单击按钮从AjaxFileUpload Control获取文件名

Question

请帮助解决这些问题;

我在Itemtemplate中有一个带有AjaxFileupload的gridview。在每行中使用AjaxFileUpload填充gridview并上载文件之后。我需要通读gridview并在bujtton点击获取文件名。请在下面找到代码

<div>
<asp:GridView ID="grdView_Requirement" runat="server" AutoGenerateColumns="False" Font-Names="Segoe UI" Font-Size="12px">
   <Columns>
      <asp:BoundField DataField="Name" HeaderText="Name" SortExpression="Name" />
           <asp:TemplateField HeaderText="Upload File">
               <ItemTemplate>
                 <asp:AjaxFileUpload ID="AjaxFileUploadNew" runat="server" Font-Names="Segoe    
                   UI" Font-Size="12px" MaximumNumberOfFiles="1" Width="388px" />
               </ItemTemplate>
           </asp:TemplateField>
           <asp:TemplateField HeaderText="">
   </Columns>
  <FooterStyle BackColor="#5D7B9D" ForeColor="#333333" />
  <HeaderStyle BackColor="#D7D9E5" ForeColor="#333333" />
  <RowStyle BackColor="#FFFFFF" ForeColor="#333333" />
  </asp:GridView>
</div>

   <asp:ToolkitScriptManager ID="ToolkitScriptManager1" runat="server">
</asp:ToolkitScriptManager>
<asp:Button ID="Button1" runat="server" Text="Button" OnClick="Button1_Click" />

上传代码

protected void AjaxFileUploadNew_UploadComplete(object sender, AjaxControlToolkit.AjaxFileUploadEventArgs e)
    {
       string strLongFilePath = e.FileName;
       string strFileExtension = System.IO.Path.GetExtension(strLongFilePath);
       char[] ch = new char[] { '\\' };
       string[] strPath = strLongFilePath.Split(ch);
       string strInternalFileName = DateTime.Now.ToFileTime().ToString() + strFileExtension;
       string strDestPath = Server.MapPath("~/UploaderTemp/");
       AjaxFileUploadNew.SaveAs(@strDestPath + strInternalFileName);
       //Save path, filename to database here
     }

请帮助在下面的事件中获取上传的文件名

 protected void Button1_Click(object sender, EventArgs e)
{
  foreach (GridViewRow row in grdView_Requirement.Rows)
    {
      //for bound field
      string str = row.Cells[0].Text.ToString();         
      //for templated control
      AjaxControlToolkit.AjaxFileUpload tb = row.FindControl("AjaxFileUploadNew") as   
      AjaxControlToolkit.AjaxFileUpload;
    }
 }

请帮助解决或提供更好的实施方法，以取得成功。