使用带有后缀的自定义id生成器时发生的Hibernate错误

Question

我试图为我的实体创建一个自定义id，但是当我试图这样做的时候，我得到了一个错误。

这是我的实体代码(部分)

@Entity
@Table(name = "notifications")
@Cache(usage = CacheConcurrencyStrategy.NONSTRICT_READ_WRITE)
public class Notifications implements Serializable {

    private static final long serialVersionUID = 1L;


    @Id
    @GeneratedValue(strategy = GenerationType.SEQUENCE , generator = "seq_notifications")
    @GenericGenerator(name = "seq_notifications",
        strategy = "com.mycompany.myapp.domain.PKeys.NotficationID")
    private String id;

下面是生成器类代码

public class NotficationID  extends SequenceStyleGenerator {
    public static final String DATE_FORMAT_PARAMETER = "dateFormat";
    public static final String DATE_FORMAT_DEFAULT = "%tY-%tm";

    public static final String NUMBER_FORMAT_PARAMETER = "numberFormat";
    public static final String NUMBER_FORMAT_DEFAULT = "%05d";

    public static final String DATE_NUMBER_SEPARATOR_PARAMETER = "dateNumberSeparator";
    public static final String DATE_NUMBER_SEPARATOR_DEFAULT = "_";

    private String format;

    @Override
    public Serializable generate(SharedSessionContractImplementor session,
                                 Object object) throws HibernateException {
        return String.format(format, LocalDate.now(), super.generate(session, object));
    }

    @Override
    public void configure(Type type, Properties params,
                          ServiceRegistry serviceRegistry) throws MappingException {
        super.configure(LongType.INSTANCE, params, serviceRegistry);

        String dateFormat = ConfigurationHelper.getString(DATE_FORMAT_PARAMETER, params, DATE_FORMAT_DEFAULT).replace("%", "%1");
        String numberFormat = ConfigurationHelper.getString(NUMBER_FORMAT_PARAMETER, params, NUMBER_FORMAT_DEFAULT).replace("%", "%2");
        String dateNumberSeparator = ConfigurationHelper.getString(DATE_NUMBER_SEPARATOR_PARAMETER, params, DATE_NUMBER_SEPARATOR_DEFAULT);
        this.format = dateFormat+dateNumberSeparator+numberFormat;
    }
}

当我试图插入一个值时，我会得到这个错误

2020-06-25 09:56:14.718 DEBUG 15572 --- [  XNIO-1 task-4] c.m.m.web.rest.NotificationsResource     : REST request to save Notifications : Notifications{id=null, notificationType='null', notificationDate='null', message='null'}
Hibernate: select next_val as id_val from seq_notifications for update
2020-06-25 09:56:14.744 ERROR 15572 --- [  XNIO-1 task-4] o.hibernate.id.enhanced.TableStructure   : could not read a hi value

java.sql.SQLSyntaxErrorException: Table 'relationships.seq_notifications' doesn't exist

2020-06-25 09:56:14.746  WARN 15572 --- [  XNIO-1 task-4] o.h.engine.jdbc.spi.SqlExceptionHelper   : SQL Error: 1146, SQLState: 42S02
2020-06-25 09:56:14.746 ERROR 15572 --- [  XNIO-1 task-4] o.h.engine.jdbc.spi.SqlExceptionHelper   : Table 'relationships.seq_notifications' doesn't exist
2020-06-25 09:56:14.753 ERROR 15572 --- [  XNIO-1 task-4] c.m.m.web.rest.NotificationsResource     : Exception in createNotifications() with cause = 'org.hibernate.exception.SQLGrammarException: error performing isolated work' and exception = 'error performing isolated work; SQL [n/a]; nested exception is org.hibernate.exception.SQLGrammarException: error performing isolated work'

注意:我使用的是jhipster

提前谢谢。

Answer 1

我想问题出在桌子上，我做了一张桌子，一切都很好，谢谢亨里克，你是对的，

我创建了一个名为“seq”的表，但是数据类型是varchar(255)，因为我的@Id是一个字符串