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databasemongodbmongodb-compass

Get documents with at least two values for a especific condition in an Array, Query in MongoDB


Suppose that you have a collection in this structure.

{ "_id" : ObjectId("5f21a77c52efa009cc9857a7"), "results" : [ 76, 72, 93 ] },
{ "_id" : ObjectId("5f21a993b4d13477dd950904"), "results" : [ 90, 94, 80 ] },
{ "_id" : ObjectId("5f21a9a0b4d13477dd950905"), "results" : [ 70, 73, 99 ] },
{ "_id" : ObjectId("5f21a9abb4d13477dd950906"), "results" : [ 91, 85, 99 ] }

I am using Mongodb compass.

I would like to find the documents that their results array contain at least two items greater than 90.

I've tried doing this:

{$or:[{$and:[ {"results.0": { $gte : 90 } },{"results.1": { $gte : 90 } } ]}, {$and: [ {"results.1": { $gte : 90 } },{"results.2": { $gte : 90 } } ]}, {$and: [ {"results.0": { $gte : 90 } },{"results.2": { $gte : 90 } } ]}]}

With this query I get this result

I would like to know if there is "an optimized query" for doing the same thing! Thanks in advance!


Solution

  • You'll have to $filter the array to contain only matching items and then check it's size:

    db.collection.aggregate([
      {
        $addFields: {
          tmp: {
            $filter: {
              input: "$results",
              as: "res",
              cond: {
                $gte: [
                  "$$res",
                  90
                ]
              }
            }
          }
        }
      },
      {
        $match: {
          "tmp.1": {
            $exists: true
          }
        }
      },
      {
        $project: {
          tmp: 0
        }
      }
    ])
    

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