0

我正在尝试在播放框架中使用 JPA（已启用进化）创建具有额外列的多对多关系，并且使用 JavaJpa hibernate-entitymanager 5.0.5 final 播放版本为 2.4.6。

这是我创建所有表的 sql 脚本：

CREATE TABLE IF NOT EXISTS `user_level` (
`level_id` bigint(20) NOT NULL AUTO_INCREMENT,
`level` int(3),
`level_name` varchar(255) NOT NULL,
 PRIMARY KEY (`level_id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

CREATE TABLE IF NOT EXISTS `menu` (
`menu_id` bigint(20) NOT NULL AUTO_INCREMENT,
`menu_name` varchar(255),
`parent` bigint(20),
 PRIMARY KEY (`menu_id`),
 CONSTRAINT `FK_PARENT_MENU` FOREIGN KEY (`parent`) REFERENCES `menu` (`menu_id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

CREATE TABLE IF NOT EXISTS `menu_level` (
`id` bigint(20) NOT NULL AUTO_INCREMENT,
`level_id` bigint(20) NOT NULL,
`menu_id` bigint(20) NOT NULL,
`privilege` int(1) DEFAULT '0',
 PRIMARY KEY (`id`),
 CONSTRAINT `FK_MENU_LEVEL_USER_LEVEL_ID` FOREIGN KEY (`level_id`) REFERENCES `user_level` (`level_id`) ON UPDATE CASCADE,
 CONSTRAINT `FK_MENU_LEVEL_MENU_ID` FOREIGN KEY (`menu_id`) REFERENCES `menu` (`menu_id`) ON UPDATE CASCADE
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

下面是我的实体类

用户级别.java

@Entity 
@Table(name = "user_level")
public class UserLevel {

  @Id
  @GeneratedValue(generator = "increment")
  @GenericGenerator(name = "increment", strategy = "increment")
  @Column(name = "level_id")
  private Long levelId;

  @Column(name = "level")
  private Integer level;

  @Column(name = "level_name")
  private String name;

  @OneToMany(fetch = FetchType.LAZY, mappedBy = "userLevel", cascade = CascadeType.ALL)
  private List<MenuLevel> menuLevels;
}

菜单.java

@Entity
@Table(name="menu")
public class Menu  {

  @Id
  @GeneratedValue(generator = "increment")
  @GenericGenerator(name = "increment", strategy = "increment")
  @Column(name = "menu_id")
  private Long menuId;

  @Column(name = "menu_name")
  private String name;

  @OneToMany(fetch = FetchType.LAZY, mappedBy = "menu", cascade = CascadeType.ALL)
  private List<MenuLevel> menuLevels;

/*parent child menu relations*/
  @OneToMany(fetch = FetchType.LAZY, mappedBy = "parent")
  private List<Menu> childMenu = new ArrayList<>();

  @JoinColumn(name="parent", referencedColumnName = "menu_id")
  @ManyToOne(fetch = FetchType.LAZY)
  private Menu parent;

}

MenuLevel.java

@Entity
@Table(name = "menu_level")
public class MenuLevel {

  @Id
  @GeneratedValue(generator = "increment")
  @GenericGenerator(name = "increment", strategy = "increment")
  public Long id;

  @JoinColumn(name = "level_id", referencedColumnName = "level_id")
  @ManyToOne(optional = false, fetch = FetchType.LAZY)
  private UserLevel userLevel;

  @JoinColumn(name = "menu_id", referencedColumnName = "menu_id")
  @ManyToOne(optional = false, fetch = FetchType.LAZY)
  private Menu menu;

  @Column(name = "privilege")
  private Integer privilege;
}

当我运行游戏时，据说

引起：org.hibernate.AnnotationException：models.MenuLevel.menu 的referencedColumnNames(menu_id) 引用models.Menu 未映射到单个属性

此时我遇到了这个问题，如果有人能指出我的错误，我很高兴，因为我已经遵循了一些关于 JPA 多对多关系的教程和说明，但仍然无法弄清楚我哪里做错了。

4

0 回答 0