How can I register a servlet with Spring Boot? - spring

Question

This code don't work
I have a web.xml that needs to be translated into a spring boot
<servlet>
<servlet-name>page-resolver</servlet-name>
<servlet-class>com.bk.ForwardServlet</servlet-class>
<init-param>
<param-name>page</param-name>
<param-value>/index.jsp</param-value>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>page-resolver</servlet-name>
<url-pattern>/index.html</url-pattern>
<url-pattern>/login</url-pattern>
<url-pattern>/qmgr/*</url-pattern>
</servlet-mapping>
and this class
#WebServlet
public class ForwardServlet extends HttpServlet {
private String page;
#Override
public void init(ServletConfig config) throws ServletException {
super.init(config);
page = config.getInitParameter("page");
}
#Override
protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
req.getRequestDispatcher(page).forward(req, resp);
}
}
I wrote code that registers this bean but it doesn't work and I don't understand why
#Configuration
public class ConfigureWeb implements ServletContextInitializer {
private void registerServlet(ServletContext servletContext) {
ServletRegistrationBean serviceServlet =
new ServletRegistrationBean(new ForwardServlet(), "/index.html","/login","/qmgr/*");
serviceServlet.addUrlMappings("/index.jsp");
}
#Override
public void onStartup(ServletContext servletContext) throws ServletException {
registerServlet(servletContext);
}
}

Answer 1

First of all check and verify that all your resources are being kept in the correct directory.For spring-boot the jsp pages are to be kept in the directory src->main->webapp->WEB-INF->view->*.jsp .
You should have the following dependencies added in pom to compile the jsp properly when using embedded tomcat :
<!-- JSTL -->
<dependency>
<groupId>javax.servlet</groupId>
<artifactId>jstl</artifactId>
</dependency>
<!-- To compile JSP files -->
<dependency>
<groupId>org.apache.tomcat.embed</groupId>
<artifactId>tomcat-embed-jasper</artifactId>
<scope>provided</scope>
</dependency>
Now we need to tell spring where the jsp files reside, we can have two approaches for this :
Add entries in application.properties
spring.mvc.view.prefix=/WEB-INF/view/
spring.mvc.view.suffix=.jsp
Here we can specify the folder where the jsp files are to be expected, as well as define the view.
or
Configure InternalResourceViewResolver to serve JSP pages
#Configuration
#EnableWebMvc
#ComponentScan
public class MvcConfiguration extends WebMvcConfigurerAdapter
{
#Override
public void configureViewResolvers(ViewResolverRegistry registry) {
InternalResourceViewResolver resolver = new InternalResourceViewResolver();
resolver.setPrefix("/WEB-INF/view/");
resolver.setSuffix(".jsp");
resolver.setViewClass(JstlView.class);
registry.viewResolver(resolver);
}
}
Now for registering servlets :
ServletRegistrationBean is used to register Servlets in Servlet 3.0 + container. We need to create a bean of ServletRegistrationBean in our JavaConfig. Find some methods of ServletRegistrationBean used to configure a Servlet.
setServlet(): Sets the servlet to be registered.
addUrlMappings(): Add URL mappings for the Servlet.
setLoadOnStartup: Sets priority to load Servlet on startup.
Suppose we have two Servlets as HelloCountryServlet and HelloStateServlet then we will register them with Spring Boot using ServletRegistrationBean as following :
#Configuration
public class WebConfig {
#Bean
public ServletRegistrationBean<HttpServlet> countryServlet() {
ServletRegistrationBean<HttpServlet> servRegBean = new ServletRegistrationBean<>();
servRegBean.setServlet(new HelloCountryServlet());
servRegBean.addUrlMappings("/country/*");
servRegBean.setLoadOnStartup(1);
return servRegBean;
}
#Bean
public ServletRegistrationBean<HttpServlet> stateServlet() {
ServletRegistrationBean<HttpServlet> servRegBean = new ServletRegistrationBean<>();
servRegBean.setServlet(new HelloStateServlet());
servRegBean.addUrlMappings("/state/*");
servRegBean.setLoadOnStartup(1);
return servRegBean;
}
}
For every Servlet we create a ServletRegistrationBean bean.
And the servlet class will look like :
public class HelloCountryServlet extends HttpServlet {
private static final long serialVersionUID = 1L;
public void doPost(HttpServletRequest request, HttpServletResponse response) throws IOException{
doGet(request,response);
}
public void doGet(HttpServletRequest request, HttpServletResponse response) throws IOException {
response.setContentType("text/html");
PrintWriter out = response.getWriter();
out.println("<h3>Hello India!</h3>");
}
}
This is one approach.
Another method to register servlets in spring-boot is using annotations : #ServletComponentScan and #WebServlet. #ServletComponentScan in Spring Boot will scan Servlets annotated with #WebServlet, Filters annotated with #WebFilter and Listeners annotated with #WebListener.The annotation #ServletComponentScan is used on JavaConfig at class level. #ServletComponentScan scans Servlets, Filters and Listeners only using an embedded web server. Find the Servlets annotated with #WebServlet.
So the earlier HelloCountryServlet servlet can be provided as :
#WebServlet(urlPatterns = "/country/*", loadOnStartup = 1)
public class HelloCountryServlet extends HttpServlet {
private static final long serialVersionUID = 1L;
public void doPost(HttpServletRequest request, HttpServletResponse response) throws IOException{
doGet(request,response);
}
public void doGet(HttpServletRequest request, HttpServletResponse response) throws IOException {
response.setContentType("text/html");
PrintWriter out = response.getWriter();
out.println("<h3>Hello India!</h3>");
}
}
and provide the ServletComponentScan annotation like :
#ServletComponentScan
#SpringBootApplication
public class SpringBootAppStarter {
public static void main(String[] args) {
SpringApplication.run(SpringBootAppStarter.class, args);
}
}

Answer 2

If you are not required to register a servlet, you could add the forwarding to the ViewControllerRegistry
#Configuration
public class WebConfig implements WebMvcConfigurer {
#Override
public void addViewControllers(ViewControllerRegistry registry) {
registry.addViewController("/index.html").setViewName("forward:/index.jsp");
registry.addViewController("/login").setViewName("forward:/index.jsp");
registry.addViewController("/qmgr/**").setViewName("forward:/index.jsp");
}
}
You could combine the three addViewControllers into one using AntPathMatcher patterns
Map a view controller to the given URL path (or pattern) in order to
render a response with a pre-configured status code and view.
Patterns like "/admin/**" or "/articles/{articlename:\w+}" are
allowed. See AntPathMatcher for more details on the syntax.
Personally I think separating them is more readable