How to get only those docs which has all the match fields in mongoose

Question

Let say I have the following collection named survey:

{ _id: 1, results: [ { product: "abc", active: true}, { product: "xyz", active: true} ] }
{ _id: 2, results: [ { product: "abc", active: false }, { product: "xyz", active: true } ] }
{ _id: 3, results: [ { product: "abc", active: false }, { product: "xyz", active: false } ] }

by doing:

db.survey.find(
   { results: { $elemMatch: { active:  false } }
)

I will get all the docs where at least one object in the array is false i.e.

{ _id: 2, results: [ { product: "abc", active: false }, { product: "xyz", active: true } ] }

{ _id: 3, results: [ { product: "abc", active: false }, { product: "xyz", active: false } ] }

but how can I get only those docs which has all the key active: false i.e:

{ _id: 3, results: [ { product: "abc", active: false }, { product: "xyz", active: false } ] }

TIA..

Answer 1

so here I got one solution, which is perfectly apt for my need:

db.getCollection('survey').aggregate([
    { "$match": { "results.active": false } },
    { "$redact": {
        "$cond": {
            "if": {
                "$allElementsTrue": {
                    "$map": {
                        "input": "$results",
                        "as": "result",
                        "in": { "$eq": [ "$$result.active", false ] }
                    }
                }               
            },
            "then": "$$KEEP",
            "else": "$$PRUNE"
        }
    }}    
])