kotlinx.serialization.SerializationException：找不到类“UnitEntity”myClassName的序列化程序

问题描述

我正在处理一个 Kotlin 多平台项目，但在保存数据库实体对象时遇到了问题。当我调用 sqldelight dao 类的插入方法时抛出异常

kotlinx.serialization.SerializationException: Serializer for class 'UnitEntity' is not found.
Mark the class as @Serializable or provide the serializer explicitly.
    at kotlinx.serialization.internal.Platform_commonKt.serializerNotRegistered(Platform.common.kt:91)
    at kotlinx.serialization.internal.PlatformKt.platformSpecificSerializerNotRegistered(Platform.kt:29)
    at kotlinx.serialization.SerializersKt__SerializersKt.serializer(Serializers.kt:59)
    at kotlinx.serialization.SerializersKt.serializer(UnkNown Source:1)
.....

那个例子类是生成的实体类

import com.example.package.core.entity.UnitEntity;
import com.example.package.core.entity.ResourceTypeEntity;

CREATE TABLE IF NOT EXISTS `ResourceEntity` (
`id` TEXT NOT NULL,`name` TEXT NOT NULL,`unitEntity` TEXT AS UnitEntity NOT NULL,`type` TEXT AS ResourceTypeEntity NOT NULL,PRIMARY KEY(`id`));

CREATE TABLE IF NOT EXISTS `ResourceTypeEntity` (
`typeId` TEXT NOT NULL,`typeName` TEXT NOT NULL,PRIMARY KEY(`typeId`)
);

CREATE TABLE IF NOT EXISTS `UnitEntity` (`unitId` TEXT NOT NULL,`unitName` TEXT NOT NULL,PRIMARY KEY(`unitId`));


save:
INSERT OR REPLACE INTO ResourceEntity
VALUES ?;

这里是 DAO 类：

 class ResourcesDaoImpl(private val dao: Resources_daoQueries) : ResourcesDao {
  override suspend fun save(resources: List<ResourceEntity>) {
    resources.forEach {
      dao.save(it)
    }
  }
}

当我尝试调用该 save 方法时，它崩溃并说将该类标记为 @Serializable，我不明白为什么会这样说，因为它是一个生成的类，我无法修改这些类。

这是我如何尝试保存这些实体的片段

val mappedData = itemsDto.map {
    ResourceEntity(
      id = it._id,name = it.name,type = ResourceTypeEntity(typeId = it.type._id,typeName = it.type.name),unitEntity = UnitEntity(unitId = it.unit._id,unitName = it.unit.unit)
    )
  }
resourcesDao.save(mappedData)

库：

val coroutinesVersion = "1.3.9-native-mt"
val ktor_version = "1.4.2"
val serializationVersion = "1.0.0-RC"
val koin_version = "3.0.0-alpha-4"
val sqlDelight = "1.4.4"

implementation("org.jetbrains.kotlinx:kotlinx-coroutines-core:$coroutinesVersion")
implementation("org.jetbrains.kotlinx:kotlinx-serialization-core:$serializationVersion")
implementation("com.squareup.sqldelight:runtime:$sqlDelight")
// sqlDelight extension
implementation("com.squareup.sqldelight:coroutines-extensions:$sqlDelight")

这是由entities

生成的实际sqlDelight和save方法

data class ResourceTypeEntity(
  val typeId: String,val typeName: String
) {
  override fun toString(): String = """
  |ResourceTypeEntity [
  |  typeId: $typeId
  |  typeName: $typeName
  |]
  """.trimMargin()
}

资源实体.class

import com.squareup.sqldelight.ColumnAdapter

data class ResourceEntity(
  val id: String,val name: String,val unitEntity: UnitEntity,val type: ResourceTypeEntity
) {
  override fun toString(): String = """
  |ResourceEntity [
  |  id: $id
  |  name: $name
  |  unitEntity: $unitEntity
  |  type: $type
  |]
  """.trimMargin()

  class Adapter(
    val unitEntityAdapter: ColumnAdapter<UnitEntity,String>,val typeAdapter: ColumnAdapter<ResourceTypeEntity,String>
  )
}

UnitEntity.class

data class UnitEntity(
  val unitId: String,val unitName: String
) {
  override fun toString(): String = """
  |UnitEntity [
  |  unitId: $unitId
  |  unitName: $unitName
  |]
  """.trimMargin()
}

保存方法

  override fun save(ResourceEntity: ResourceEntity) {
    driver.execute(-1658323214,"""
     |INSERT OR REPLACE INTO ResourceEntity
     |VALUES (?,?,?)
       """.trimMargin(),7) {
          bindString(1,ResourceEntity.id)
          bindString(2,ResourceEntity.name)
          bindString(3,database.ResourceEntityAdapter.unitEntityAdapter.encode(ResourceEntity.unitEntity))
      bindString(4,database.ResourceEntityAdapter.typeAdapter.encode(ResourceEntity.type))
    }
    notifyQueries(-1658323214,{database.resources_daoQueries.getResources +
        database.resources_daoQueries.getResourceByManufacturerId})
  }

解决方法

您可以将注释添加到 ResourceEntity：在 @Serializable(with = UnitEntitySerializer::class) 字段上方添加一个 UnitEntity。

然后你需要定义 UnitEntitySerializer 来实际帮助序列化它。这就是我对 question/ answer 中的 UUID 所做的。但是，我不确定 UnitEntity 是什么，所以我无法帮你序列化它/写 UnitEntitySerializer