JSP在虚拟连接的设置下如何读出SERVLET?

Question

将你的E:\test的WEB-INF的web.xml修改成如下形式：
<?xml version="1.0" encoding="ISO-8859-1"?>
<!DOCTYPE web-app
    PUBLIC "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN"
    "http://java.sun.com/dtd/web-app_2_3.dtd">
    
<web-app>
  <display-name>Welcome to Tomcat</display-name>
  <description>
     Welcome to Tomcat
  </description>   <servlet-mapping>
    <servlet-name>invoker</servlet-name>
    <url-pattern>/servlet/*</url-pattern>
  </servlet-mapping>
   
</web-app>
就可以运行你的servlet了，url为：
localhost:8080/test/servlet/yourServletClassName

Answer 1

如下格式修改:
<?xml version="1.0" encoding="ISO-8859-1"?>
<!DOCTYPE web-app PUBLIC
   "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN"
   "http://java.sun.com/dtd/web-app_2_3.dtd"><web-app>
  <description>The standard web descriptor for the html adaptor</description>
  <servlet>
    <servlet-name>HelloServlet</servlet-name>
    <servlet-class>test.Hello.HelloServlet</servlet-class>
  </servlet>  <servlet-mapping>
    <servlet-name>HelloServlet</servlet-name>
    <url-pattern>/HelloServlet</url-pattern>
  </servlet-mapping>
</web-app>

Answer 2

如下格式修改:
<?xml version="1.0" encoding="ISO-8859-1"?>
<!DOCTYPE web-app PUBLIC
   "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN"
   "http://java.sun.com/dtd/web-app_2_3.dtd"><web-app>
  <description>The standard web descriptor for the html adaptor</description>
  <servlet>
    <servlet-name>HelloServlet</servlet-name>
    <servlet-class>test.Hello.HelloServlet</servlet-class>
  </servlet>  <servlet-mapping>
    <servlet-name>HelloServlet</servlet-name>
    <url-pattern>/HelloServlet</url-pattern>
  </servlet-mapping>
</web-app>