중첩 예외는 org.hibernate.exception.SQLGrammarException : ResultSet, Spring4, Hibernate4를 추출 할 수 없습니다.

Question

해당 프로젝트가 Spring4 및 hibernate4를 사용하여 결과를 추출 할 수 없으므로 발음을 실행하는 데 문제가 있습니다. 데이터베이스를 crearted했습니다. 평소와 같이 연결했습니다.하지만 오류를 반환합니다. 어떻게 해결할 수 있습니까?

HomeController 코드

@Controller 
public class HomeController { 

@Autowired 
private UserDao userDAO; 

@RequestMapping(value="/") 
public ModelAndView home(){ 
List<User> listUser=userDAO.list(); 
ModelAndView model=new ModelAndView("home"); 
model.addObject("userList",listUser); 
return model; 
} 

UserDAOImplementation 코드

public class UserDAOImpl implements UserDAO { 

private SessionFactory sessionFactory; 
public UserDAOImpl (SessionFactory sessionFactory){ 

    this.sessionFactory=sessionFactory; 
} 

@Override 
@Transactional 
public List<User> list() { 

    @SuppressWarnings("unchecked") 
    List<User>listUser =(List<User>)sessionFactory.getCurrentSession().createCriteria(User.class).setResultTransformer(Criteria.DISTINCT_ROOT_ENTITY).list(); 

    return listUser; 
} 
} 

User.hbm.xml 코드

<?xml version="1.0" encoding="UTF-8"?> 
<!DOCTYPE hibernate-mapping PUBLIC 
    "-//Hibernate/Hibernate Mapping DTD 3.0//EN" 
    "http://www.hibernate.org/dtd/hibernate-mapping-3.0.dtd"> 
<hibernate-mapping package="com.spring4hibernate4test.org.model"> 
<class name="User" table="USERS" > 
<id name="id" column="USER_ID"> 
<generator class="native"/> 
</id> 
<property name="username" column="USERNAME" /> 
<property name="password" column="PASSWORD" /> 
<property name="email" column="EMAIL" /> 
</class> 

</hibernate-mapping> 

서블릿의 context.xml 코드

<?xml version="1.0" encoding="UTF-8"?> 
<beans xmlns:mvc="http://www.springframework.org/schema/mvc" 
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
xmlns="http://www.springframework.org/schema/beans" 
xmlns:context="http://www.springframework.org/schema/context" 
xmlns:tx="http://www.springframework.org/schema/tx" 
xsi:schemaLocation="http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc.xsd 
    http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd 
    http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx.xsd 
    http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context.xsd"> 


<!-- DispatcherServlet Context: defines this servlet's request-processing infrastructure --> 

<!-- Enables the Spring MVC @Controller programming model --> 
<mvc:annotation-driven /> 

<!-- Handles HTTP GET requests for /resources/** by efficiently serving up static resources in the ${webappRoot}/resources directory --> 
<mvc:resources mapping="/resources/**" location="/resources/" /> 

<!-- Resolves views selected for rendering by @Controllers to .jsp resources in the /WEB-INF/views directory --> 
<bean class="org.springframework.web.servlet.view.InternalResourceViewResolver"> 
    <property name="prefix" value="/WEB-INF/views/" /> 
    <property name="suffix" value=".jsp" /> 
</bean> 

<context:component-scan base-package="com.spring4hibernate4test.org" /> 



<bean id="dataSource" class="org.apache.commons.dbcp2.BasicDataSource" destroy-method="close"> 
    <property name="driverClassName" value="com.mysql.jdbc.Driver"/> 
    <property name="url" value="jdbc:mysql://localhost:3306/usersdb"/> 
    <property name="username" value="root"/> 
    <property name="password" value="root"/> 
</bean> 

<bean id="sessionFactory" 
    class="org.springframework.orm.hibernate4.LocalSessionFactoryBean"> 
    <property name="dataSource" ref="dataSource" /> 
    <property name="configLocation" value="classpath:hibernate.cfg.xml" /> 
</bean> 

<tx:annotation-driven /> 
<bean id="transactionManager" 
    class="org.springframework.orm.hibernate4.HibernateTransactionManager"> 
    <property name="sessionFactory" ref="sessionFactory" /> 
</bean> 

<bean id="userDao" class="com.spring4hibernate4test.org.dao.UserDAOImpl"> 
    <constructor-arg> 
     <ref bean="sessionFactory" /> 
    </constructor-arg> 
</bean> 

내가 기억으로

근본 원인

org.springframework.web.util.NestedServletException: Request processing failed; nested exception is org.hibernate.exception.SQLGrammarException: could not extract ResultSet 
org.hibernate.exception.SQLGrammarException: could not extract ResultSet 
com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: Table 'usersdb.USERS' doesn't exist 

Answer 1

는 MySQL의 테이블/열 이름은 리눅스 환경에 민감한 기본 케이스로, 그래서 그에 따라 User.hbm.xml를 업데이트하려고합니다.