休眠不自动生成表

Posted 2023-04-14

【中文标题】休眠不自动生成表

【英文标题】：Hibernate not auto generating tables

【发布时间】：2017-02-25 16:04:02

【问题描述】：

年年

我正在制作一个简单的 Spring MVC Web 应用程序。我可以很好地运行应用程序，但如果我尝试登录，我会得到以下信息（我删除了所有我认为不相关的消息）：

INFO: HHH000412: Hibernate Core 5.2.6.Final
org.hibernate.cfg.Environment <clinit>
INFO: HHH000206: hibernate.properties not found
INFO: HCANN000001: Hibernate Commons Annotations5.0.1.Final
org.hibernate.annotations.common.reflection.java.JavaReflectionManager <clinit>
INFO: HHH000400: Using dialect: org.hibernate.dialect.mysqlDialect
INFO: HHH000400: Using dialect: org.hibernate.dialect.MySQLDialect
org.springframework.security.authentication.InternalAuthenticationServiceException: PreparedStatementCallback; bad SQL grammar [SELECT username, password, enabled FROM Users WHERE username=?]; nested exception is java.sql.SQLSyntaxErrorException: Table 'ElsLoggerSchema.Users' doesn't exist

我使用 Spring Security 对用户进行身份验证。我希望 Hibernate 自动生成表，我的模式存在但它没有表。这是spring security的配置：

  @Configuration
    @EnableGlobalMethodSecurity
    @EnableWebSecurity
    @Import(SpringConfiguration.class)
    public class SecurityContext extends WebSecurityConfigurerAdapter 

        @Autowired
        private DataSource dataSource;  

        // authorizeRequests() -> use-expresions = "true"
        @Override
        protected void configure(HttpSecurity http) throws Exception 

            http.authorizeRequests()
                .antMatchers("/createaccount","/error", "/register", "/login", "/newaccount", "/resources/**").permitAll()
                .antMatchers("/**", "/*", "/").authenticated()
                .anyRequest().authenticated()
                    .and()
                .formLogin().loginPage("/login").defaultSuccessUrl("/dashboard").loginProcessingUrl("/j_spring_security_check")
                .usernameParameter("username").passwordParameter("password").failureUrl("/login?error=true")
                    .and()
                .logout().logoutUrl("/logout").logoutSuccessUrl("/login").invalidateHttpSession(true);
        


        // Equivalent of jdbc-user-service in XML
        @Autowired
        public void configureGlobal(AuthenticationManagerBuilder auth) throws Exception

            auth.jdbcAuthentication().dataSource(dataSource)
                .usersByUsernameQuery("SELECT username, password, enabled FROM Users WHERE username=?")
                .authoritiesByUsernameQuery("SELECT username, authority FROM authorities where username=?");
        

    

我的 hibernate 持久化配置就在这里，它包括设置 Hibernate 属性：

    @Configuration
    @EnableTransactionManagement
    @PropertySource( "/WEB-INF/persistence-mysql.properties" )
    @ComponentScan( "com.LearnersLogger" )
    @Import(SpringConfiguration.class)
    public class PersistenceConfig 

        @Autowired
        private Environment env;

        @Autowired
        private DataSource dataSource;

        @Bean(name="sessionFactory")
        public LocalSessionFactoryBean sessionFactory() 
            LocalSessionFactoryBean sessionFactory = new LocalSessionFactoryBean();
            sessionFactory.setDataSource(dataSource);
            sessionFactory.setPackagesToScan("com.LearnersLogger");
            try 
                sessionFactory.setHibernateProperties(hibernateProperties());
             catch (Exception e)
                System.out.println("Error with instantiating session factory");
                throw new RuntimeException(e);
            

            return sessionFactory;
        


        @Bean
        @Autowired
        public HibernateTransactionManager transactionManager(SessionFactory sessionFactory)

            HibernateTransactionManager htm = new HibernateTransactionManager();
            htm.setSessionFactory(sessionFactory);

            return htm;
        

        @Bean
        @Autowired
        public HibernateTemplate getHibernateTemplate(SessionFactory sessionFactory)
            HibernateTemplate hibernateTemplate = new HibernateTemplate(sessionFactory);
            return hibernateTemplate;
        

        public Properties hibernateProperties() 

            Properties properties = new Properties();
            properties.put("hibernate.hbm2dll.auto", this.env.getProperty("hibernate.hbm2ddl.auto"));
            properties.put("hibernate.dialect", this.env.getProperty("hibernate.dialect"));
            properties.put("hibernate.show", this.env.getProperty("hibernate.show_sql"));

            return properties;
        

        @Bean
        public PersistenceExceptionTranslationPostProcessor exceptionTranslation()
            return new PersistenceExceptionTranslationPostProcessor();
        

    

我的用户模型如下所示：

@Entity
@Table(name = "Users")
public class User implements Serializable

    /**
     * 
     */
    private static final long serialVersionUID = 5729727545813781294L;

    public User()

    

    // various attributes typical to a User object model

我尝试更改 hibernate.hbm2ddl.auto 以更新、创建和创建删除，但没有效果。 我的问题是，我遗漏了什么或者可能导致我的应用程序没有自动生成表的问题？

【参考方案1】：

这里有错别字，prop name应该是hibernate.hbm2ddl.auto:

properties.put("hibernate.hbm2dll.auto", this.env.getProperty("hibernate.hbm2ddl.auto"));

【讨论】：

好吧。该死。那解决了它。你知道什么好笑吗？我记得在 *** 上遇到了一个线程，其中 OP 犯了同样的错误，而解决方案正是这个错字。我对此嗤之以鼻，不考虑我会犯同样错误的任何机会，然后继续下一个线程。非常感谢您为我发现它xD

不客气。 :) 该属性的 JPA 版本更长，但可能更难打错：jpaProperties.put("javax.persistence.schema-generation.database.action", "create");

【参考方案2】：

你可以试试这个说法：properties.put("hibernate.hbm2ddl.auto", "create"); 来解决你的问题