急!!!我是个struts的初学者,一天一个程序都没调试出来,急死我了.请大家帮帮忙!大恩不言谢!

我是struts初学者，做了个登录，可怎么也运行不了。以下是目录结构：
TryStruts
| src
| app
  | RegisterForm.java
  | RegisterAction.java
| TryStrus
| WEB-INF
   |success.html
   |failure.html
   |struts-config.xml
   |struts-html.tld
   |web.xml
| Register.jsp开发工具:MyEclipse, Tomcat5.5, jdk6.0
Register.jsp源代码如下：
<%@ page language="java" import="java.util.*" pageEncoding="gbk"%>
<%@ taglib uri="WEB-INF/struts-html.tld" prefix="html"%>
<%
String path = request.getContextPath();
String basePath = request.getScheme()+"://"+request.getServerName()+":"+request.getServerPort()+path+"/";
%><!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"> <html:form action="/Register.do">
   rUsername<html:text property="username"/><br>
   enter password<html:password property="password1"/><br>
   re-enter password<html:password property="password1"/><br>
   <html:submit value="Register"/>
</html:form>
Struts-config.xml代码如下：
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE struts-config PUBLIC "-//Apache Software Foundation//DTD Struts Configuration 1.2//EN" "http://struts.apache.org/dtds/struts-config_1_2.dtd"><struts-config>
  <data-sources />
  <form-beans>
   <form-bean name="registerForm" type="app.RegisterForm"/>
  </form-beans>
  <global-exceptions />
  <global-forwards />
  <action-mappings>
   <action path="/Register" type="app.RegisterForm" name="registerForm">
   <forward name="success" path="/success.html"/>
   <forward name="failure" path="/failure.html"/>
   </action>
  </action-mappings>
  <message-resources parameter="ly.ApplicationResources" />
</struts-config>
RegisterAction.java代码如下：
package app;
import org.apache.struts.action.*;
import javax.servlet.http.*;
import java.io.*;public class RegisterAction extends Action {
public ActionForward perform (ActionMapping mapping,ActionForm form,HttpServlet request,HttpServlet response) {
RegisterForm rf = (RegisterForm)form;
String username = rf.getUsername();
String password1 = rf.getPassword1();
String password2 = rf.getPassword2();

if(password1.equals(password2)) {

return mapping.findForward("success");

}
return mapping.findForward("failure");
}
}
RegisterForm.java代码如下：
package app;
import org.apache.struts.action.*;
public class RegisterForm extends ActionForm {
String username;
String password1;
String password2;

public RegisterForm (String username,String password1,String password2) {
this.username = username;
this.password1 = password1;
this.password2 = password2;
}

public void setUsername(String username) {
this.username = username;
}

public void setPassword1(String password1) {
this.password1 = password1;
}

public void setPassword2(String password2) {
this.password2 = password2;
}

public String getUsername() {
return this.username;
}

public String getPassword1() {
return this.password1;
}

public String getPassword2() {
return this.password2;
}
}
其中web.xml中部分代码：

<servlet-mapping>
    <servlet-name>action</servlet-name>
    <url-pattern>*.do</url-pattern>
  </servlet-mapping>
  <welcome-file-list>
    <welcome-file>Register.jsp</welcome-file>
  </welcome-file-list>

我在浏览器中输入: http://localhost:8080/TryStruts/试了好几遍，出错如下：
HTTP Status 500 -
________________________________________
type Exception report
message
description The server encountered an internal error () that prevented it from fulfilling this request.
exception
org.apache.jasper.JasperException: Exception in JSP: /Register.jsp:107:
8: <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
9:
10:  <html:form action="/Register.do">
11:    rUsername<html:text property="username"/><br>
12:    enter password<html:password property="password1"/><br>
13:    re-enter password<html:password property="password1"/><br>
Stacktrace:
org.apache.jasper.servlet.JspServletWrapper.handleJspException(JspServletWrapper.java:467)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:371)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:315)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:265)
javax.servlet.http.HttpServlet.service(HttpServlet.java:803)
root cause
javax.servlet.ServletException: Cannot retrieve mapping for action /Register
org.apache.jasper.runtime.PageContextImpl.doHandlePageException(PageContextImpl.java:846)
org.apache.jasper.runtime.PageContextImpl.handlePageException(PageContextImpl.java:779)
org.apache.jsp.Register_jsp._jspService(Register_jsp.java:88)
org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:98)
javax.servlet.http.HttpServlet.service(HttpServlet.java:803)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:328)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:315)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:265)
javax.servlet.http.HttpServlet.service(HttpServlet.java:803)
root cause
javax.servlet.jsp.JspException: Cannot retrieve mapping for action /Register
org.apache.struts.taglib.html.FormTag.lookup(FormTag.java:810)
org.apache.struts.taglib.html.FormTag.doStartTag(FormTag.java:506)
org.apache.jsp.Register_jsp._jspx_meth_html_005fform_005f0(Register_jsp.java:104)
org.apache.jsp.Register_jsp._jspService(Register_jsp.java:78)
org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:98)
javax.servlet.http.HttpServlet.service(HttpServlet.java:803)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:328)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:315)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:265)
javax.servlet.http.HttpServlet.service(HttpServlet.java:803)
note The full stack trace of the root cause is available in the Apache Tomcat/5.5.23 logs.
________________________________________
Apache Tomcat/5.5.23

