我在这里有一个完整的示例应用程序:https://github.com/mrmojor/hibernateonkotlin
此代码基于此Blogpost:https://kotlinexpertise.com/hibernate-with-kotlin-spring-boot/
问题是,尽管懒惰从集成测试完美地工作,但调试器中有一个例外:测试例外
当我从控制器中运行相同的代码时,也没有例外,整个实体被加载:控制器没有例外
怎么可能?非常感谢您的帮助!
我无论如何都会发布代码片段:
Abstractjpapersistable.kt
import org.springframework.data.domain.Persistable
import org.springframework.data.util.ProxyUtils
import java.io.Serializable
import javax.persistence.GeneratedValue
import javax.persistence.Id
import javax.persistence.MappedSuperclass
import javax.persistence.Transient
/**
* Abstract base class for entities. Allows parameterization of id type, chooses auto-generation and implements
* [equals] and [hashCode] based on that id.
*
* This class was inspired by [org.springframework.data.jpa.domain.AbstractPersistable], which is part of the Spring Data project.
*/
@MappedSuperclass
abstract class AbstractJpaPersistable<T : Serializable> : Persistable<T> {
companion object {
private val serialVersionUID = -5554308939380869754L
}
@Id
@GeneratedValue
private var id: T? = null
override fun getId(): T? {
return id
}
/**
* Must be [Transient] in order to ensure that no JPA provider complains because of a missing setter.
*
* @see org.springframework.data.domain.Persistable.isNew
*/
@Transient
override fun isNew() = null == getId()
override fun toString() = "Entity of type ${this.javaClass.name} with id: $id"
override fun equals(other: Any?): Boolean {
other ?: return false
if (this === other) return true
if (javaClass != ProxyUtils.getUserClass(other)) return false
other as AbstractJpaPersistable<*>
return if (null == this.getId()) false else this.getId() == other.getId()
}
override fun hashCode(): Int {
return 31
}
}
person.kt:
import javax.persistence.CascadeType
import javax.persistence.Entity
import javax.persistence.FetchType
import javax.persistence.ManyToOne
import javax.persistence.OneToMany
@Entity
class Person(
val name: String,
@ManyToOne(cascade = [(CascadeType.ALL)], fetch = FetchType.EAGER)
val street: Street
) : AbstractJpaPersistable<Long>()
@Entity
class Address(
val zipCode: String,
val city: String
) : AbstractJpaPersistable<Long>()
@Entity
class Street(
@OneToMany(cascade = [(CascadeType.ALL)], fetch = FetchType.LAZY)
val adresses: MutableSet<Address>
) : AbstractJpaPersistable<Long>()
人格统治:
import com.kotlinexpertise.hibernatedemo.model.Person
import org.springframework.data.jpa.repository.JpaRepository
interface PersonRepository : JpaRepository<Person, Long>
PersonService:
import com.kotlinexpertise.hibernatedemo.model.Person
import com.kotlinexpertise.hibernatedemo.repository.PersonRepository
import org.springframework.stereotype.Service
@Service
class PersonService(val personRepository: PersonRepository) {
fun savePerson(person: Person) {
personRepository.saveAndFlush(person)
}
}
解决方案:
今年春季是什么
此属性应设置为false:
spring.jpa.open-in-view=false
这不是Kotlin问题,而是春季问题。
懒惰依赖于它具有可用连接的事实。
连接由Hibernate的EntityManager管理。
但是您的调试器在完全不同的线程上运行,因此它无法访问EntityManager。因此例外。