SimpleJdbcTemplate not serializable - spring

Question

I use JSF and Spring. The JSF beans must be serializable by design. The JSF beans get injected with Spring beans that handle the db access. Thus these beans must be serializable as well. However simpleJdbcTemplate is not serializable so I get stack traces when restarting the servlet and sessions are lost. If I declare simpleJdbcTemplate to be transient, I don't get stack traces about "not serializable" but then simpleJdbcTemplate is sometimes NULL after a restart, which is even worse.
So what can I do so that the spring beans get injected with a "fresh" simpleJdbcTemplate when the servlet is restarted?
The classes look like this:
#Repository
public class Users implements Serializable
{
#Autowired
private SimpleJdbcTemplate simpleJdbcTemplate;
// ...
}
#ManagedBean
#SessionScoped
public class ThisBean implements Serializable
{
#ManagedProperty(value = "#{users}")
private Users users;
public void setUsers(Users users)
{
this.users = users;
}
// ...
}
Now some config files, just so that you see I didn't miss something standard.
The web.xml has this:
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<listener>
<listener-class>org.springframework.web.util.Log4jConfigListener</listener-class>
</listener>
<listener>
<listener-class>org.springframework.web.context.request.RequestContextListener</listener-class>
</listener>
<listener>
<listener-class>Listeners.SessionListener</listener-class>
</listener>
The applicationContext.xml has this:
<context:annotation-config />
<context:component-scan base-package="SpringDB"/>
<jee:jndi-lookup id="dataSource" jndi-name="java:comp/env/jdbc/mysql"/>
<bean id="simpleJdbcTemplate" class="org.springframework.jdbc.core.simple.SimpleJdbcTemplate">
<constructor-arg>
<ref bean="dataSource"/>
</constructor-arg>
</bean>
faces-config.xml has this:
<application>
<variable-resolver>org.springframework.web.jsf.DelegatingVariableResolver</variable-resolver>
<application>
<message-bundle>jsf</message-bundle>
</application>
</application>

Answer 1

Why don't you try to pass in the DataSource reference instead of the JdbcTemplate. The common practice is to inject the data source and create the JdbcTemplate within the setter for the data source. You'll find the following example in the spring documentation:
public class JdbcCorporateEventDao implements CorporateEventDao {
private JdbcTemplate jdbcTemplate;
public void setDataSource(DataSource dataSource) {
this.jdbcTemplate = new JdbcTemplate(dataSource);
}
// JDBC-backed implementations of the methods on the CorporateEventDao follow...
}
You can also use #Autowired on the setDataSource method if you don't want to set up your beans with XML. For further details have a look at the JdbcTemplate best practices section of the excellent spring documentation.

Answer 2

Add scope="prototype" definition for simpleJDBCTemplate bean. Then Spring Framework will create new instance for every servlet instance. More info about bean scopes: http://static.springsource.org/spring/docs/2.5.x/reference/beans.html#beans-factory-scopes