내가 구글에 대한 이유를 찾았지만 내가 어디
나를 인도 해주십시오 박히이 예외
org.springframework.dao.InvalidDataAccessResourceUsageException: could not insert: [com.xgrid.evaltask.Entities.User]; SQL [insert into User (UserName, passwd, ID) values (?, ?, ?)]; nested exception is org.hibernate.exception.SQLGrammarException: could not insert: [com.xgrid.evaltask.Entities.User]
을 나는 봄과 최대 절전 모드를 통합하려고하지만 코드를 실행하고 내를 제출할 때 다음이 발생합니다 여기에 잘못된
는 applicationcontext.xml
봄 + 최대 절전 모드 confuguration
<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:aop="http://www.springframework.org/schema/aop"
xmlns:context="http://www.springframework.org/schema/context"
xmlns:jee="http://www.springframework.org/schema/jee"
xmlns:tx="http://www.springframework.org/schema/tx"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:p="http://www.springframework.org/schema/p"
xsi:schemaLocation="http://www.springframework.org/schema/aop
http://www.springframework.org/schema/aop/spring-aop-3.0.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context-3.0.xsd
http://www.springframework.org/schema/jee
http://www.springframework.org/schema/jee/spring-jee-3.0.xsd
http://www.springframework.org/schema/tx
http://www.springframework.org/schema/tx/spring-tx-3.0.xsd
http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-3.0.xsd">
<context:component-scan base-package="com.xgrid.evaltask"/>
<bean id="dataSource"
class="org.springframework.jdbc.datasource.DriverManagerDataSource"
p:driverClassName="com.microsoft.sqlserver.jdbc.SQLServerDriver"
p:url="jdbc:sqlserver://DBURL:1433;databaseName=Test_DB"
p:username="sampleUser"
p:password="@123456asdfgh" />
<tx:annotation-driven transaction-manager="transactionManager" />
<bean id="sessionFactory" class="org.springframework.orm.hibernate3.annotation.AnnotationSessionFactoryBean"
p:dataSource-ref="dataSource"
p:packagesToScan="com.xgrid.evaltask"
/>
<context:annotation-config />
<bean id="transactionManager" class="org.springframework.orm.hibernate3.HibernateTransactionManager"
p:sessionFactory-ref="sessionFactory" />
</beans>
입니다 그리고 여기에 여기에있는 hibernate.cfg.xml
<hibernate-configuration> <session-factory> <property name="hibernate.dialect">org.hibernate.dialect.SQLServerDialect</property> <!-- Enable this to see the SQL statements in the logs--> <property name="show_sql">true</property> <!-- This will drop our existing database and re-create a new one. Existing data will be deleted! --> <property name="hbm2ddl.auto">create</property> </session-factory> </hibernate-configuration>
내 DAO 클래스 0123입니다
@Repository("authorizeDao")
public class AuthorizeDaoImpl implements AuthorizeDao {
@Autowired
HiberAdd hiberAdd;
@Override
public boolean doAuthorize(Authentication authentication) {
System.out.println("Name: " + authentication.getUserName());
System.out.println("PW: " + authentication.getPassWord());
User user = new User();
user.setId(1212);
user.setName(authentication.getUserName());
user.setPasswd(authentication.getPassWord());
hiberAdd.add(user);
return true;
}
}
는이 내 CRUD 클래스
`@Service("hiberAdd")
@Transactional
public class HiberAdd {
@Resource(name="sessionFactory")
private SessionFactory sessionFactory;
public void add(User user) {
// Retrieve session from Hibernate
Session session = sessionFactory.getCurrentSession();
// Save
session.save(user);
System.out.println("Saved ......................... ");
}
}`
이
`@Entity
@Table(name="User")
public class User implements Serializable{
@Id
@Column(name="ID")
private Integer id;
@Column(name="UserName")
private String name;
@Column(name="passwd")
private String passwd;
//getters setters
사용자 테이블 구조 란 무엇입니까? –
ID (int), 이름 (varchar), 암호 (varchar) – Despicable
더 많은 오류 메시지를 붙여 넣을 수 있습니까? –