SpringMVC访问不到路径

Question

localhost:8080/工程名称/user/add 试试

Answer 1

还是不行啊
后台tomcat报这个
2014-05-23 11:15:40 040 [] [DEBUG] org.springframework.web.servlet.DispatcherServlet.doService(DispatcherServlet.java:843) - DispatcherServlet with name 'dispatcher' processing GET request for [/ifr_status/]
2014-05-23 11:15:40 040 [] [WARN] org.springframework.web.servlet.DispatcherServlet.noHandlerFound(DispatcherServlet.java:1118) - No mapping found for HTTP request with URI [/ifr_status/] in DispatcherServlet with name 'dispatcher'
2014-05-23 11:15:40 040 [] [DEBUG] org.springframework.web.servlet.FrameworkServlet.processRequest(FrameworkServlet.java:991) - Successfully completed request
2014-05-23 11:15:42 042 [] [DEBUG] org.springframework.web.servlet.DispatcherServlet.doService(DispatcherServlet.java:843) - DispatcherServlet with name 'dispatcher' processing GET request for [/user/add]
2014-05-23 11:15:42 042 [] [WARN] org.springframework.web.servlet.DispatcherServlet.noHandlerFound(DispatcherServlet.java:1118) - No mapping found for HTTP request with URI [/user/add] in DispatcherServlet with name 'dispatcher'
2014-05-23 11:15:42 042 [] [DEBUG] org.springframework.web.servlet.FrameworkServlet.processRequest(FrameworkServlet.java:991) - Successfully completed request

Answer 2

类不需要加@requestMapping()注解吧。
方法上面加注解，应该是这样子的吧@RequestMapping("/xxx.do")

Answer 3

Controller可以加@requestMapping()注解的，表示这个控制器里面的方法前面都+这个/user

Answer 4

<web-app version="2.4"
xmlns="http://java.sun.com/xml/ns/j2ee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee 
http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener>
    <display-name>Spring MVC Application</display-name>
    <!-- 设置Spring容器加载配置文件路径 -->
    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>classpath:spring-context.xml</param-value>
    </context-param>    <!-- 配置Spring核心控制器 -->
    <servlet>
        <servlet-name>dispatcher</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <init-param>
            <param-name>contextConfigLocation</param-name>
            <param-value>classpath*:mvc-*.xml</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>    <servlet-mapping>
        <servlet-name>dispatcher</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>      <welcome-file-list>
          <welcome-file>aaa.jsp</welcome-file>
      </welcome-file-list></web-app>

Answer 5

我刚才试验了一下 
注释掉
<!--配置spirngMVC-web的静态资源，不配会当成控制器-->
    <mvc:resources mapping="/assets/**" location="assets/"/>
    <mvc:resources mapping="/codefans.net/**" location="codefans.net/"/>
    <mvc:resources mapping="/Css/**" location="Css/"/>
    <mvc:resources mapping="/font/**" location="codefans.net/"/>
    <mvc:resources mapping="/Images/**" location="Images/"/>
    <mvc:resources mapping="/Js/**" location="Js/"/>
    <mvc:resources mapping="/Menu/**" location="Menu/"/>
    <mvc:resources mapping="/Node/**" location="Node/"/>
    <mvc:resources mapping="/pages/**" location="pages/"/>
    <mvc:resources mapping="/Public/**" location="Public/"/>
    <mvc:resources mapping="/Role/**" location="Role/"/>
    <mvc:resources mapping="/User/**" location="User/"/>
注释掉这一部分就可以了但不知道原因

Answer 6

我也碰到过这个问题，在web.xml添加
<servlet-mapping>
<servlet-name>default</servlet-name>
<url-pattern>/resources/*</url-pattern>
</servlet-mapping>
里面配置这个，替代spring配置文件里的
<mvc:resources mapping="" location=""/>
就可以解决。