JPA @ManyToMany使用@OrderBy订购失败

时间:2014-10-08 08:36:14

标签: java hibernate jpa

当JPA尝试选择AdmUser实体时,我有sql错误:

ERROR: column locations1_.name does not exist. 

我的实体有什么问题吗?我的AdmUser实体:

@Entity
@Table(name = "ADM_USERS")
@SequenceGenerator(name = "ADM_USER_SEQ", sequenceName = "ADM_USER_SEQ", allocationSize = 1)
public class AdmUser implements EntityInt, Serializable {

    private static final long serialVersionUID = 786L;

    @Id
    @Column(nullable = false)
    @GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "ADM_USER_SEQ")
    private Long id;
(...)
    @ManyToMany(cascade = CascadeType.MERGE, fetch = FetchType.EAGER)
    @JoinTable(name = "loc_locations_adm_users", joinColumns = @JoinColumn(name = "id_user", referencedColumnName="id"), 
                inverseJoinColumns = @JoinColumn(name = "id_location"))
    @OrderBy("name")
    private Set<LocLocation>        locations;
(...)
}

我的LocLocation实体:

@Entity
@Table(name = "loc_locations", schema = "public")
@SequenceGenerator(name = "LOC_LOCATIONS_SEQ", sequenceName = "LOC_LOCATIONS_SEQ", allocationSize = 1)
public class LocLocation implements EntityInt, java.io.Serializable {

    private static final long serialVersionUID = 1L;

    @Id
    @Column(name = "id", unique = true, nullable = false)
    @GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "LOC_LOCATIONS_SEQ")
    private Long id;

    @Column(nullable = false, unique = true, length = 200)
    private String name;
(...)

    @ManyToMany(cascade = CascadeType.REFRESH, fetch = FetchType.LAZY, mappedBy="locations")
    private List<AdmUser>       users;
}

现在 - 当JPA尝试选择AdmUser实体时,我有sql错误。 JPA生成的查询看起来:

 select
  admuser0_.id as id1_2_0_,
  admuser0_.actived as actived2_2_0_,
  admuser0_.admin as admin3_2_0_,
  admuser0_.allow_ip as allow_ip4_2_0_,
  admuser0_.created as created5_2_0_,
  admuser0_.deleted as deleted6_2_0_,
  admuser0_.id_domain as id_doma16_2_0_,
  admuser0_.email as email7_2_0_,
  admuser0_.language as language8_2_0_,
  admuser0_.login as login9_2_0_,
  admuser0_.name as name10_2_0_,
  admuser0_.passwd as passwd11_2_0_,
  admuser0_.phone as phone12_2_0_,
  admuser0_.picture as picture13_2_0_,
  admuser0_.surname as surname14_2_0_,
  admuser0_.theme as theme15_2_0_,
  locations1_.id_user as id_user1_2_1_,
  loclocatio2_.id as id_locat2_6_1_,
  loclocatio2_.id as id1_17_2_,
  loclocatio2_.description as descript2_17_2_,
  loclocatio2_.name as name3_17_2_
  from
        public.ADM_USERS admuser0_
   left outer join
        public.loc_locations_adm_users locations1_
            on admuser0_.id=locations1_.id_user
   left outer join
        public.loc_locations loclocatio2_
            on locations1_.id_location=loclocatio2_.id
   where
        admuser0_.id=1
   order by
        locations1_.name

locations1_.name分的订单,但应为loclocatio2_.name。我的实体有什么问题吗?

2 个答案:

答案 0 :(得分:0)

您在该字段的一侧有一个Set。因此没有“排序”(除了hashCode()gves之外)。如果你想订购,请使用List(这是Java,真的与JPA无关)。

你似乎也错过了那个M-N的非所有者方面的“mappedBy”。

答案 1 :(得分:0)

@OrderBy适用于ManyToMany。还有我在问题中提供的结构。问题是我的查询,JPA没有用它来管理。遗憾。