java - org.hibernate.HibernateException : createSQLQuery is not valid without active transaction

Question

当我想通过 hibernate 连接到我的数据库时，我遇到了这个异常，我尝试了很多在互联网上找到的东西，但没有任何帮助，我的一些文件:带连接的 dao 类:

@Repository
public class UserDaoImpl implements UserDao {

    @Autowired
    SessionFactory sessionFactory;
//the problem with query is here
    public List<User> getAllUsers() {
        return sessionFactory.getCurrentSession().createSQLQuery("SELECT * FROM user").list();
    }

}

web.xml:

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee"
         xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
         xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">


    <display-name>Archetype Created Web Application</display-name>

    <servlet>
        <servlet-name>dispatcher</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <servlet-mapping>
        <servlet-name>dispatcher</servlet-name>
        <url-pattern>*.htm</url-pattern>
    </servlet-mapping>

    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>classpath:context.xml</param-value>
    </context-param>

    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener>

</web-app>

我的servlet:

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
       xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
       xmlns:context="http://www.springframework.org/schema/context"
       xmlns:mvc="http://www.springframework.org/schema/mvc"
       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/mvc
           http://www.springframework.org/schema/mvc/spring-mvc-3.1.xsd">

    <context:annotation-config />
    <context:component-scan base-package="com.lime" />

    <mvc:annotation-driven />
    <mvc:default-servlet-handler />

    <bean id="viewResolver" class="org.springframework.web.servlet.view.InternalResourceViewResolver">
        <property name="viewClass" value="org.springframework.web.servlet.view.JstlView" />
        <property name="prefix" value="/WEB-INF/" />
        <property name="suffix" value=".jsp" />
    </bean>

</beans>

和context.xml:

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
       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.xsd">

    <bean id="sessionFactory" class="org.springframework.orm.hibernate4.LocalSessionFactoryBean">
        <property name="hibernateProperties">
            <props>
                <prop key="hibernate.dialect">org.hibernate.dialect.PostgreSQL9Dialect</prop>
                <prop key="hibernate.show_sql">true</prop>
                <prop key="hibernate.current_session_context_class">thread</prop>
                <prop key="hibernate.connection.driver_class">org.postgresql.Driver</prop>
                <prop key="hibernate.connection.url">jdbc:postgresql://localhost:5432/come_to_blog_db</prop>
                <prop key="hibernate.connection.username">postgres</prop>
                <prop key="hibernate.connection.password">admin</prop>
            </props>
        </property>
        <property name="annotatedClasses">
            <list>
                <value>com.lime.model.User</value>
            </list>
        </property>
    </bean>

</beans>

Answer 1

试试这个

@Repository
public class UserDaoImpl implements UserDao {

@Autowired
SessionFactory sessionFactory;
//the problem with query is here
public List<User> getAllUsers() {
    Session session=null;
    try 
    {
    Session session = sessionFactory.openSession();
    return session.createSQLQuery("SELECT * FROM user").list();
    }
    catch(Exception e)
    {
     //Logging
    }
    finally
    {
        if(session !=null && session.isOpen)
        {
          session.close();
          session=null;
        }
    }
}

}

更新

使用 genericDAO，它获取需要使用 openSession() 显式打开的当前 session ，而 getCurrentSession() 只是将其附加到当前 session 。据作者介绍

GenericDAO makes the assumption that you will be handling transactions externally to the DAO