Strut框架报错

Question

报错：严重: Servlet /11.01 threw load() exception
javax.servlet.UnavailableException: Missing configuration resource for path /WEB-INF/struts-config.xml
我已经配置了struts-config.xml在web-inf下了，为什么还说找不到，求大神指教

Answer 1

struts-config.xml放哪其实是参考web.xml。再好好检查下把？
或者你把工程的路径结构和web.xml贴上来，大家看看。

Answer 2

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" version="2.5" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee   http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
  <servlet>
    <servlet-name>action</servlet-name>
    <servlet-class>org.apache.struts.action.ActionServlet</servlet-class>
    <init-param>
      <param-name>config</param-name>
      <param-value>/WEB-INF/struts-config.xml</param-value>
    </init-param>
    <init-param>
      <param-name>debug</param-name>
      <param-value>3</param-value>
    </init-param>
    <init-param>
      <param-name>detail</param-name>
      <param-value>3</param-value>
    </init-param>
    <load-on-startup>0</load-on-startup>
  </servlet>
  <servlet>
    <description>This is the description of my J2EE component</description>
    <display-name>This is the display name of my J2EE component</display-name>
    <servlet-name>InsertUserinfp</servlet-name>
    <servlet-class>com.action.InsertUserinfp</servlet-class>
  </servlet>
  <servlet>
    <description>This is the description of my J2EE component</description>
    <display-name>This is the display name of my J2EE component</display-name>
    <servlet-name>insertUserInfp</servlet-name>
    <servlet-class>com.action.insertUserInfp</servlet-class>
  </servlet>
  <servlet>
    <description>This is the description of my J2EE component</description>
    <display-name>This is the display name of my J2EE component</display-name>
    <servlet-name>InsertUserInfp</servlet-name>
    <servlet-class>com.action.InsertUserInfp</servlet-class>
  </servlet>  <servlet-mapping>
    <servlet-name>action</servlet-name>
    <url-pattern>*.do</url-pattern>
  </servlet-mapping>
  <servlet-mapping>
    <servlet-name>InsertUserinfp</servlet-name>
    <url-pattern>/InsertUserinfp</url-pattern>
  </servlet-mapping>  <servlet-mapping>
    <servlet-name>InsertUserInfp</servlet-name>
    <url-pattern>/InsertUserInfp</url-pattern>
  </servlet-mapping>
  <welcome-file-list>
    <welcome-file>index.jsp</welcome-file>
  </welcome-file-list>
</web-app>
你看下，谢了哈

Answer 3

struts.xml这个文件的文件名称最好不好改变，其次struts.xml如果在src下默认可以找到，如果在其它路劲下，如：WebRoot下，则需要你在web.xml里面配置路劲！

Answer 4

那你是怎么配置的 呢？
我有两种方法：
<init-param>      
        <param-name>config</param-name>      
        <param-value>/WEB-INF/struts-config.xml</param-value>
</init-param> 
或者
<filter>
        <filter-name>struts2</filter-name>
        <filter-class>org.apache.struts2.dispatcher.ng.filter.StrutsPrepareAndExecuteFilter</filter-class>
        <init-param>   
              <param-name>filterConfig</param-name>   
              <param-value>/WEB-INF/struts-config.xml</param-value>   
          </init-param>
    </filter>

Answer 5

嗯，<param-value>里面保存路径，两边最好不要加空格，换行之类的。。东东西西