Question

我是Hibernate的新手，并且一直在努力解决hibernate问题。我得到org.Hibernate.QueryException: not an association: id。

这是正确运行的查询，我正在尝试实现。可以说我有带架子的房间。每个方框都包含一个项目。我只有盒子的条形码，我想得到那个房间里的所有物品。

select * from items where location_id in 
   (select id from storage where parent_id in
   (select id from storage where parent_id in 
   (select id from storage where barcode like 'some_box_barcode_here')))

这就是我在JPA中组织它的方式（我得到了例外）：

DetachedCriteria subCriteriaRoom = DetachedCriteria.forClass(Storage.class, "storageRoom")
            .createAlias("id", "roomId")
            .createAlias("index", "roomIndex")
            .add(Restrictions.eq("roomIndex", barcode));
subCriteriaRoom.setProjection(Projections.property("roomId"));

DetachedCriteria subCriteriaShelf = DetachedCriteria.forClass(Storage.class, "storageShelf")
            .createAlias("id", "shelfId")
            .createAlias("parent", "shelfParent")
            .add(Restrictions.in("shelfParent", subCriteriaRoom));
subCriteriaShelf.setProjection(Projections.property("shelfId"));

DetachedCriteria subCriteriaBox = DetachedCriteria.forClass(Storage.class, "storageBox")
            .createAlias("id", "boxId")
            .createAlias("parent", "boxParent")
            .add(Restrictions.in("boxParent", subCriteriaShelf));
subCriteriaBox.setProjection(Projections.property("boxId"));

Criteria criteria = session.createCriteria(Item.class, "item");
criteria.createAlias("storage", "storage")
            .add(Subqueries.propertyEq("location", subCriteriaBox));

List list = criteria.list();

有关表格的一些相关信息：

1

@Entity
@Table(name = "storage")
public class Storage implements DataModel {
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO, generator = "sequence")
    private long id;

    @ManyToOne(fetch = FetchType.EAGER)
    @JoinColumn(name = "parent_id")
    private Storage parent;

    @Column(name = "barcode", length = 31, nullable = true)
    private String barcode;

2

@Entity
@Table(name = "items")
public class Items implements DataModel {
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO, generator = "sequence")
    private long id;

    @OneToOne(fetch = FetchType.EAGER) //the foreign key to storage.id
    @JoinColumn(name = "location_id")
    private Storage location;

任何提示都会非常感激。

P.S。我需要这么大的查询的原因是，现在房间在数据库中成倍增加，因为一些错误将暂时无法修复。

org.Hibernate.QueryException：不是表id的关联

0 个答案: