方法1：
直接用
xxx=upload.form("imgname")
其他方法：
form里的值不能Request，但是querystring可以，可以试试这样
<script language="javascript">
function uploadData(){
frm1.action+='&path='+frm1.sPath.value;
//alert(frm1.action);
frm1.submit();
}
</script>
<%
if Request("action") = "upload" then
sPath = Request("path")
call UploadFunction()
end if
%>
<form name="frm1" action="testup.asp?action=upload" method="post" enctype="multipart/form-data">
<input type="text" name="sPath"><br>
<input type="file" name="txtFile" class="topinput" id="txtFile"><br>
<input type="button" value="上传文件" class="topinput" onclick="uploadData();">
</form>
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另外，aspupload里有个move方法，不知道是否适用
Set Upload = Server.CreateObject("Persits.Upload.1")
Upload.Save (myPath)
For Each FileObj In Upload.Files
filename = FileObj.Path
Next
Set f1 = fso.GetFile(filename)
For Each Item In Upload.Form
arr(i) = Item.Value
arr(i) = Replace(arr(i)‚ "'"‚ "''")
i = i + 1
Next
f1.Move (myPath & "\" & arr(0) & "\ExcelToSTOCK.xls")
