EntityManager查詢出現問題

Question

由於某種原因，從方法運行此方法時未得到結果。

@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應該正確，因為它在HSQL數據庫管理器中執行並返回正確的值。 但是java函數會在查詢時停止運行。 它不會拋出任何錯誤，只是停止。 我沒有主意，應該去哪里看？ （其他具有count和select的類似方法都可以正常工作）。

使用HSQLDB和Hibernate。

發現引發了以下錯誤：

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]

但這並不能為我解決問題，因為RATING屬性是在表和實體中定義的...

@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（）中傳遞JP-QL。 但是似乎您正在使用帶有r.RATING，r.POINT_ID之類的值的本機SQL，該值可能不在Java實體中。 將其替換為等效的Java實體變量，可以為pointId

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

如果要使用本機sql，則可以使用em.createNativeQuery（）。

Answer 2

此問題最有可能是由大寫鎖定的屬性名稱引起的： RATING ， POINT_ID 。

嘗試用在Ratings類中使用的替代它們，可能是：

Select AVG(r.rating) from Ratings r WHERE r.point.id = :point_id