Struts 2 + urlrewrite 的问题，急，急，急！

最近项目中要用到urlrewrite的技术，可是我按照网上找的资料配置，
当我直接访问 http://localhost:8080/Mysite/html 时确实可以直接链接到 admin/目录下的index.jsp文件，该文件是用户输入用户名密码的表单，但是就在登录的时候即提交表单时，因为要跳转至一个action，结果就不成功，我就郁闷了，各位大哥帮帮忙看看那里的问题……

web.xml代码如下：

<!-- URL Rewrite Filter -->

  <filter>

  	<filter-name>UrlRewriteFilter</filter-name>

  	<filter-class>org.tuckey.web.filters.urlrewrite.UrlRewriteFilter</filter-class>

	<init-param>

		<param-name>logLevel</param-name>

		<param-value>WARN</param-value>

	</init-param>

  </filter>

  <filter-mapping>

  	<filter-name>UrlRewriteFilter</filter-name>

  	<url-pattern>/*</url-pattern>

  </filter-mapping>

  

  <!-- Struts 2 Filter  -->

  <filter>

  	<filter-name>struts2</filter-name>

  	<filter-class>org.apache.struts2.dispatcher.FilterDispatcher</filter-class>

  </filter>

  <filter-mapping>

  	<filter-name>struts2</filter-name>

  	<url-pattern>/*</url-pattern>

  </filter-mapping>



<welcome-file-list>

    <welcome-file>index.jsp</welcome-file>

</welcome-file-list>

WEB-INF/urlrewrite.xml代码如下：

<urlrewrite>



	<rule>

		<note>

			The rule means that requests to /test/status/ will be redirected to

			/rewrite-status

			the url will be rewritten.

        </note>

		<from>/test/status/</from>

		<to type="redirect">%{context-path}/rewrite-status</to>

	</rule>



	<outbound-rule>

        <note>

            The outbound-rule specifies that when response.encodeURL is called (if you are using JSTL c:url)

            the url /rewrite-status will be rewritten to /test/status/.



            The above rule and this outbound-rule means that end users should never see the

            url /rewrite-status only /test/status/ both in thier location bar and in hyperlinks

            in your pages.

        </note>

        <from>/rewrite-status</from>

        <to>/test/status/</to>

    </outbound-rule>

	

	<rule>

		<from>^/html</from>  

		<to>/admin/</to>  

	</rule>

	

	<rule>

		<from>^/html/user/login/index.html</from>  

		<to>/admin/userInfo_userLogin.action</to>  

	</rule>



</urlrewrite>

admin/index.jsp 代码如下：

<!--

原本应该是：

<form name="loginForm" onsubmit="return checkLogin()" action="userInfo_userLogin.action" method="post">



 -->

<form name="loginForm" onsubmit="return checkLogin()" action="user/login/index.html" method="post">

		    <table width="100%" border="0" cellspacing="0" cellpadding="0">

				<tr>

		        <td width="31%" height="35" class="login-text02">用户名：<br /></td>

		        <td width="69%"><input name="loginName" id="loginName" type="text" size="29" value="admin"  style="width:200px;"/></td>

		      </tr>

		      <tr>

		        <td height="35" class="login-text02">密  码：<br /></td>

		        <td><input name="passWord" id="passWord" type="password" size="30" value="123456" style="width:200px;"/></td>

		      </tr>

		      <tr>

		        <td height="35"> </td>

		        <td><input type="submit" class="right-button01" value="确认登陆"/>      

		          <input type="reset" class="right-button02" value="重 置" /></td>

		      </tr>

		      <tr>

		        <td height="35" colspan=2 align="center">            

		        <span style="color:red;font-size:14px;font-weight:bold;" id="errorMsg">${loginMsg} <% session.removeAttribute("loginMsg"); %> </span></td>

		      </tr>

		     

		    </table>

    </form>

原本访问 http://localhost:8080/MySite/admin 直接进入登录表单录入界面，在提交表单后，正确跳转后url地址为：
http://localhost:8080/MySite/admin/userInfo_userLogin.action 进入跳转后的页面。

而现在访问 http://localhost:8080/MySite/html 也可以同时进入登录表单录入界面，但是在提交表单后，url地址为：
http://localhost:8080/MySite/user/login/index.html 同时页面报404错误，
说找不到 HTTP Status 404 - /MySite/user/login/index.html

究竟是什么问题，拜托各位大哥帮帮忙解决一哈，谢谢了！！！！