Question

我目前正在使用Java EE / EclipseLink和postgreSQL处理Web应用程序。关于我的实体，我特别管理“Project”，“User”和“Right”（访问权限，如读取，写入，删除...）。这些实体之间的关系是：项目可以拥有多个对项目具有不同权限的用户。因此，我获得了三重关联：Project + User + Right。

在这种关联的持久性过程中，我正面临一个烦人的问题。当我坚持一个项目及其信息时，一切都很好（它在数据库中，我可以在我的应用程序中使用它），我也坚持权利和用户。然后，我想在它们之间添加一个关联，所以我创建一个名为ProjectUserRight的关联实体，我从现有实体设置项目，用户和权利，但是当我坚持它时，我得到以下例外：

org.springframework.web.util.NestedServletException: Request processing failed; nested exception is javax.persistence.RollbackException: Exception [EclipseLink-4002] (Eclipse Persistence Services - 2.3.2.v20111125-r10461): org.eclipse.persistence.exceptions.DatabaseException
Internal Exception: org.postgresql.util.PSQLException: ERREUR: une valeur NULL viole la contrainte NOT NULL de la colonne « id_project »
Error Code: 0
Call: INSERT INTO project_user_right (id_right, id_project, id_user) VALUES (?, ?, ?)
bind => [3 parameters bound]
Query: InsertObjectQuery(ProjectUserRight@192db555)

这是课程项目：

@Entity
@Table(name="project")
public class Project implements Serializable {
private static final long serialVersionUID = 1L;

@Id
@GeneratedValue(strategy=GenerationType.IDENTITY)
@Column(name="id_project", unique=true, nullable=false)
private Integer idProject;


   //bi-directional many-to-one association to ProjectUserRight
@OneToMany(mappedBy="project", cascade={CascadeType.ALL})
private Set<ProjectUserRight> projetUtilDroits;


  ...

班级用户：

 @Entity
 @Table(name="user")
 public class User implements Serializable {
private static final long serialVersionUID = 1L;

@Id
@GeneratedValue(strategy=GenerationType.IDENTITY)
@Column(name="id_user", unique=true, nullable=false)
private Integer idUser;


@Column(name="nom_utilisateur", nullable=false, length=50)
private String userName;

   //bi-directional many-to-one association to ProjectUserRight
@OneToMany(mappedBy="user")
private Set<ProjectUserRight> projetUtilDroits;

   ...

班级权利：

   @Entity
   @Table(name="right")
   public class Right implements Serializable {
private static final long serialVersionUID = 1L;

@Id
@GeneratedValue(strategy=GenerationType.IDENTITY)
@Column(name="id_right", unique=true, nullable=false)
private Integer idRight;

@Column(name="type_right", nullable=false, length=10)
private String typeRight;

//bi-directional many-to-one association to ProjectUserRight
@OneToMany(mappedBy="right")
private Set<ProjectUserRight> projetUtilDroits;

    ...

关联类：

 @Entity
 @Table(name="project_user_right")
 public class ProjectUserRight implements Serializable {
private static final long serialVersionUID = 1L;

@EmbeddedId
private ProjectUserRightPK id;

//bi-directional many-to-one association to Right
    @ManyToOne(cascade={CascadeType.MERGE, CascadeType.REFRESH})
@JoinColumn(name="id_right", insertable=false, updatable=false)
private Right right;

//bi-directional many-to-one association to Project
@ManyToOne(cascade={CascadeType.MERGE, CascadeType.REFRESH})
@JoinColumn(name="id_project", insertable=false, updatable=false)
private Project project;

//bi-directional many-to-one association to User
@ManyToOne(cascade={CascadeType.MERGE, CascadeType.REFRESH})
@JoinColumn(name="id_user", insertable=false, updatable=false)
private User user;

ProjectUserRightPK（评论后编辑）：

@Embeddable
public class ProjectUserRightPK implements Serializable {
//default serial version id, required for serializable classes.
private static final long serialVersionUID = 1L;

@Column(name="id_project", unique=true, nullable=false)
private Integer idProject;

@Column(name="id_right", unique=true, nullable=false)
private Integer idRight;
    ...//getters and setters
@Column(name="id_user", unique=true, nullable=false)
private Integer idUser;

我编写了以下代码来保持关联：

 Project project = getService(Projet.class).getFromID(projectId);//retrieve the existing project from database with id

 User user = getService(Utilisateur.class).getFromID(userId);//retrieve the existing user from database with id

 Right right = getService(Right.class).getFromID(rightId);//retrieve the existing right from database with id
 if(project !=null && user!=null && right!=null){
   ProjectUserRight pud = new ProjectUserRight();
   pud.setRight(right);
   pud.setProject(project);
   pud.setUser(user);
   project.getProjectUserRights().add(pud);
   right.getProjectUserRights().add(pud);
   user.getProjectUserRights().add(pud);
   service(ProjetUtilDroit.class).persist(pud);//call the persist function on this entity (works fine with all other entities)
   }

我也尝试合并项目而不是持久化关联，但我得到了相同的错误...（我还检查了id项目并且设置正确）我想知道它是否是错误注释的关联但是我找不到改变它的正确方法。

所以，我真的需要你的帮助！： - ）

提前感谢您的建议！

Answer 1

您错过了@MapsId注释：

指定ManyToOne或OneToOne关系属性提供EmbeddedId主键（属性）的映射在EmbeddedId主键中，或者是一个简单的主键父实体。 value元素指定a中的属性关系属性对应的复合键。

@MapsId("idRight")
@ManyToOne(cascade={CascadeType.MERGE, CascadeType.REFRESH})
@JoinColumn(name="id_right")
private Right right;

@MapsId("idProject")
@ManyToOne(cascade={CascadeType.MERGE, CascadeType.REFRESH})
@JoinColumn(name="id_project")
private Project project;

@MapsId("idUser")
@ManyToOne(cascade={CascadeType.MERGE, CascadeType.REFRESH})
@JoinColumn(name="id_user")
private User user;

持久性三重关联JPA / EclipseLink

1 个答案: