How to ignore a field from serializing when using circe in scala

Question

I am using circe in scala and have a following requirement :

Let's say I have some class like below and I want to avoid password field from being serialised then is there any way to let circe know that it should not serialise password field?

In other libraries we have annotations like @transient which prevent field from being serialised ,is there any such annotation in circe?

case class Employee(                   
                     name: String,
                     password: String)

https://circe.github.io/circe/codecs/custom-codecs.html – Dylan May 13 '22 at 11:20 — Dylan, May 13 '22 at 11:20

gatear · Accepted Answer · 2022-05-17T13:26:14.713

1

You could make a custom encoder that redacts some fields:

implicit val encodeEmployee: Encoder[Employee] = new Encoder[Employee] {
  final def apply(a: Employee): Json = Json.obj(
    ("name", Json.fromString(a.name)),
    ("password", Json.fromString("[REDACTED]")),
  )
}

LATER UPDATE

In order to avoid going through all fields contramap from a semiauto/auto decoder:

import io.circe.generic.semiauto._

implicit val encodeEmployee: Encoder[Employee] =
  deriveEncoder[Employee]
    .contramap[Employee](unredacted => unredacted.copy(password = "[REDACTED]"))

edited May 17 '22 at 13:26

answered May 13 '22 at 12:17

gatear

946
2
10

if I have 100 fields in the class, do I need to specify how to handle each field? – Hitesh May 14 '22 at 11:34
@Hitesh i've updated my answer with another approach – gatear May 17 '22 at 07:23

Daniel · Answer 2 · 2022-07-09T07:42:51.470

Although @gatear's answer is useful, it doesn't actually answer the question.

Unfortunately Circe (at least until version 0.14.2) does not have annotations to ignore fields. So far there is only a single annotation (@JsonKey) and this is used to rename field names.

In order to ignore a field when serialising (which Circe calls encoding) you can avoid that field in the Encoder implementation.

So instead of including the password field:

implicit val employeeEncoder: Encoder[Employee] =
  Encoder.forProduct2("name", "password")(employee => (employee.name, employee.password))

you ommit it:

implicit val employeeEncoder: Encoder[Employee] =
  Encoder.forProduct1("name")(employee => (u.name))

Alternatively what I've been using is creating a smaller case class which only includes the fields I'm interested in. Then I let Circe's automatic derivation kick in with io.circe.generic.auto._:

import io.circe.generic.auto._
import io.circe.syntax._

case class EmployeeToEncode(name: String)

// Then given an employee object:
EmployeeToEncode(employee.name).asJson

score 1 · Answer 3 · answered Oct 24 '22 at 06:18

1

deriveEncoder.mapJsonObject(_.remove("password"))

answered Oct 24 '22 at 06:18

FunctorPrototype

1,173
2
12
24

How to ignore a field from serializing when using circe in scala

3 Answers3