Redirecting from a legacy Servlet to Spring 3 and from Spring 3 to a legacy Servlet - spring

Question

I'm learning Spring by integrating Spring 3 into a legacy Servlet application and gradually converting the legacy app over.
A web.xml *-servlet.xml similar to the ones I am using are posted below. Basically things are set up such that retrieving a string like "search", Spring will route it to a Controller and the view resolver will convert "search" into "/jsp/search.jsp"
I ran into problems doing a response.sendRedirect("search") from a legacy Servlet and a legacy ServletFilter To Spring. The URL came out correctly, but I got a blank page despite System.out.println() calls indicating that the JSP was reached. No error messages from Spring and the browser only told me something went wrong with the redirect.
I fixed that problem by forwarding, instead of redirecting from the legacy Servlets and ServletFilters:
request.getRequestDispatcher("search").forward(request,response);
getServletConfig().getServletContext().getRequestDispatcher("search").forward(request,response);
Going in the OTHER direction from a new JSP done in Spring 3 to a legacy Servlet, I have buttons on screens so I just used a javascript call to "location.href=/helloworld". I will need to send some parameters, so I will likely convert those buttons into submitting tiny HTML forms.
What I am wondering is, is there a better approach to getting Spring 3 and the Legacy Servlets communicating in a better way.
Legacy Servlet => Spring 3
and
Spring 3 => Legacy Servlet
web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app id="WebApp_ID" version="3.0"
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_3_0.xsd">
<display-name>Acme</display-name>
<!--welcome-file-list>
<welcome-file>/login</welcome-file>
</welcome-file-list-->
<servlet>
<servlet-name>acme</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>acme</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
<!-- Help Find The Spring Config Files -->
<listener>
<listener-class>
org.springframework.web.context.ContextLoaderListener
</listener-class>
</listener>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
/WEB-INF/nsd-servlet.xml,
/WEB-INF/nsd-security.xml
</param-value>
</context-param>
<!-- Spring Security -->
<filter>
<filter-name>springSecurityFilterChain</filter-name>
<filter-class>org.springframework.web.filter.DelegatingFilterProxy</filter-class>
</filter>
<filter-mapping>
<filter-name>springSecurityFilterChain</filter-name>
<url-pattern>/*</url-pattern>
</filter-mapping>
<!-- Integrate A Legacy Screen Done With A Servlet -->
<servlet>
<servlet-name>HelloWorldServlet</servlet-name>
<servlet-class>
com.legacy.HelloWorldServlet
</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>HelloWorldServlet</servlet-name>
<url-pattern>/helloworldservlet</url-pattern>
</servlet-mapping>
</web-app>
My acme-servlet.xml
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:context="http://www.springframework.org/schema/context"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:mvc="http://www.springframework.org/schema/mvc"
xsi:schemaLocation="http://www.springframework.org/schema/mvc
http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd
http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context-3.0.xsd">
<context:component-scan base-package="com.acme.controller" />
<mvc:resources mapping = "/**" location = "/,file:/apps1/bea/user_projects/domains/acme/common/,file:/c:/weblogic_common_files/acme/"/>
<mvc:annotation-driven/>
<bean
class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name = "prefix" value = "/jsp/"/>
<property name = "suffix" value = ".jsp"/>
</bean>
</beans>

Answer 1

When you use response.sendRedirect(String url) you are essentially passing in a URL that you want to redirect to.
sendRedirect observes the following rules:
/*
* Destination, it can be any relative or context specific path.
* if the path starts without '/' it is interpreted as relative to the current request URI.
* if the path starts with '/' it is interpreted as relative to the context.
*/
String destination ="/jsp/destination.jsp";
response.sendRedirect(response.encodeRedirectURL(destination));
When you type in "search", this is relative to the current request URI. Thus, unless your current request was to /acme/something, and assuming /acme/search is your servlet, the request will fail.
However, if you put the path relative to the root, using /acme/search, then the request will work from any context.
With that said, I'm not convinced this is really the best approach, as a redirect involves sending a response back to the client browser telling it to once again fetch content from yet another URL. This seems like a wasted trip to and from the server.
A better method may be to just wire in your servlets into Spring using something like Spring MVC. It's pretty flexible in that you can wrap new controller classes around your existing servlets and then invoke them directly by passing in the HttpServletRequest and HttpServletResponse objects. Once done, you can then slowly eliminate anything redundant with a working, efficient system instead of one chained together with redirects.
package samples;
public class SampleController extends AbstractController {
private int cacheSeconds;
// for wiring in cacheSeconds
public void setCacheSeconds(int cacheSeconds) {
this.cacheSeconds = cacheSeconds;
}
public int getCacheSeconds() {
return cacheSeconds;
}
public ModelAndView handleRequestInternal(
HttpServletRequest request,
HttpServletResponse response) throws Exception {
// you now have a Spring Controller wired in, and you could delegate to
// your legacy servlet in this manner
YourServlet servlet = new YourServlet();
servlet.doGet(request, response);
ModelAndView mav = new ModelAndView("hello");
mav.addObject("message", "Hello World!");
return mav;
}
}
<bean id="sampleController" class="samples.SampleController">
<property name="cacheSeconds" value="120"/>
</bean>
Note that manually instantiating a servlet is not really considered a good practice or good design. Instead, this is more of a stepping stone to get your Spring Controllers wired in and connected so that you can then systematically refactor your code and eliminate the legacy servlets by moving your business logic to service layers.
This approach jives with your plan to systematically migrate to and learn more about Spring, tackle some technical debt, all while still working on business goals related to your site.