MongoDB How to compare fields in a single document? [duplicate] - mongodb

Question

This question already has answers here:
MongoDb query condition on comparing 2 fields
(4 answers)
Closed 4 years ago.
Given the collection "example":
{_id: 1, prediction: "H", result: "A"}
{_id: 2, prediction: "H", result: "H"}
{_id: 3, prediction: "A", result: "A"}
What do I need to do to find records where the prediction and result values match? ie. documents 2 and 3?
Finding all the "H" predictions is:
db.example.find( { prediction: "H" } )
But I need to replace the literal "H" with the value in the result field of the same document.
Edit: Sorry, I should have said that I was using Mongo 3.6.

Answer 1

You should be able to do this with an aggregation query, try out the $redact stage:
db.test.aggregate(
[
{ $redact: {
$cond: {
if: { $eq: [ "$prediction", "$result" ] },
then: "$$DESCEND",
else: "$$PRUNE"
}
}
}
]
);
This will yield the result:
{ "_id" : 2, "prediction" : "H", "result" : "H" }
{ "_id" : 3, "prediction" : "A", "result" : "A" }
More info on redact can be found here - https://docs.mongodb.com/manual/reference/operator/aggregation/redact/#redact-aggregation

Answer 2

You can use agregation
db.example.aggregate(
[
{
$project:
{
_id: 1,
prediction: 1,
result: 1,
compare: { $eq: [ "$prediction", "$result" ] }
}
},
{
$match:
{
compare: true
}
}
]
)

Answer 3

If you are using 3.6, this will work. refer this
db.getCollection('TEST').find( { $where: function() {
return this.prediction == this.result
} })
Result :
{
"_id" : 2,
"prediction" : "H",
"result" : "H"
}
{
"_id" : 3,
"prediction" : "A",
"result" : "A"
}