启动了之后范文路径404

来源:4-1 Shiro集成Spring

kkdj1476238393

2019-07-24 16:52

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启动成了  访问不到是怎么回事

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  • 慕慕0467570
    2019-07-31 16:12:40
    <?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

    qq_Bra...

    解决了, 谢谢

    2020-02-11 11:05:33

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