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启动了之后范文路径404

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启动成了  访问不到是怎么回事

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<?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

0 回复 有任何疑惑可以回复我~
#1

qq_Brave_Cui_0

解决了, 谢谢
2020-02-11 回复 有任何疑惑可以回复我~

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启动了之后范文路径404

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