在IE上运行结果:
HTTP Status 404 - /ch02/SimpleServlet--------------------------------------------------------------------------------type Status reportmessage /ch02/SimpleServletdescription The requested resource (/ch02/SimpleServlet) is not available.
--------------------------------------------------------------------------------Apache Tomcat/6.0.26我用的软件是eclipse tomcat程序如下:
package test0;import java.io.IOException;
import java.io.PrintWriter;import javax.servlet.ServletConfig;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;@SuppressWarnings("serial")
public class SimpleServlet extends HttpServlet
{
public void init(ServletConfig config) 
      throws ServletException
      {
              super.init(config);
          }
    public void doGet(
          HttpServletRequest request,
          HttpServletResponse response)throws ServletException,IOException
          {
           doPost(request,response);
          }
    public void doPost(
       HttpServletRequest request,
       HttpServletResponse response)throws ServletException,IOException
       {
           response.setContentType("text/html");
           PrintWriter out = response.getWriter();
           out.println("<html>");
           out.println("<head><title>Simple Servlet</title></head>");
           out.println("<body>");
           out.println("<center><font size=6 color=red>");
           out.println("<Your address is" + "request.getRemoteAddr()"+"\n");
           out.println("</center></font></body>");
           out.println("</html>");
           out.close();
       }
}
class 的路径:D:\java\eclips\ch01\ch02\WEB-INF\classes\test0\SimpleServlet.classxml代码如下:
<?xml version="1.0" encoding="ISO-8859-1"?>
<web-app 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"
   version="2.5">
   <servlet>
        <servlet-name>SimpleServlet</servlet-name>
        <servlet-class>test0.SimpleServlet</servlet-class>
   </servlet>
   <servlet-mapping>
        <servlet-name>SimpleServlet</servlet-name>
        <url-pattern>/GetAddress</url-pattern>
   </servlet-mapping>
</web-app>
xml的路径:D:\java\eclips\ch01\ch02\WEB-INF\web.xml
请高手帮忙指点下,非常感谢!

解决方案 »

  1.   


    <?xml version="1.0" encoding="ISO-8859-1"?>
    <web-app 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"
      version="2.5">
      <servlet>
      <servlet-name>SimpleServlet</servlet-name>
      <servlet-class>test0.SimpleServlet</servlet-class>
      </servlet>
      <servlet-mapping>
      <servlet-name>SimpleServlet</servlet-name>
      <url-pattern>/GetAddress</url-pattern>//那你调用servlet就调用这个url了/GetAddress,建议一致
      </servlet-mapping>
    </web-app>
      

  2.   

    页面里面 import 你的类没
    再就是楼上说的
      

  3.   


    <?xml version="1.0" encoding="ISO-8859-1"?>
    <web-app 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"
      version="2.5">
      <servlet>
      <servlet-name>SimpleServlet</servlet-name>
      <servlet-class>test0.SimpleServlet</servlet-class>
      </servlet>
      <servlet-mapping>
      <servlet-name>SimpleServlet</servlet-name>
      <url-pattern>/GetAddress</url-pattern>//那你调用servlet就调用这个url了/GetAddress,建议一致
      </servlet-mapping>
    </web-app>
      

  4.   

    路径: 写 <url-pattern>/GetAddress</url-pattern> 里的/GetAddress而不是/SimpleServlet
      

  5.   


    在IE中访问:
    http://localhost:8080/项目名/GetAddress.do
    就OK了!
      

  6.   

    把/SimpleServlet改为/GetAddress  /GetAddress才是路径