MongoDB kotlin driver: Serializer for class 'ObjectId' is not found

Question

I'm using the kotlin driver and MongoDB is not allowing me to make my data class serializable so far.

Here's the class, before I add what I've tried:

@Serializable
data class Word(
    @SerialName("_id")
    @Contextual
    val id: ObjectId,
    val pos: String,
    val word: String,
    val hyphenation: String? = null,
    val related: List<String>? = null,
    val derived: List<String>? = null,
    val senses: List<Sense>,
    val forms: VerbConjugation?,
    val searchSet: Set<String>
)

Without the @Serializable annotation added, MongoDB does not give any errors.

When added @Serializable with @Contextual as the official website suggests, it crashes with:

kotlinx.serialization.SerializationException: Serializer for class 'ObjectId' is not found. Please ensure that class is marked as '@Serializable' and that the serialization compiler plugin is applied.

Making the id nullable does not change anything. I've been stuck here for hours. Please help!

Probably worth checking why you want to make it Serializable. IME, Java serialisation isn't used much these days; for Mongo it's much more usual to let Spring/Mongo do its thing, and/or to serialise to Json instead. — gidds
– gidds, Commented Aug 11, 2024 at 18:17
@gidds Unfortunately I'm using Ktor, which doesn't integrate in the sense Spring does. I ended up doing a separate serializable class WordResponse. — Shazniq
– Shazniq, Commented Aug 11, 2024 at 19:15

Abhijith mogaveera · Accepted Answer · 2025-03-10 17:45:31Z

0

Not a solid answer just a workaround some how `ObjectIdSerializer` working fine in spring project but not in ktor

@Serializable(with = ObjectIdSerializer2::class)
val organisationId: ObjectId?,


@ExperimentalSerializationApi
object ObjectIdSerializer2 : KSerializer<ObjectId> {
    
    override val descriptor: SerialDescriptor = PrimitiveSerialDescriptor("ObjectIdSerializer2", PrimitiveKind.STRING)

    override fun serialize(encoder: Encoder, value: ObjectId) {
        if(encoder is BsonEncoder){
            encoder.encodeObjectId(value)
        }else {
            encoder.encodeString(value.toHexString())
        }
    }

    override fun deserialize(decoder: Decoder): ObjectId {
        if(decoder is BsonDecoder){
            return decoder.decodeObjectId()
        }
        return ObjectId(decoder.decodeString())
    }
}

answered Mar 10 at 17:45

Abhijith mogaveera

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