本文介绍了org.hibernate.HibernateException：如果没有活动事务，createSQLQuery无效的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
当我想通过休眠连接到我的数据库时，我收到了这个异常，我尝试了很多我在互联网上发现的东西，但没有任何帮助，我的一些文件：
dao类连接：
@Repository
public class UserDaoImpl implements UserDao {
@Autowired
SessionFactory sessionFactory;
//查询的问题在这里
public List< User> getAllUsers（）{
return sessionFactory.getCurrentSession（）。createSQLQuery（SELECT * FROM user）。list（）;
}
}
web.xml：
<？xml version =1.0encoding =UTF-8？>
< web-app version =2.5xmlns =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>原型创建的Web应用程序< / display-name>
< servlet>
< servlet-name>调度程序< / servlet-name>
< servlet-class> org.springframework.web.servlet.DispatcherServlet< / servlet-class>
1< / load-on-startup>
< / servlet>
< servlet-mapping>
< servlet-name>调度程序< / 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 =1.0encoding =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 =viewResolverclass =org.springframework.web.servlet.view.InternalResourceViewResolver>
< property name =viewClassvalue =org.springframework.web.servlet.view.JstlView/>
< property name =prefixvalue =/ WEB-INF //>
< property name =suffixvalue =。jsp/>
< / bean>
< / beans>
和context.xml：
<？xml version =1.0encoding =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 =sessionFactoryclass =org.springframework.orm.hibernate4.LocalSessionFactoryBean>
< property name =hibernateProperties>
<道具>
< 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>
< /道具>
< / property>
< property name =annotatedClasses>
< list>
< value> com.lime.model.User< / value>
< / list>
< / property>
< / bean>
< / beans>
解决方案
b
@Repository
public class UserDaoImpl implements UserDao {
$ b $ @Autowired
SessionFactory sessionFactory;
//查询的问题在这里
public List< User> getAllUsers（）{
Session session = null;
尝试
{
Session session = sessionFactory.openSession（）;
return session.createSQLQuery（SELECT * FROM user）。list（）;
catch（例外e）
{
//记录
}
终于
{
if（session！= null&& session.isOpen）
{
session.close（）;
session = null;
}
}
}
}
使用genericDAO获取当前会话需要使用openSession（）明确打开，而getCurrentSession（）只需要附加一个使用openSession（）的 p 更新它到当前会话。根据作者
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:
@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>
my 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:
<?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>
解决方案
Just try with this
@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
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
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