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我目前正在使用 Java EE/EclipseLink 和 postgreSQL 开发一个 Web 应用程序。关于我的实体，我特别管理“项目”、“用户”和“权限”（读取、写入、删除等访问权限...）。这些实体之间的关系是：一个项目可以有多个对项目具有不同权限的用户。因此，我得到了三重关联：项目+用户+权利。

在这个协会的坚持中，我遇到了一个烦人的问题。当我用它的信息持久化一个项目时，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），我也拥有持久的权限和用户。然后，我想在它们之间添加一个关联，所以我创建了一个名为 ProjectUserRight 的关联实体，我从现有实体中设置了项目、用户和权限，但是当我坚持它时，我得到以下异常：

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)

这是类项目：

@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;


  ...

类用户：

 @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;

   ...

右类：

   @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;

    ...

和关联类：

 @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;

和 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保留关联，但我得到了同样的错误......（我还检查了 id 项目并且它设置正确）我想知道它是否是错误注释的关联但我可以'找不到合适的方法来改变它。

所以，我真的需要你的帮助！:-)

提前感谢您的建议！

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1 回答 1

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你错过了@MapsId注释：

指定为 EmbeddedId 主键、EmbeddedId 主键中的属性或父实体的简单主键提供映射的 ManyToOne 或 OneToOne 关系属性。value 元素指定关系属性对应的复合键中的属性。

@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;
于 2012-07-13T11:22:22.007 回答