Question

我需要将SQL本机查询结果作为映射，如别名作为键和列值作为Spring启动JPA中的值例如：

查询SELECT u.first_name as firstName, u.last_name as lastName from user u返回两行。 JPA中的实体经理应返回

[{
"firstName": "XXX",
"lastName": "YYY"
}, {
"firstName": "AAA",
"lastName": "BBB"
}]

Answer 1

您可以创建扩展pojo类的存储库，并制定查询方法以提供所需的输出

<强> users.java

@Entity
@Table(name = "USERS")
@EntityListeners(UsersEntityListener.class)
@NamedQueries({
    @NamedQuery(name = "User.findAll", query = "SELECT u FROM User u")})
public class User extends Domain implements Serializable {

    private static final long serialVersionUID = 1L;
    @Basic(optional = false)
    @Column(name = "FIRST_NAME", nullable = false)
    private String firstName;
    @Basic(optional = false)
    @Column(name = "LAST_NAME", nullable = false)
    private String lastName;
    @Column(name = "MIDDLE_NAME")
    private String middleName;
    @Basic(optional = false)
    @Column(name = "GENDER", nullable = false)
    private String gender;
    @Basic(optional = false)
    @Column(name = "MOBILE", nullable = false)
    private String mobile;
    @Basic(optional = false)
    @Column(name = "EMAIL", nullable = false)
    private String email;
    @Basic(optional = false)
    @Column(name = "USERNAME", nullable = false)
    private String username;
    @Basic(optional = false)
    @Column(name = "PASSWORD", nullable = false)
    private String password;
    @Column(name = "URL")
    private String url;

    public User() {
    }

//*************Getters and Setters******************/
}

<强> UserRepository.java

@Repository
@RepositoryRestResource
    public interface UserRepository extends JpaRepository<User,Integer> {

     @Query("select u.firstName,u.lastName from Users u ")
        List<BuyingRequirement> find();

    }

现在您可以直接使用此方法：

@RestController
@RequestMapping("/api/v1")
public class xyz{

@Autowired
UserRepository ur;

 @GetMapping(value = "/user")
    public List<Users> getNameandLastName(){
        List user= ur.find();

        return user;
    }

}

这将在键值对中提供所需的结果列表

Answer 2

您可以尝试以下代码：

import org.hibernate.internal.SessionImpl;
import java.sql.Connection;


Connection connection = entitymanager.unwrap(SessionImpl.class).connection();
PreparedStatement ps = connection.prepareStatement(procedure);`

ResultSet rs = ps.executeQuery();

在Spring boot JPA中获取Map作为结果集

2 个答案: