我在web.xml中有一个简单的servlet配置:
<servlet>
<servlet-name>appServlet</servlet-name>
<servlet-class>org.atmosphere.cpr.MeteorServlet</servlet-class>
<init-param>
<param-name>org.atmosphere.servlet</param-name>
<param-value>org.springframework.web.servlet.DispatcherServlet</param-value>
</init-param>
<init-param>
<param-name>contextClass</param-name>
<param-value>
org.springframework.web.context.support.AnnotationConfigWebApplicationContext
</param-value>
</init-param>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>net.org.selector.animals.config.ComponentConfiguration</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
<async-supported>true</async-supported>
</servlet>
<servlet-mapping>
<servlet-name>appServlet</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
如何为SpringBootServletInitializer重写它?
答案 0 :(得分:24)
如果我从表面上看你的问题(你想要一个复制现有应用的SpringBootServletInitializer
),我想它会是这样的:
@Configuration
public class Restbucks extends SpringBootServletInitializer {
protected SpringApplicationBuilder configure(SpringApplicationBuilder builder) {
return builder.sources(Restbucks.class, ComponentConfiguration.class);
}
@Bean
public MeteorServlet dispatcherServlet() {
return new MeteorServlet();
}
@Bean
public ServletRegistrationBean dispatcherServletRegistration() {
ServletRegistrationBean registration = new ServletRegistrationBean(dispatcherServlet());
Map<String,String> params = new HashMap<String,String>();
params.put("org.atmosphere.servlet","org.springframework.web.servlet.DispatcherServlet");
params.put("contextClass","org.springframework.web.context.support.AnnotationConfigWebApplicationContext");
params.put("contextConfigLocation","net.org.selector.animals.config.ComponentConfiguration");
registration.setInitParameters(params);
return registration;
}
}
有关详细信息,请参阅docs on converting an existing app。
但是,使用Tomcat和Spring中的本机Websocket支持(而不是使用Atmosphere),最好不要使用Tomcat和Spring中的本机Websocket支持(例如,参见websocket sample和guide)。