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spring – kotlin中的数字不可序列化

@H_502_0@

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 : PagingAndSortingRepositoryaram("name") name: String): List

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?’

最佳答案
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.

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