java - Spring MVC - DispatcherServlet 通过注释

Question

我得到了 Spring MVC 应用程序。它在 Tomcat 7 上运行。现在我已经在我的 web.xml 文件中得到了这部分:

<servlet>
        <servlet-name>Spring MVC Dispatcher Servlet</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <init-param>
            <param-name>contextConfigLocation</param-name>
            <param-value>
                /WEB-INF/app-config.xml
            </param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>Spring MVC Dispatcher Servlet</servlet-name>
        <url-pattern>*.htm</url-pattern>
    </servlet-mapping>

有没有办法通过注解来初始化它？我有一个 MainSettings.java 类，其中所有 bean 均由 @Bean 注释初始化。那么我该如何初始化DispatherServlet呢？

Answer 1

这是带有注释的示例。希望这对您有帮助。

public class ApplicationInitializer implements WebApplicationInitializer {

    //Called first when the application starts loading.
    public void onStartup(ServletContext servletContext)
            throws ServletException {
        System.out.println("Inside application initializer...");

        //Registering the class that incorporates the annotated DispatcherServlet configuration of spring
        AnnotationConfigWebApplicationContext rootContext = new AnnotationConfigWebApplicationContext();
        rootContext.register(DispatcherConfig.class);

        //Adding the listener for the rootContext
        servletContext.addListener(new ContextLoaderListener(rootContext));

        //Registering the dispatcher servlet mappings.
        ServletRegistration.Dynamic dispatcher = servletContext.addServlet("dispatcher", new DispatcherServlet(rootContext));
        dispatcher.setLoadOnStartup(1);
        dispatcher.addMapping("/");
    }

}