请教tomcat下的设置 (救!)

web.xml正确格式如下,你可以参考一下
<?xml version="1.0" encoding="ISO-8859-1"?>
<!DOCTYPE web-app PUBLIC
          "-//Sun Microsystems, Inc.//DTD Web Application 2.2//EN"
          "http://java.sun.com/j2ee/dtds/web-app_2_2.dtd">
<web-app>
  <display-name>VB</display-name>
    <servlet>
        <servlet-name>VBDemoServlet</servlet-name>
        <servlet-class>com.sunnitech.VirtualBankDemo.VBDemoServlet</servlet-class>
    </servlet>
    <servlet>
        <servlet-name>ServletTest</servlet-name>
        <servlet-class>com.sunnitech.VirtualBankDemo.ServletTest</servlet-class>
    </servlet>
    <servlet-mapping>
        <servlet-name>VBDemoServlet</servlet-name>
        <url-pattern>/do</url-pattern>
    </servlet-mapping>
    <servlet-mapping>
        <servlet-name>ServletTest</servlet-name>
        <url-pattern>/sa</url-pattern>
    </servlet-mapping>
</web-app>

解决方案 »

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加句试试<servlet-mapping>
        <servlet-name>servlet名字</servlet-name>
        <url-pattern>servlet所在的文字以及名字(相对classes目录)</url-pattern>
</servlet-mapping>
不需要在examples目录下
自建目录请注册Webapp的目录格式。
简单的servlet在下,可编译成.class 放在webapps\examples\WEB-INF\classes
package myservlet;
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;
public class FirstServlet extends HttpServlet{
   public void doGet(HttpServletRequest req, HttpServletResponse resp)
     throws ServletException, IOException{
      resp.setContentType("text/html;charset=GB2312");
      PrintWriter out = resp.getWriter();
    out.println("<html><head><title>第一个Servlet</title></head><body>");
    out.println("<h2>这是个简单的servlet实例</h2>");

   //make sure method="post" so that the servlet service method
   //calls doPost in the response to this form submit
    out.println(
        "<form method='post' action ='" + req.getContextPath( ) + "' >");
    out.println("<h4>输入你的名字然后提交:</h4>");
    out.println("Your name: ");
    out.println("<input type=text name=userName >");
    out.println("<input type=submit value=Submit >");
    out.println("</form>");
    out.println("</body></html>");
    out.close();
    }//doGet

  public void doPost(HttpServletRequest req,
    HttpServletResponse resp) throws ServletException,
    java.io.IOException {
    req.setCharacterEncoding("GBK");
    String userName= req.getParameter("userName");
    resp.setContentType("text/html;charset=GB2312");
    PrintWriter out = resp.getWriter( );

    //Begin assembling the HTML content
    out.println("<b>欢迎您:</b>"+userName);
    out.close();
    }

    public String getServletInfo() {
     return "这是个简单servlet程序" ;
    }
   public void init(ServletConfig cfg) throws ServletException{
      super.init(cfg);
      }
      public void destroy(){
      super.destroy();
      }
      }在WEB-INF下的web.xml里部署<servlet>
        <servlet-name>
            FirstServlet
        </servlet-name>
<display-name>FirstServlet</display-name>
        <servlet-class>
            myservlet.FirstServlet
        </servlet-class>    </servlet>
<servlet-mapping>
        <servlet-name>FirstServlet</servlet-name>
        <url-pattern>/FirstServlet</url-pattern>
    </servlet-mapping>
运行则出现如下错误:
description The server encountered an internal error () that prevented it from fulfilling this request.exception javax.servlet.ServletException: Wrapper cannot find servlet class myservlet.FirstServlet or a class it depends on
at org.apache.catalina.core.StandardWrapper.loadServlet(StandardWrapper.java:844)
at org.apache.catalina.core.StandardWrapper.allocate(StandardWrapper.java:621)
at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:163)
~~~~~~~~~~~...................
谁帮帮忙了,大老!
放在webapps\examples\WEB-INF\classes\myservlet下
谢谢 lspy (晓剑),在这里是不是package myservlet的原因, 所以要放在myservlet.还请问为什么一定要在examples下运行,自己创建个目录怎不可以,我也在conf目录下的server中设置了content,可每次开了startup,为什么就是一闪就没有了?