Question

我在kotlin中有2个数据类，每个类都有相互参照。 Profile和Kweet（代码位于底部）。当使用EntityManager获取其中一个实体时，它可以成功获取单个对象。然而，它永远不会返回，因为JPA一直在后台获取递归关系。

调用ProfileDao.getById或ProfileDao.getByScreenname时会出现此问题。

Profile.kt

@Entity(name = "profile")
@NamedQueries(
(NamedQuery(name = "Profile.getByScreenName", query = "select p from profile p where p.screenname LIKE :screenname")),
(NamedQuery(name = "Profile.getAll", query = "select p from profile p"))
)
data class Profile(
@Id
@GeneratedValue
var id: Int? = null,

var screenname: String,

var created: Timestamp
) {

@OneToMany(mappedBy = "profile", fetch = FetchType.LAZY, cascade = [CascadeType.DETACH])
var kweets: List<Kweet> = emptyList()

@ManyToMany(fetch = FetchType.LAZY, cascade = [CascadeType.DETACH])
@JoinTable(
    name = "liked_kweets",
    joinColumns = [(JoinColumn(name = "profile_id", referencedColumnName = "id"))],
    inverseJoinColumns = [(JoinColumn(name = "kweet_id", referencedColumnName = "id"))]
)
var likes: List<Kweet> = emptyList()

@ManyToMany(fetch = FetchType.LAZY, cascade = [CascadeType.DETACH])
@JoinTable(
    name = "follows",
    joinColumns = [(JoinColumn(name = "follower_id", referencedColumnName = "id"))],
    inverseJoinColumns = [(JoinColumn(name = "followed_id", referencedColumnName = "id"))]
)
var follows: List<Profile> = emptyList()

@ManyToMany(mappedBy = "follows", fetch = FetchType.LAZY, cascade = [CascadeType.DETACH])
var followers: List<Profile> = emptyList()
}

Kweet.kt

@Entity(name = "kweet")
@NamedQuery(name = "Kweet.getAll", query = "select k from kweet k")
data class Kweet(
@Id
@GeneratedValue()
var Id: Int? = null,
var created: Timestamp,
var message: String,
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "profile_id")
@JsonBackReference
var profile: Profile,
@ManyToMany(mappedBy = "likes", fetch = FetchType.LAZY)
@JsonBackReference
var likedBy: List<Profile> = emptyList()
)

ProfileDao.kt

@Stateless
class ProfileDao {
@PersistenceContext
lateinit var em: EntityManager

fun getById(id: Int) = em.find(Profile::class.java, id)

fun getAll(): List<Profile> = em.createNamedQuery("Profile.getAll", Profile::class.java).resultList

fun getByScreenname(name: String) = em.createNamedQuery("Profile.getByScreenName", Profile::class.java)
    .setParameter("screenname", name)
    .resultList
    .firstOrNull()

fun create(profile: Profile) = em.persist(profile)

fun follow(follower: Profile, leader: Profile) {
    follower.follows += leader
    leader.followers += follower
    em.persist(follower)
    em.persist(leader)
}
}

更新：添加DTO并将其标记为正确打开可解决递归错误。例如：

@Open
class ProfileFacade(
private val profile: Profile
) : Serializable {
var screenname: String
    get() = profile.screenname
    set(value) {
        profile.screenname = value
    }

var kweets: List<SimpleKweetFacade>
    get() = profile.kweets.map { SimpleKweetFacade(it) }
    set(value) = Unit

var follows: List<String>
    get() = profile.follows.map { it.screenname }
    set(value) = Unit

var created: Timestamp
    get() = profile.created
    set(value) = Unit
}

@Open注释是一个简单的annotation class Open()，然后由gradle处理以添加open和noarg构造函数

Answer 1

您应该使用DTO表示前端的数据，或者使用来自子对象的@JsonIgnore父引用（而不是@JsonBackReference）。

使用DTO可能是一个更明智的选择，因为您可以将前端表示与后端模型分离，这使您在两个层中都具有灵活性（即，更改一个层不会破坏/可能在另一个层中引入错误）。

JPA在Kotlin和Glassfish中以双向关系无限递归

1 个答案: