HQL查询如何返回实体？

Question

@Entity
@Table(name="customer")
public class Customer {

    @Id
    @Column(name="Customer_Code")
    private String customer_Code;

    @Column(name="Customer_Name")
    private String customer_Name;
}


 @Entity
@Table(name="project")
public class Project {

    @Id
    @Column(name="Project_Code")
    public String project_Code;

    @Column(name="Project_Customer")
    public String project_Customer;

    @Column(name="Project_Description")
    public String project_Description;

    @Column(name="Project_Pastel_Prefix")
    public String project_Pastel_Prefix;

    @Column(name="Project_Name")
    public String project_Name;
}

// need to inject the session factory
    @Autowired
    private SessionFactory sessionFactory;

    @Override
    public List<Customer> getCustomers() {

        // get the current hibernate sessio
    Session currentSession = sessionFactory.getCurrentSession();

    // create a query  ... sort by last name
    Query<Customer> theQuery = 
            currentSession.createQuery("Query goes here",
                                        Customer.class);

    // execute query and get result list
    List<Customer> customers = theQuery.getResultList();

        // return the results       
        return customers;
    }

null

这似乎意味着我不像使用简单的“From customer”查询那样获取customer实体的实例。如果是这种情况，如何返回客户实体？如果没有，那我做错了什么？

Answer 1

当执行customer.*时，查询应该选择customer_code和customer_name，这是两个string对象。但结果是您需要一个customer实体。

使您的查询选择customer对象，如下所示。

从Customer as cus，Project as pro中选择distincent cus，cus.customer_code=pro.project_customer order by