Java org.hibernate.HibernateException: createSQLQuery 在没有活动事务的情况下无效

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

提问by Charlie Harper

I'm getting this exception when i want to connect to my database via hibernate, i was trying a lot of things i found on the internet but nothing helped, some of my files: dao class with connection:

当我想通过休眠连接到我的数据库时，我遇到了这个异常，我尝试了很多我在互联网上找到的东西，但没有任何帮助，我的一些文件：带有连接的 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:

网页.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>

my servlet:

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>

and context.xml:

和上下文.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>

采纳答案by Amogh

Just try with this

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;
        }
    }
}

}

Update

更新

with genericDAO it gets the current session which needs to be explicitly open using openSession(), while getCurrentSession() just attaches it to the current session. According to the author

使用 genericDAO，它获取需要使用 openSession() 显式打开的当前会话，而 getCurrentSession() 只是将它附加到当前会话。据作者说

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

GenericDAO 假设您将在 DAO 外部处理交易