我发现kotlin中的数字不可序列化.
>第一个问题
Device.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
我尝试编译此代码时出错,因为kotlin.Long不是Serializable：
Error:(14,72) Kotlin: Type argument is not within its bounds: should
be subtype of ‘java.io.Serializable?’
>第二个问题
我尝试使用java.lang.Long时遇到同样的错误：
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",java.lang.Long?> {
public fun findByName(Param("name") name: String): List
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?’
至于Kotlin 1.0 Beta 1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.
