JPA 无限递归与 Kotlin 和 Glassfish 中的双向关系

Question

我在 kotlin 中有 2 个数据类，每个数据都相互引用。 Profile 和 Kweet（代码在底部）。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 return 这个，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 并将其标记为打开正确解决了递归错误。示例：

@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 在前端表示数据或 @JsonIgnore 来自子对象的父引用（而不是 @JsonBackReference）。

使用 DTO 可能是更明智的选择，因为您可以将前端表示与后端模型分离，这为您在两层提供了灵活性（即更改一层不会 break/possibly 在另一层引入错误).

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