2回答

• 

  weixin_慕先生7407711

  2019-01-27 16:56:25

  <web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
           xmlns="http://java.sun.com/xml/ns/javaee"
            xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
           id="WebApp_ID" version="3.0">
           
      <context-param>  
          <param-name>contextConfigLocation</param-name>  
          <param-value>classpath*:spring.xml</param-value>  
      </context-param>
           
      <listener>  
          <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>  
      </listener>
      
      <listener>
            <listener-class>org.springframework.web.context.request.RequestContextListener</listener-class>
      </listener>
      
       <filter>
          <filter-name>shiroFilter</filter-name>
          <filter-class>org.springframework.web.filter.DelegatingFilterProxy</filter-class>
      </filter>
      
      <filter-mapping>
          <filter-name>shiroFilter</filter-name>
          <url-pattern>/*</url-pattern>
      </filter-mapping>
      
      <!-- The front controller of this Spring Web application, responsible for handling all application requests -->
      <servlet>
          <servlet-name>springDispatcherServlet</servlet-name>
          <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
          <init-param>
              <param-name>contextConfigLocation</param-name>
              <param-value>classpath*:spring-mvc.xml</param-value>
          </init-param>
          <load-on-startup>1</load-on-startup>
      </servlet>
  
      <!-- Map all requests to the DispatcherServlet for handling -->
      <servlet-mapping>
          <servlet-name>springDispatcherServlet</servlet-name>
          <url-pattern>/</url-pattern>
      </servlet-mapping>
      
      
  </web-app>

  
  

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• 

  gallinipper

  2018-11-23 15:03:24

  这个web.xml是自己添加的模板(idea),网上有很多教程的。

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