How do I use Spring/JUnit to verify a controller is not sending me to a non-existent view? - spring

Question

I'm using Spring 3.1.1.RELEASE and JUnit 4.11. I setup my JUnit tests like so
#RunWith(SpringJUnit4ClassRunner.class)
#ContextConfiguration({ "classpath:test-context.xml" })
public class MySpringTest
{
protected MockHttpServletRequest request;
protected MockHttpServletResponse response;
protected MockHttpSession session;
#Autowired
protected RequestMappingHandlerAdapter handlerAdapter;
#Autowired
protected RequestMappingHandlerMapping handlerMapping;
When testing controllers, I have this line to verify that the view the controller's method is returning is the right view …
import static org.springframework.test.web.ModelAndViewAssert.assertViewName;
...
final ModelAndView mav = submitMyForm(…);
assertViewName(mav, "folder/myView");
...
protected ModelAndView submitMyForm(… params ...) throws Exception {
request = new MockHttpServletRequest();
response = new MockHttpServletResponse();
request.setRequestURI("/myurl");
request.setMethod("POST");
request.addParameter("param1", param1);
...
final Object handler = handlerMapping.getHandler(request).getHandler();
return handlerAdapter.handle(request, response, handler);
}
My question is, once I verify the view returned my the controller is the expected view, how do I verify it won't result in a 404? The main problem I'm gaving now is testing whether or not the view actually maps to an underlying page in my WAR file.

Answer 1

why don't use spring-mvc-test and do something like this ?
#Autowired
private ViewResolver viewResolver;
// code
View view = viewResolver.resolveViewName(viewName, locale);
//assert view not null
or something like this, in wich you can check both if the view is ok and the returned status (is status 200/404?)
(more code here: http://goo.gl/fMqBsl)
#Test
public void indexTest() throws Exception {
mockMvc.perform(get("/")).andDo(print())
.andExpect(handler().handlerType(MainController.class))
.andExpect(handler().methodName("index"))
.andExpect(view().name("index"))
.andExpect(forwardedUrl("/WEB-INF/tiles/template.jsp"))
.andExpect(status().isOk());
}

Answer 2

i am using standard jsp view
basically, you need to know the view resolver(s). can a specific view be resolved? that means, if you DON'T have a file called abc.xml, it might still be a valid view.
for simplicity sake, lets assume that we have only one view resolver, and, its
"org.springframework.web.servlet.view.UrlBasedViewResolver"
and here is the bean definition
spring 3.2.4 documentation pdf, page 477
<bean id="viewResolver"
class="org.springframework.web.servlet.view.UrlBasedViewResolver">
<property name="viewClass" value="org.springframework.web.servlet.view.JstlView"/>
<property name="prefix" value="/WEB-INF/jsp/"/>
<property name="suffix" value=".jsp"/>
</bean>
eg: the view name "page1" => /WEB-INF/jsp/page1.jsp and "admin/page2" => /WEB-INF/jsp/admin/page2.jsp
using this, you can Inject the view resolved to your junit test using #Autowired and/or #Qualifier
then read the "prefix" and suffix value and find the full path like "src/main/webapp/" + prefix + viewname + suffix
and check if the file exists.
you may have multiple view resolvers, so you may want to inject the context and handle the view => filename resolution using a strategy pattern.
something like
foreach resolver
{
if i can resolve the view to a file (resolver type, viewname)
return the physical filename
else
try next resolver
}