2回答

• 

  victoriaMay

  2018-05-24 10:11:59

  已采纳

  右键创建不了，我试过~

  
  

  1 0

  等一束花開3...

  非常感谢！

  2018-05-24 14:56:17

  共 1 条回复 >

• 

  victoriaMay

  2018-05-24 10:11:17

  

  file—>Project Structure 

  如图所示

  但是你得自己写内容

  <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>
  
      <!--spring 监听器-->
      <listener>
          <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
      </listener>
  
      <listener>
          <listener-class>org.springframework.web.context.request.RequestContextListener</listener-class>
      </listener>
  
      <!--shiro Filter-->
      <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>
  
      <!-- 解析springMvc -->
      <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>
  
      <!-- 拦截器 拦截所有 -->
      <servlet-mapping>
          <servlet-name>springDispatcherServlet</servlet-name>
          <url-pattern>/</url-pattern>
      </servlet-mapping>
  
  </web-app>

  
  

  0 6