求助，急！！如何在同一个Form中点击一次提交就可以把普通的input type="text"和type="file"的值都提交

Question

问题如题。
我的代码如下:
提交数据的页面upload.jsp,我使用的是Struts框架
<html:form action="/upLoad" enctype="multipart/form-data">
<input name="file" id="file1" type="file">
<input type="text" name="year" >
<html:submit/><html:cancel/>
</html:form>接受处理的类为：
public ActionForward execute(
ActionMapping mapping,
ActionForm form,
HttpServletRequest request,
HttpServletResponse response) throws Exception {
.....
.....
}急啊，大虾救命......

Answer 1

我查了很多网页，都说用SmartUpload，但是我调不通代码。
接受处理的Action为：
public class UpLoadAction extends Action {
public ActionForward execute(
ActionMapping mapping,
ActionForm form,
HttpServletRequest request,
HttpServletResponse response) throws Exception {
UpLoadForm upLoadForm = (UpLoadForm) form;
ServletNewList snl = new ServletNewList();
snl.doPost(request,response);
                  return null;
         }
}另一个类为：
public class ServletNewList extends HttpServlet{
public void doPost(HttpServletRequest request,
HttpServletResponse response) 
throws ServletException, IOException{
    SmartUpload temp=new SmartUpload();
    //必须先初始化！！
    servletUpload su = new servletUpload();
temp.initialize(su.getServletConfig(),request,response);
String str;
PrintWriter out=response.getWriter();
try 
{
temp.upload() ;
    //str--获得文本框txt中的值
String tempStr=temp.getRequest().getParameter("T_wfsj_yyyy");
request.setCharacterEncoding("gbk");
str=new String(tempStr.getBytes("gbk"),"ISO8859-1");
System.out.println("T_wfsj_yyyy is " + str);
com.jspsmart.upload.File myFile = temp.getFiles().getFile(0);
//str2--获得fImage中指定的上传文件的路径
String str2 = myFile.getFilePathName();
System.out.println("path: " + str2);
} 
catch (SmartUploadException e) 
{
e.printStackTrace();
}
catch (IOException e)
{
e.printStackTrace();
} 
catch (ServletException e) 
{
e.printStackTrace();
}

}
在jspSmartUpload.jar包中各个类都在com.jspsmart.upload包中，包括File.class,Files.class,Request.class,serletUpload.class,SmartUpload.class,
SmartUploadException.class报错为：
2006-4-11 11:01:57 org.apache.catalina.core.StandardWrapperValve invoke
严重: Servlet.service() for servlet action threw exception
java.lang.NoClassDefFoundError: servletUpload
at com.spacen.utilities.ServletNewList.doPost(ServletNewList.java:24)
at com.spacen.struts.action.UpLoadAction.execute(UpLoadAction.java:60)
at org.apache.struts.action.RequestProcessor.processActionPerform(RequestProcessor.java:484)
at org.apache.struts.action.RequestProcessor.process(RequestProcessor.java:274)
at org.apache.struts.action.ActionServlet.process(ActionServlet.java:1482)
at org.apache.struts.action.ActionServlet.doPost(ActionServlet.java:525)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:709)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:802)
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:252)
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:173)
at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:214)
at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:178)
at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:126)
at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:105)
at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:107)
at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:148)
at org.apache.coyote.http11.Http11Processor.process(Http11Processor.java:825)
at org.apache.coyote.http11.Http11Protocol$Http11ConnectionHandler.processConnection(Http11Protocol.java:738)
at org.apache.tomcat.util.net.PoolTcpEndpoint.processSocket(PoolTcpEndpoint.java:526)
at org.apache.tomcat.util.net.LeaderFollowerWorkerThread.runIt(LeaderFollowerWorkerThread.java:80)
at org.apache.tomcat.util.threads.ThreadPool$ControlRunnable.run(ThreadPool.java:684)
at java.lang.Thread.run(Thread.java:595)

Answer 2

使用smartupdate好像可以，他本身提供了接收的方法：su.getRequest.getParameter("name");你去查查，我记不清楚了