404 error redirect in Spring with java config

Question

As you know, in XML, the way to configure this is:

    404
    /my-custom-page-not-fo         


        

Answer 1

The solution proposed in comments above really works:

@Override
protected void customizeRegistration(ServletRegistration.Dynamic registration)
{
  registration.setInitParameter("throwExceptionIfNoHandlerFound", "true");
}

Answer 2

Use code-based Servlet container initialization as described in the doc and override registerDispatcherServlet method to set throwExceptionIfNoHandlerFound property to true:

public class WebAppInitializer extends AbstractAnnotationConfigDispatcherServletInitializer {
    @Override
    protected Class<?>[] getRootConfigClasses() {
        return null;
    }

    @Override
    protected Class<?>[] getServletConfigClasses() {
        return new Class[] { WebConfig.class };
    }

    @Override
    protected String[] getServletMappings() {
        return new String[] { "/" };
    }

    @Override
    protected void registerDispatcherServlet(ServletContext servletContext) {
        String servletName = getServletName();
        Assert.hasLength(servletName, "getServletName() may not return empty or null");

        WebApplicationContext servletAppContext = createServletApplicationContext();
        Assert.notNull(servletAppContext,
            "createServletApplicationContext() did not return an application " +
                    "context for servlet [" + servletName + "]");

        DispatcherServlet dispatcherServlet = new DispatcherServlet(servletAppContext);

        // throw NoHandlerFoundException to Controller
        dispatcherServlet.setThrowExceptionIfNoHandlerFound(true);

        ServletRegistration.Dynamic registration = servletContext.addServlet(servletName, dispatcherServlet);
        Assert.notNull(registration,
            "Failed to register servlet with name '" + servletName + "'." +
                    "Check if there is another servlet registered under the same name.");

        registration.setLoadOnStartup(1);
        registration.addMapping(getServletMappings());
        registration.setAsyncSupported(isAsyncSupported());

        Filter[] filters = getServletFilters();
        if (!ObjectUtils.isEmpty(filters)) {
            for (Filter filter : filters) {
                registerServletFilter(servletContext, filter);
            }
        }

        customizeRegistration(registration);
    }
}    

Then create an exception handler:

@ControllerAdvice
public class ExceptionHandlerController {
    @ExceptionHandler(Exception.class)
    public String handleException(Exception e) {
        return "404";// view name for 404 error
    }   
}

Don't forget about using @EnableWebMvc annotation on your Spring configuration file:

@Configuration
@EnableWebMvc
@ComponentScan(basePackages= {"org.project.etc"})
public class WebConfig extends WebMvcConfigurerAdapter {
    ...
}

Answer 3

Simple answer for 100% free xml:

1. Set properties for DispatcherServlet

   public class SpringMvcInitializer extends AbstractAnnotationConfigDispatcherServletInitializer {
   
       @Override
       protected Class<?>[] getRootConfigClasses() {
           return new Class[] { RootConfig.class  };
       }
   
       @Override
       protected Class<?>[] getServletConfigClasses() {
           return new Class[] {AppConfig.class  };
       }
   
       @Override
       protected String[] getServletMappings() {
           return new String[] { "/" };
       }
   
       @Override
       protected void customizeRegistration(ServletRegistration.Dynamic registration) {
           boolean done = registration.setInitParameter("throwExceptionIfNoHandlerFound", "true"); // -> true
           if(!done) throw new RuntimeException();
       }
   
   }
   

2. Create @ControllerAdvice:

   @ControllerAdvice
   public class AdviceController {
   
       @ExceptionHandler(NoHandlerFoundException.class)
       public String handle(Exception ex) {
           return "redirect:/404";
       }
   
       @RequestMapping(value = {"/404"}, method = RequestMethod.GET)
       public String NotFoudPage() {
           return "404";
   
       }
   }