$match": { "field2": { "$elemMatch": { "$nin": ["1.1.1","1.1.4"] } } } }

but right thing is just to use

{ "$match": { "field2": {                 "$nin": ["1.1.1","1.1.4"] } } }

Don't overestimate $elemMatch :)

Answer 1

Actually all you need is a $nin in a $match:

{ "$match": { "field2": { "$nin": ["1.1.1","1.1.4"] } } }

It's also likely that there is even a better approach to getting that result before what you have reached in the pipeline, but since your question does not explain that then there is no way to tell.