Question

我通过@Configuration类使用Spring MVC：

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

在我的web.xml中，我创建了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>

我还创建了一个DispatcherServlet，如下所示：

<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。但是您必须告诉Dispatcher不要查找默认文件：

<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>

仅供参考：事实上，在Servlet 3中也不再需要web.xml。

使用java Configuration类时如何正确配置Spring MVC webapp？

1 个答案: