"명명 된 쿼리의 오류 : User.findByUserNameAndPassword"- 해결할 수 없음

Question

이것은 내가 게시 한 첫 번째 질문이므로 모든 규칙이나 예의 범절을 위반하면 사전에 사과드립니다. 나는 다음과 같은 예외를 제거 할 수없는 것 :

Caused by: javax.persistence.PersistenceException: [PersistenceUnit: default] Unable to build EntityManagerFactory 
at org.hibernate.ejb.Ejb3Configuration.buildEntityManagerFactory(Ejb3Configuration.java:915) 
at org.hibernate.ejb.HibernatePersistence.createContainerEntityManagerFactory(HibernatePersistence.java:74) 
at org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean.createNativeEntityManagerFactory(LocalContainerEntityManagerFactoryBean.java:257) 
at org.springframework.orm.jpa.AbstractEntityManagerFactoryBean.afterPropertiesSet(AbstractEntityManagerFactoryBean.java:310) 
at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.invokeInitMethods(AbstractAutowireCapableBeanFactory.java:1514) 
at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1452) 
... 79 more 

Caused by: org.hibernate.HibernateException: Errors in named queries: User.findByUserNameAndPassword 
at org.hibernate.impl.SessionFactoryImpl.<init>(SessionFactoryImpl.java:426) 
at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1872) 
at org.hibernate.ejb.Ejb3Configuration.buildEntityManagerFactory(Ejb3Configuration.java:906) 

내 명명 된 질의는 다음과 같습니다 가지고있는 엔티티 (! 나하려고)

@Entity 
@Table(name="USERS") 
@NamedQueries 
({ 
    @NamedQuery(name="User.findByUserNameAndPassword", query="SELECT u FROM User u WHERE u.username = :username AND u.password = :password") 
}) 
public class User implements Serializable { 
private static final long serialVersionUID = 1L; 

@Id 
@SequenceGenerator(name="USERS_ID_GENERATOR", sequenceName="PERSONAL.GLOBALSEQUENCE") 
@GeneratedValue(strategy=GenerationType.SEQUENCE, generator="USERS_ID_GENERATOR") 
private long id; 

private String notes; 

@Column(name="PASSWORD") 
private String password; 



@Column(name="USERNAME") 
private String username; 
... 

내가 실행하고 UserServiceImpl 클래스에서이 쿼리 같은

:

@Transactional(readOnly=true) 
public User authenticate(String userName, String password) { 
    List<User> usersList = em.createQuery("User.findByUserNameAndPassword", User.class).setParameter("username", userName).setParameter("password", password).getResultList(); 
    User firstUserFromList = usersList.get(0); 
    return firstUserFromList; 
} 

내가 그럴 필요있다 많은 일들이 있었지만 잠시 동안이 일에 매달 렸습니다. 도움이나 안내는 정말 감사하겠습니다.

내 applicationContext.xml 파일은 다음과 같습니다

환호 :

 

<beans xmlns="http://www.springframework.org/schema/beans" 
xmlns:context="http://www.springframework.org/schema/context" 
xmlns:jpa="http://www.springframework.org/schema/data/jpa" 
xmlns:mvc="http://www.springframework.org/schema/mvc" 
xmlns:tx="http://www.springframework.org/schema/tx" 
xmlns:jee="http://www.springframework.org/schema/jee" 
xmlns:jdbc="http://www.springframework.org/schema/jdbc" 
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
xsi:schemaLocation=" 
    http://www.springframework.org/schema/beans  
    http://www.springframework.org/schema/beans/spring-beans-3.1.xsd 
    http://www.springframework.org/schema/context 
    http://www.springframework.org/schema/context/spring-context-3.1.xsd 
    http://www.springframework.org/schema/data/jpa 
    http://www.springframework.org/schema/data/jpa/spring-jpa-1.0.xsd 
    http://www.springframework.org/schema/mvc 
    http://www.springframework.org/schema/mvc/spring-mvc-3.1.xsd 
    http://www.springframework.org/schema/tx 
    http://www.springframework.org/schema/tx/spring-tx-3.1.xsd 
    http://www.springframework.org/schema/jee 
    http://www.springframework.org/schema/jee/spring-jee.xsd 
    http://www.springframework.org/schema/jdbc 
    http://www.springframework.org/schema/jdbc/spring-jdbc-3.1.xsd">  

<context:component-scan base-package="com.transience.sandbox" /> 
<mvc:annotation-driven /> 
<tx:annotation-driven /> 

<mvc:resources mapping="/static_resources/**" location="/static_resources/" /> 
<bean 
    class="org.springframework.web.servlet.view.InternalResourceViewResolver"> 
    <property name="prefix"> 
     <value>/WEB-INF/pages/</value> 
    </property> 
    <property name="suffix"> 
     <value>.jsp</value> 
    </property> 
</bean> 

<bean id="transactionManager" class="org.springframework.orm.jpa.JpaTransactionManager"> 
    <property name="entityManagerFactory" ref="emf"/> 
</bean> 

<bean id="emf" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean"> 
    <property name="dataSource" ref="dataSource" /> 
    <property name="jpaVendorAdapter"> <bean class="org.springframework.orm.jpa.vendor.HibernateJpaVendorAdapter" /></property> 
    <property name="packagesToScan" value="com.transience.sandbox.domain"/> 

    <property name="jpaProperties"> 
     <props> 
      <prop key="hibernate.query.factory_class">org.hibernate.hql.classic.ClassicQueryTranslatorFactory</prop> 
      <prop key="hibernate.dialect">org.hibernate.dialect.Oracle10gDialect</prop> 
      <prop key="hibernate.max_fetch_depth">3</prop> 
      <prop key="hibernate.jdbc.fetch_size">50</prop> 
      <prop key="hibernate.jdbc.batch_size">10</prop> 
      <prop key="hibernate.show_sql">true</prop> 
     </props>   
    </property> 
</bean>  

<jee:jndi-lookup id="dataSource" jndi-name="oracleXEDS"/> 
<jpa:repositories base-package="com.transience.sandbox.domain" entity-manager-factory-ref="emf" transaction-manager-ref="transactionManager"/> 

</beans> 

내가 Spring (core, webmvc, context, orm, jdbc, tx) 3.1, Hibernate 3.6.8, Spring-data-jpa 1.2.0, Weblogic 12c 및 OracleXE을 사용하고 있습니다. 그리고 메이븐.

Answer 1

em.createQuery() 대신 em.createNamedQuery()을 사용해보세요.

Answer 2

거의 초기 관찰 : 당신은 오라클에서 예약어 USER로 엔티티를 정의

. 따라서 쿼리 SELECT u FROM User u ...이 (가) 네이티브 쿼리로 변환되면 문제가 발생할 수 있습니다.

적절한 로깅 수준 &을 실행하여 생성 된 쿼리를 조사하십시오. 또한 예약어/키워드 사용을 피하십시오. & 더 의미있는 규칙을 지정하십시오.