How can I register a secondary servlet with Spring Boot?

Question

I have an extra servlet I need to register in my application. However with Spring Boot and its Java Config, I can\'t just add servlet mappings in a web.xml file

Answer 1

Also available in the BeanDefinitionRegistryPostProcessor

package bj;

import org.springframework.beans.BeansException;
import org.springframework.beans.factory.config.ConfigurableListableBeanFactory;
import org.springframework.beans.factory.support.BeanDefinitionRegistry;
import org.springframework.beans.factory.support.BeanDefinitionRegistryPostProcessor;
import org.springframework.beans.factory.support.RootBeanDefinition;
import org.springframework.boot.SpringApplication;
import org.springframework.boot.autoconfigure.SpringBootApplication;
import org.springframework.boot.web.servlet.ServletRegistrationBean;

import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import java.io.IOException;

@SpringBootApplication
class App implements BeanDefinitionRegistryPostProcessor {
    public static void main(String[] args) {
        SpringApplication.run(App.class, args);
    }

    @Override
    public void postProcessBeanDefinitionRegistry(BeanDefinitionRegistry registry) throws BeansException {
        registry.registerBeanDefinition("myServlet", new RootBeanDefinition(ServletRegistrationBean.class,
                () -> new ServletRegistrationBean<>(new HttpServlet() {
                    @Override
                    protected void service(HttpServletRequest req, HttpServletResponse resp) throws IOException {
                        resp.getWriter().write("hello world");
                    }
                }, "/foo/*")));
    }

    @Override
    public void postProcessBeanFactory(ConfigurableListableBeanFactory beanFactory) {
    }
}

Answer 2

You can register multiple different servlet with different ServletRegistrationBean like @Bean in Application class and you can register a servlet has multiple servlet mapping;

   @Bean
   public ServletRegistrationBean axisServletRegistrationBean() {
      ServletRegistrationBean registration = new ServletRegistrationBean(new AxisServlet(), "/services/*");
      registration.addUrlMappings("*.jws");
      return registration;
   }

   @Bean
   public ServletRegistrationBean adminServletRegistrationBean() {
      return new ServletRegistrationBean(new AdminServlet(), "/servlet/AdminServlet");
   }

Answer 3

If you're using embedded server, you can annotate with @WebServlet your servlet class:

@WebServlet(urlPatterns = "/example")
public class ExampleServlet extends HttpServlet

From @WebServlet:

Annotation used to declare a servlet.

This annotation is processed by the container at deployment time, and the corresponding servlet made available at the specified URL patterns.

And enable @ServletComponentScan on a base class:

@ServletComponentScan
@EntityScan(basePackageClasses = { ExampleApp.class, Jsr310JpaConverters.class })
@SpringBootApplication
public class ExampleApp 

Please note that @ServletComponentScan will work only with embedded server:

Enables scanning for Servlet components (filters, servlets, and listeners). Scanning is only performed when using an embedded web server.

More info: The @ServletComponentScan Annotation in Spring Boot

Answer 4

This way worked for me, having a servlet called WS01455501EndpointFor89

@Bean
public ServletRegistrationBean<WS01455501EndpointFor89> servletRegistrationBeanAlt(ApplicationContext context) {
    ServletRegistrationBean<WS01455501EndpointFor89> servletRegistrationBean = new ServletRegistrationBean<>(new WS01455501EndpointFor89(),
            "/WS01455501Endpoint");
    servletRegistrationBean.setLoadOnStartup(1);
    return servletRegistrationBean;
}

Answer 5

We can also register the Servlet as follow way:

@Configuration
public class ConfigureWeb implements ServletContextInitializer, EmbeddedServletContainerCustomizer {

  @Override
  public void onStartup(ServletContext servletContext) throws ServletException {
      registerServlet(servletContext);
  }

  private void registerServlet(ServletContext servletContext) {
      log.debug("register Servlet");
      ServletRegistration.Dynamic serviceServlet = servletContext.addServlet("ServiceConnect", new ServiceServlet());

      serviceServlet.addMapping("/api/ServiceConnect/*");
      serviceServlet.setAsyncSupported(true);
      serviceServlet.setLoadOnStartup(2);
  }
}

Answer 6

Just add a bean for the servlet. It'll get mapped to /{beanName}/.

@Bean
public Servlet foo() {
    return new FooServlet();
}