解决方案 »

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你的JSP页面的错误
提示告诉你了 acation=""里的do去掉因为你在xml里的PATH属性不一样我感觉是这样好久没用struts 不过错误提示告诉你很详细了
是这样的，在你的配置文件里写了：
<action path="/Register" type="app.RegisterForm" name="registerForm">
表示配置了一个action的映射路径为/Register，而你在JSP里写的路径是/Register.do，这样不匹配所以出错了。
楼主的配置是没有问题的，请你确认一下WEB-INF目录下有没有生成的classes文件夹，在其中有没有相应的.class文件？
Struts-config.xml中<action path="/Register" type="app.RegisterAction" name="registerForm">
你写成 type="app.RegisterForm"
<action path="/Register" type="app.RegisterForm" name="registerForm">
这句有问题， type指的是action类的全名，不是form的全名，你好好看看例子吧
改完这行还是不行呀<action path="/Register" type="app.RegisterAction" name="registerForm">
我tomcat现一启动就会出问题,这是怎么回事呢.我是MyEclipse的初学者,原一直用lomboz,请大家帮帮我呀,我特别急!没成就感!!
tomcat报错如下:
2007-7-17 9:39:48 org.apache.struts.action.ActionServlet handleConfigException
严重: Parsing error processing resource path
java.net.UnknownHostException: struts.apache.org
at java.net.PlainSocketImpl.connect(PlainSocketImpl.java:177)
at java.net.Socket.connect(Socket.java:519)
at java.net.Socket.connect(Socket.java:469)
at sun.net.NetworkClient.doConnect(NetworkClient.java:157)
at sun.net.www.http.HttpClient.openServer(HttpClient.java:382)
at sun.net.www.http.HttpClient.openServer(HttpClient.java:509)
at sun.net.www.http.HttpClient.<init>(HttpClient.java:231)
at sun.net.www.http.HttpClient.New(HttpClient.java:304)
at sun.net.www.http.HttpClient.New(HttpClient.java:316)
at sun.net.www.protocol.http.HttpURLConnection.getNewHttpClient(HttpURLConnection.java:813)
at sun.net.www.protocol.http.HttpURLConnection.plainConnect(HttpURLConnection.java:765)
at sun.net.www.protocol.http.HttpURLConnection.connect(HttpURLConnection.java:690)
at sun.net.www.protocol.http.HttpURLConnection.getInputStream(HttpURLConnection.java:934)
at com.sun.org.apache.xerces.internal.impl.XMLEntityManager.setupCurrentEntity(XMLEntityManager.java:973)
at com.sun.org.apache.xerces.internal.impl.XMLEntityManager.startEntity(XMLEntityManager.java:905)
at com.sun.org.apache.xerces.internal.impl.XMLEntityManager.startDTDEntity(XMLEntityManager.java:872)
at com.sun.org.apache.xerces.internal.impl.XMLDTDScannerImpl.setInputSource(XMLDTDScannerImpl.java:282)
at com.sun.org.apache.xerces.internal.impl.XMLDocumentScannerImpl$DTDDispatcher.dispatch(XMLDocumentScannerImpl.java:1021)
at com.sun.org.apache.xerces.internal.impl.XMLDocumentFragmentScannerImpl.scanDocument(XMLDocumentFragmentScannerImpl.java:368)
at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(XML11Configuration.java:834)
at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(XML11Configuration.java:764)
at com.sun.org.apache.xerces.internal.parsers.XMLParser.parse(XMLParser.java:148)
at com.sun.org.apache.xerces.internal.parsers.AbstractSAXParser.parse(AbstractSAXParser.java:1242)
at org.apache.commons.digester.Digester.parse(Digester.java:1548)
at org.apache.struts.action.ActionServlet.parseModuleConfigFile(ActionServlet.java:1006)
at org.apache.struts.action.ActionServlet.initModuleConfig(ActionServlet.java:955)
at org.apache.struts.action.ActionServlet.init(ActionServlet.java:470)
at javax.servlet.GenericServlet.init(GenericServlet.java:212)
at org.apache.catalina.core.StandardWrapper.loadServlet(StandardWrapper.java:1139)
好像是说你的配置文件有问题，无法解析，你仔细检查一下web.xml和struts-config.xml，另外你的form类里面没有reset和validate方法，不知道有没有影响，我记得，这两个方法好像是必需的。
enter password<html:password property="password1"/><br/>
   re-enter password<html:password property="password1"/><br/>
  这个地方是不是设计的有问题？
对啦,还想问一下MyEclipse中在tomcat刚刚启动时,它就会查找程序中的错误吗,可是我还没有:
http://localhost:8080/呢呀,是这样的吧?是Deploy的原因吗?Deploy到底有什么用呢
你的程序要放到服务起目录中才能运行,myeclipse的deploy是把的程序放到服务器下,不用手动的配置拉