我有@ManyToMany注释的父和子实体:
@Entity
@Table(name = "parent")
public class Parent {
@Id
@GenericGenerator(name = "uuid-gen", strategy = "uuid2")
@GeneratedValue(generator = "uuid-gen",strategy=GenerationType.IDENTITY)
private String id;
@ManyToMany(fetch = FetchType.LAZY)
@JoinTable(name = "parents_childs",
joinColumns = {@JoinColumn(name = "parent_id", nullable = false, updatable = false)},
inverseJoinColumns = {@JoinColumn(name = "child_id", nullable = false, updatable = false)})
private List<Child> childs;
}
和儿童实体:
@Entity
@Table(name="child")
public class Child {
@Id
@GenericGenerator(name = "uuid-gen", strategy = "uuid2")
@GeneratedValue(generator = "uuid-gen",strategy=GenerationType.IDENTITY)
private String id;
}
我的任务是找到包含具有特定ID的Child的所有父母。我尝试以这种方式在我的存储库中执行此操作:
@Query("select p from Parent p where p.childs.id = :childId and --some other conditions--")
@RestResource(path = "findByChildId")
Page<Visit> findByChild(@Param("childId") final String childId, final Pageable pageable);
例外:
java.lang.IllegalArgumentException: org.hibernate.QueryException: illegal attempt to dereference collection [parent0_.id.childs] with element property reference [id] [select p from Parent p where p.childs.id = :childId and --some other conditions--]
我知道可以解决将_
添加到findByChilds_Id
等方法名称(如here)的问题,但我无法找到如何写入@Query
注释。
如何使用JPQL编写它?
答案 0 :(得分:3)
我找到了解决方案:
@Query("select p from Parent p join p.childs c where c.id = : childId and --some other conditions--")
答案 1 :(得分:0)
您实际上根本不需要使用@Query,您可以使用 Spring Data JPA Repositories 中的派生查询方法并通过子属性查找父对象。在您的场景中,存储库中 find 方法的正确名称应该是:
Page<Visit> findByChildId(String id);