java - EntityManager 查询出现问题

Question

出于某种原因，我在从方法运行此方法时没有得到结果。

@SuppressWarnings("unchecked")
public Object[] getPointRaiting(Long id) {
    EntityManager em = createEntityManager();
    em.getTransaction().begin();
    Query allPointsQuery = em
            .createQuery("Select AVG(r.RATING) from Ratings r WHERE r.POINT_ID = :point");
    allPointsQuery.setParameter("point", id);
    Object[] rating = (Object[]) allPointsQuery.getSingleResult();
    em.getTransaction().commit();
    em.close();
    closeEntityManager();
    return rating;
}

SQL 应该是正确的，因为它在 HSQL 数据库管理器中执行并返回正确的值。但是java函数在查询时停止运行。它不会抛出任何错误只是停止。我没有想法，我应该去哪里寻找？ (其他类似的计数和选择方法都可以正常工作)。

使用 HSQLDB 和 Hibernate。

发现抛出如下错误:

org.hibernate.QueryException: could not resolve property: RATING of: kaart.entities.Ratings [Select AVG(r.RATING) from kaart.entities.Ratings r WHERE r.POINT_ID = :point]

但这并不能解决我的问题，因为 RATING 属性是在表和实体中定义的......

@Entity @Table(name = "RATINGS") 
public class Ratings implements Serializable {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;

    @ManyToOne
    private Point point;

    @ManyToOne
    private User user;

    @Column(name = "RATING")
    private int rating;

    private static final long serialVersionUID = 1L;

    public Ratings() {
        super();
    }

    public Ratings(Point point, User user, int rating) {
        this.point = point;
        this.user = user;
        this.rating = rating;
    }

    /*all getters and setters here*/}


@Entity
@Table(name = "POINT")
public class Point implements Serializable {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;

    @OneToMany(mappedBy = "point")
    private List<Category> pointsByCategory;

    @OneToMany(mappedBy = "point")
    private List<Ratings> pointRatings;

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

    @Column(name = "LOCATION")
    private String location;

    @Column(name = "DESCRIPTION")
    private String description;

    @Column(name = "LINK")
    private String link;

    @ManyToOne
    private User user;
    private static final long serialVersionUID = 1L;

    public Point() {
        super();
    }

    public Point(String name, String location, String description, String link, User user) {
        this.name = name;
        this.location = location;
        this.description = description;
        this.link = link;
        this.user = user;
    } /* getters and setters*/

Answer 1

您只能在 em.createQuery() 中传递 JP-QL。但似乎您正在使用 native SQL 以及 r.RATING、r.POINT_ID 等值，这些值可能不在 Java 实体中。将其替换为等效的 java 实体变量，可以是 pointId

em.createQuery("Select AVG(r.RATING) from Ratings r WHERE r.POINT_ID = :point");

如果你想使用原生sql，可以使用em.createNativeQuery()。