Tomcat的web.XML的简单问题,困惑了我一个晚上.求教各位.
我今天刚开始学习JSP的.按照教程上的,在webapps下新建了myapp,目录结构:
myapp
----WEB-INF
----web.XML
----classes
----test
----Test.class


web.XML中的内容如下:
<?xml version="1.0" encoding="gb2312"?>
<web-app>
<display-name>My Web Application</display-name>
<description>
A application for test.
</description>
<servlet>
<servlet-name>Test</servlet-name>
<servlet-class>test.Test</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Test</servlet-name>
<url-pattern>/test</url-pattern>
</servlet-mapping>
</web-app>

但是访问http://localhost:8080/myapp/test的时候会提示"The requested resource (/myapp/test) is not available."

但是我把web.XML这个文件放在Tomcat目录下的conf文件夹中,就可以访问成功.是不是我的映射错了呢?请教各位,麻烦大家帮帮忙.先谢了:)

[解决办法]

XML code

 
  
  
  <?xml version="1.0" encoding="UTF-8"?> 
  <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"> 
  <display-name >My Web Application  </display-name >  
     <description >  
     A application for test.  
     </description >  
     <servlet >  
       <servlet-name >Test  </servlet-name >  
       <servlet-class >test.Test  </servlet-class >  
     </servlet >  
     <servlet-mapping >  
       <servlet-name >Test  </servlet-name >  
       <url-pattern >/test  </url-pattern >  
     </servlet-mapping >  
  </web-app>