启动了之后范文路径404
启动成了 访问不到是怎么回事
启动成了 访问不到是怎么回事
2019-07-24
<?xml version="1.0" encoding="UTF-8"?> <web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"> <context-param> <param-name>contextConfigLocation</param-name> <param-value>classpath*:spring/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> <servlet> <servlet-name>spring-mvc</servlet-name> <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class> <init-param> <param-name>contextConfigLocation</param-name> <param-value>classpath*:spring/spring-mvc.xml</param-value> </init-param> <load-on-startup>1</load-on-startup> </servlet> <servlet-mapping> <servlet-name>spring-mvc</servlet-name> <url-pattern>/</url-pattern> </servlet-mapping> </web-app>
web.xml
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