【发布时间】:2021-09-03 14:54:17
【问题描述】:
我试图在我的 intelliJ spring boot jpa 应用程序中创建一个用户,但是发生了这个错误
Servlet.service() 用于 servlet [dispatcherServlet] 的上下文中的路径 [] 抛出异常 [请求处理失败;嵌套异常是 javax.persistence.PersistenceException: org.hibernate.exception.SQLGrammarException:无法提取 ResultSet] 有根本原因
oracle.jdbc.OracleDatabaseException: ORA-02289: 序列不存在
所以我尝试了这个我找到但仍然不起作用的解决方案
@SequenceGenerator(name = "SEQ_CADASTRO", sequenceName = "SEQ_CADASTRO", allocationSize = 1, initialValue = 1)
@GeneratedValue(generator = "SEQ_CADASTRO", strategy = GenerationType.SEQUENCE)
现在在我的应用程序中编写代码
@Id
@Column(name = "UTILIZADOR_ID")
@SequenceGenerator(name = "SEQ_CADASTRO", sequenceName = "SEQ_CADASTRO", allocationSize = 1, initialValue = 1)
@GeneratedValue(generator = "SEQ_CADASTRO", strategy = GenerationType.SEQUENCE)
private BigInteger utilizadorId;
这是我创建的Bll
public static void createUtilizador(SpringWebMVC.s2.DAL.Utilizador uti){
if(factory == null)
factory = Persistence.createEntityManagerFactory(PERSISTENCE_UNIT_NAME);
if (em == null) em = factory.createEntityManager();
em.getTransaction().begin();
em.persist(uti);
em.getTransaction().commit();
}
【问题讨论】:
-
你能发布你的 application.properties 吗?
标签: java spring-boot hibernate jpa intellij-idea