当我尝试使用HQL映射对象时出现“找不到查询类的持久类”错误

如何解决当我尝试使用HQL映射对象时出现“找不到查询类的持久类”错误

我想使用HQL构造函数方法将自定义对象直接从数据库（ApplicationStatus映射到Status）。但是查询结果为空，并且Hibernate返回如下消息。

2020-08-14 17:22:45.224  WARN 6668 --- [nio-9090-exec-2] o.hibernate.hql.internal.QuerySplitter   : HHH000183: no persistent classes found for query class: select new com.tt.practices6.map.Status(stat.id,stat.status,stat.application.name,stat.application.production.name,stat.log,stat.requestTime,stat.checked,stat.idUserChecked) from com.tt.practices6.model.ApplicationStatus stat

这是我的休眠配置xml文件：

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE hibernate-configuration PUBLIC
        "-//Hibernate/Hibernate Configuration DTD 3.0//EN"
        "https://hibernate.org/dtd/hibernate-configuration-3.0.dtd">
<hibernate-configuration>
    <session-factory>
        <property name="hibernate.connection.driver_class">org.postgresql.Driver</property>
        <property name="hibernate.connection.password">passwd</property>
        <property name="hibernate.connection.url">jdbc:postgresql://localhost:5432/postgres</property>
        <property name="hibernate.connection.username">user</property>
        <property name="hibernate.dialect">org.hibernate.dialect.PostgreSQL9Dialect</property>

        <property name="hibernate.current_session_context_class">thread</property>
        <property name="hibernate.show_sql">true</property>

        <mapping class="com.tt.practices6.model.ApplicationStatus"/>
        <mapping class="com.tt.practices6.model.Application"/>
        <mapping class="com.tt.practices6.model.Production"/>
        <mapping class="com.tt.practices6.map.Status"/>
    </session-factory>
</hibernate-configuration>

查询方法：

    @Transactional
    @GetMapping("/applicationsStatus/map/list")
    public ResponseEntity<List<Status>> getAllMappedApplicationsStatus() {

        SessionFactory sessionFactory = HibernateUtil.getSessionFactory();
        Session session = sessionFactory.getCurrentSession();
        Transaction tx = session.beginTransaction();

        Query<Status> statusesQuery = session.createQuery(
                "select new com.tt.practices6.map.Status(stat.id," +
                        "stat.application.name," +
                        "stat.log,stat.idUserChecked) " +
                        "from com.tt.practices6.model.ApplicationStatus stat"
        );

        System.out.println(statusesQuery.getResultList());
        List<Status> statuses = new ArrayList<>(statusesQuery.list());

        tx.rollback();
        sessionFactory.close();

        return ResponseEntity.ok().body(statuses);
    }

ApplicationStatus实体：

@Entity
@Table(name = "applications_status")
public class ApplicationStatus {

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

    private boolean status;

    @ManyToOne(fetch = LAZY)
    @JoinColumn(name="guid_application",referencedColumnName = "guid")
    @JsonBackReference
    private Application application;

    private String log;

    @DateTimeFormat(iso = DateTimeFormat.ISO.DATE_TIME)
    private LocalDateTime requestTime;

    private boolean checked;

    @Column(name = "id_user_checked")
    private Long idUserChecked;

    public ApplicationStatus() {
    }

   // setters and getters

状态对象：

public class Status {

    private Long id;

    private boolean status;

    private String applicationName;

    private String productionName;

    private String log;

    private LocalDateTime requestTime;

    private boolean checked;

    private Long idUserChecked;

    public Status(Long id,boolean status,String applicationName,String productionName,String log,LocalDateTime requestTime,boolean checked,Long idUserChecked) {
        this.id = id;
        this.status = status;
        this.applicationName = applicationName;
        this.productionName = productionName;
        this.log = log;
        this.requestTime = requestTime;
        this.checked = checked;
        this.idUserChecked = idUserChecked;
    }

// setters and getters

解决方法

在@Table注释处检查表名称，也许所有大写字母都可以使用。也许您也需要提供架构。

,

状态类未使用任何@Entity注释声明。但是为它<mapping class="com.tt.practices6.map.Status"/>定义了一个映射。 因此，Hibernate将期望一个实体状态。

出于您的目的，您可以首先查询ApplicationStatus。然后，将值提取到Query<ApplicationQuery>中之后，您可以创建一个自定义对象。否则，请删除xml中的映射并尝试您的代码。