Question

我试图从表访问者Postgresql数据库中获取行。但我不需要所有列，但目前执行自定义查询，它不起作用。

EntityManager em = emf.createEntityManager();

Query query = em.createQuery("SELECT visitor_id, first_name, last_name, email, company, country_id FROM Visitor", Visitor.class);
List<Visitor> result = query.getResultList();
return result;

错误控制台显示我：

org.hibernate.exception.SQLGrammarException：无法执行查询 org.postgresql.util.PSQLException：找不到列名城市在此ResultSet中。

SELECT visitor_id, city, company, email, email_confirmed, first_name, 
       identification_number, identification_other, image, is_active, 
       job_title, last_name, mobile_number, password, phone_number, 
       province, country_id, created_by, identification_type_id
FROM visitor;

所有栏目都在那里。所以我所做的是调用函数来获取所有行并将其设置在另一个名为bindRegisterToDashboard（）;的函数中。它的函数只从表中设置六列到对象visitorDashBoardViewModel，以便在视图上显示它。

//service file code
for (Visitor item: getVisitors()) {
     visitorDashBoardViewModel.add(bindRegisterToDashboard(item);
}
return visitorDashBoardViewModel;

访客模特：

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private long visitorId;

    private String identificationOther;
    private long identificationNumber;
    private String firstName;
    private String lastName;
    private String email;
    private boolean emailConfirmed;
    private String phoneNumber;
    private String mobileNumber;
    private String password;
    private String province;
    private String city;
    private String company;
    private String jobTitle;
    private Boolean isActive;
    private byte[] image;


    @ManyToOne(cascade = CascadeType.ALL, fetch = FetchType.LAZY)
    @JoinColumn(name="identification_type_id")
    private IdentificationType identificationType;

    @ManyToOne(cascade = { CascadeType.PERSIST, CascadeType.MERGE }, fetch = FetchType.LAZY)
    @JoinColumn(name = "country_id", nullable = false)
    private Country country= new Country();

Answer 1

您的查询未返回访问者的实例。它返回对象数组。所以它不应该是TypedQuery<Visitor>，而只是Query。

Query query = em.createQuery("SELECT v.visitorId, v.firstName, v.lastName, v.email, v.company, v.country.id FROM Visitor v");
List<Object[]> rows = query.getResultList();
// now, in each Object[], visitor_id is at index 0, first_name at index 1, etc.

另外，请尊重Java命名约定：id，firstName，lastName等

如果您希望查询返回Visitor的实例，则应为

TypedQuery query = em.createQuery("SELECT v from Visitor v");
List<Visitor> visitors = query.getResultList();

创建自定义查询以获取Hibernate中的特定列

1 个答案: