spring - kotlin 中的数字不可序列化

Question

我发现 kotlin 中的数字不可序列化。

第一个问题

设备.kt:

package test.domain

import javax.persistence.*

Entity public class Device {
    public Id GeneratedValue var id: Long = -1
    public var name: String = ""
    ...
}

DeviceRestRepository.kt:

package test.domain

import org.springframework.data.repository.PagingAndSortingRepository
import org.springframework.data.repository.query.Param
import org.springframework.data.rest.core.annotation.RepositoryRestResource

RepositoryRestResource(collectionResourceRel = "device", path = "device")
public trait DeviceRestRepository : PagingAndSortingRepository<Device, Long?> {
    public fun findByName(Param("name") name: String): List<Device>
}

当我尝试编译此代码时出现错误，因为 kotlin.Long 不是 可序列化 :

Error:(14, 72) Kotlin: Type argument is not within its bounds: should be subtype of 'java.io.Serializable?'

第二个问题

当我尝试使用 时，我遇到了同样的错误java.lang.Long :

DeviceRestRepository.kt:

package test.domain

import org.springframework.data.repository.PagingAndSortingRepository
import org.springframework.data.repository.query.Param
import org.springframework.data.rest.core.annotation.RepositoryRestResource

RepositoryRestResource(collectionResourceRel = "device", path = "device")
public trait DeviceRestRepository : PagingAndSortingRepository<Device, java.lang.Long?> {
    public fun findByName(Param("name") name: String): List<Device>
}

Warning:(14, 72) Kotlin: This class shouldn't be used in Kotlin. Use kotlin.Long instead.

Error:(14, 72) Kotlin: Type argument is not within its bounds: should be subtype of 'java.io.Serializable?'

Answer 1

从 Kotlin 1.0 Beta 1 开始，原始类型是可序列化的:

Int is Serializable

Now the type Int and other basic types are Serializable on the JVM. This should help many frameworks.

从:
http://blog.jetbrains.com/kotlin/2015/10/kotlin-1-0-beta-candidate-is-out/
因此，您不再有任何问题。