我如何在mongodb上执行“ $ all $ in”？

Question

I have the following collection

>db.prueba.find({})
{ "_id" : "A", "requi" : null }
{ "_id" : "B", "requi" : null }
{ "_id" : "C", "requi" : [ "A" ] }
{ "_id" : "D", "requi" : [ "A", "B" ] }
{ "_id" : "E", "requi" : [ "C" ] }
{ "_id" : "F", "requi" : [ "B", "D"] }
{ "_id" : "G", "requi" : [ "F" ] }

I need each element of the requi field to be in the following array. in this case, the array has only one element

["A", "D"]

When I use the operator $all $in returns the following

 >db.prueba.find({requi:{$elemMatch:{$in:['A','D']}}})
{ "_id" : "C", "requi" : [ "A" ] }
{ "_id" : "D", "requi" : [ "A", "B" ] }
{ "_id" : "F", "requi" : [ "B", "D" ] }

the query must returns only document, because 'B' not exists in the array ["A" , "D"]

{ "_id" : "C", "requi" : [ "A" ] }

please, help me

Answer 1

You can use $setIsSubset to check whether the given array is set of the requi array and then $redact to eliminate the non-matched ones.

db.collection.aggregate([
  { "$match": { "requi": { "$ne": null } } },
  { "$redact": {
    "$cond": {
      "if": { "$eq": [{ "$setIsSubset": ["$requi", ["A", "D"]] }, true] },
      "then": "$$DESCEND",
      "else": "$$PRUNE"
    }
  }}
])