Re: The original name of the uploaded file

  •  06-01-2011, 10:34 AM

    Re: The original name of the uploaded file

    Dear illaki,
     
    Please test the following snippet, it returns the original file name:
     

    <%@ Language="VBScript" %>
    <!-- #include file="aspuploader/include_aspuploader.asp" -->
    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
    <html xmlns="http://www.w3.org/1999/xhtml">
    <head>
     <title>
      Form - Multiple files upload
     </title>
     <link href="demo.css" rel="stylesheet" type="text/css" />
     
     <script type="text/javascript">
     function CuteWebUI_AjaxUploader_OnPostback() {
      //submit the form after the file have been uploaded:
      document.forms[0].submit();
     }
     </script>
    </head>
    <body>
     <div class="demo">                       
            <h2>Selecting multiple files for upload</h2>
      <p>ASP Uploader allows you to select multiple files and upload multiple files at once.</p>
      
       <!-- do not need enctype="multipart/form-data" -->
       <form id="form1" method="POST">
       <%
       Dim uploader
       Set uploader=new AspUploader
       uploader.MaxSizeKB=10240
       uploader.Name="myuploader"
       uploader.InsertText="Upload File (Max 10M)"
       uploader.MultipleFilesUpload=true
       %>
       <%=uploader.GetString() %>
       </form>
       
       <br/><br/>
    <%

    If Request.Form("myuploader")&""<>"" Then

     Dim list,i
     
     'Gets the GUID List of the files based on uploader name
     list=Split(Request.Form("myuploader"),"/")

     For i=0 to Ubound(list)
      if i>0 then
       Response.Write("<hr/>")
      end if
      Dim mvcfile
      
      'get the uploaded file based on GUID
      Set mvcfile=uploader.GetUploadedFile(list(i))

      Response.Write("<div style='font-family:Fixedsys'>")
      Response.Write("Uploaded File:")
       'Gets the name of the file.
      Response.Write("<br/>FileName: ")
      Response.Write(mvcfile.FileName)
      'Gets the size of the file.
      Response.Write("<br/>FileSize: ")
      Response.Write(mvcfile.FileSize)
      'Gets the temp file path.
      Response.Write("<br/>FilePath: ")
      Response.Write(mvcfile.FilePath)
      Response.Write("</div>")
      'Copys the uploaded file to a new location.   
            'mvcfile.CopyTo("/uploads")           
            'Moves the uploaded file to a new location.   
            'mvcfile.MoveTo("/uploads")  

     Next
    End If
    %> 
      
      <script type='text/javascript'>
     function CuteWebUI_AjaxUploader_OnTaskComplete(task)
     {
      alert(task.FileName + " is uploaded!");
     }
     </script> 
     </div>
    </body>
    </html>

    Thanks for asking
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