MongoDB - compare fields of array objects

Question

I need to find all documents where an element of an array in that document matches a condition of it's fields:

{
  doc_id: 0,
  array_field: [
    { id: 0, min: 1, max: 2 },
    { id: 1, min: 1, max: 1 },
    ...
  ]
}
...

I need all documents where any object in array_field has a max != min.

I've been using $elemMatch to query the array elements (which works fine by itself), but I can't seem to make it work with an $expr:

{
  array_field: {
    $elemMatch: {
        $expr: { $ne: [ "$min", "$max" ]}
    }
  }
}

How do I correctly compare the fields within the array object?

s7vr · Accepted Answer · 2018-10-25 17:30:05Z

3

You can't use aggregation operators with query operators. You can use $setDifference to compare min with max array and return if there are elements in the min array which are not in max array.

Use $expr with $setDifference. $expr allows use of aggregation expressions in the regular find query.

Something like

db.collection.find({
    "$expr":{
      "$ne":[
        {"$size":{"$setDifference":["$array_field.min","$$array_field.max"]}},
        0
      ]
    }
  }
)

You can also look at here to return when there is matching elements in arrays.

Comparing two object arrays and check if they have common elements

edited Oct 25, 2018 at 17:30

answered Oct 25, 2018 at 16:33

s7vr

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Comments

Kuldeep Mishra · Accepted Answer · 2018-10-25 13:24:50Z

1

db.collection.aggregate([
{"$match":{
"$expr":{
  "$eq":[
    {"$size":{"$setIntersection":["$min","$max"]}},
    0
  ]
}
}},
    {"$project":{"_id":1}}
])

$setIntersection to compare the FirstArrays Names with SecondArrays Names and return array of common names documents followed by $size and $redact and compare result with 0 to keep and else remove the document.

answered Oct 25, 2018 at 13:24

Kuldeep Mishra

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