MongoDB Only show equal fields when they exist-CodePudding

How do you only show those that exist?

This is my current code: db.collection.find({ $expr: {$eq: ["fieldA", "fieldB"]}, {"_id":0, "fieldC":1})

I have been trying things like the below but unsure of the correct syntax

db.collection.find({ $expr: {$eq: {$exists: ["fieldA", "fieldB"]}}, {"_id":0, "fieldC":1})

db.collection.find({ $expr: {$eq: ["fieldA": {$exists: true}, "fieldB": {$exists: true}]}}, {"_id":0, "fieldC":1})

CodePudding user response：

I don't think you need to use $expr for this query.

db.collection.find({ fieldA: { $exists: true }, fieldB: { $exists: true }, _id: 0, fieldC: 1 })

Reference

$exists - MongoDB Documentation

CodePudding user response：

Maybe you are looking for this:

db.collection.aggregate([
{
 $match: {
    $and: [
      {
        "fieldA": {
          $exists: true
        }
      },
      {
        "fieldB": {
          $exists: true
        }
      },
      {
        $expr: {
           $eq: [
              "$fieldA",
              "$fieldB"
           ]
         }
       }
      ]
     }
    }
 ])

Explained: and:[ {exist(A)} ,{exist(B)} ,{A=B?} ]

playground