如何在使用java配置类时正确配置Spring MVC webapp？

Question

我通过@Configuration类使用Spring MVC：

@Configuration 
@EnableWebMvc 
public class WebConfiguration extends WebMvcConfigurerAdapter { 
    // more stuff 
} 

在我的web.xml我创建的ApplicationContext：

<context-param> 
    <param-name>contextClass</param-name> 
    <param-value>org.springframework.web.context.support.AnnotationConfigWebApplicationContext</param-value> 
</context-param> 

<context-param> 
    <param-name>contextConfigLocation</param-name> 
    <param-value>my.package.WebConfiguration</param-value> 
</context-param> 

，我也创建一个DispatcherServlet的，如下：

<servlet> 
    <servlet-name>mywebapp</servlet-name> 
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class> 
    <load-on-startup>1</load-on-startup> 
</servlet> 

<servlet-mapping> 
    <servlet-name>mywebapp</servlet-name> 
    <url-pattern>/</url-pattern> 
</servlet-mapping> 

为了让dispatcherServlet工作，我需要一个mywebapp-servlet.xml现在，它是空的。我真的需要mywebapp-servlet.xml文件吗？

Answer 1

您不需要任何XML文件。但是你必须告诉调度员未查找默认文件：

<listener> 
    <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class> 
</listener> 

<context-param> 
    <param-name>contextClass</param-name> 
    <param-value>org.springframework.web.context.support.AnnotationConfigWebApplicationContext</param-value> 
</context-param> 

<context-param> 
    <param-name>contextConfigLocation</param-name> 
    <param-value>my.pack.WebConfiguration</param-value> 
</context-param> 

<servlet> 
    <servlet-name>mywebapp</servlet-name> 
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class> 
    <init-param> 
     <param-name>contextConfigLocation</param-name> 
     <param-value></param-value> 
    </init-param> 
    <load-on-startup>1</load-on-startup> 
</servlet> 

<servlet-mapping> 
    <servlet-name>mywebapp</servlet-name> 
    <url-pattern>/</url-pattern> 
</servlet-mapping> 

只是为了您的信息：事实上，在Servlet的3 web.xml中不再需要了。