ASP.NET,AJAX文件上传的疑问。

ASP.NET,AJAX文件上传的疑问

原本以为用这种方式，把此方法放到一个Httphandler中，然后从JAVASCRIPT访问Httphandler来实现无刷新的文件上传。没想到，这样同样是不行的，只能在本机测试通过，如果放到网上，或者从其他地方访问就会提示无法找到文件，可见在ASP.NET中它只能用于上传服务器端的文件。 

之前在网站也搜索到一篇文章是写怎么在Httphandler中来实现文件上传的，是国外的一个BLOG，没怎么看明白，貌似还借助于一个applet。后来由于没资源我也没去试过那方法到底是否可行了。

我现在还真的不知道在ASP.NET怎么能够纯手工的实现AJAX的无刷新上传文件，因为从JavaScript从提交表单访问Httphandler，在里面根本无法获取到表单的信息,HttpFileCollection,HttpPostedFile,里面都是空的。我觉得ASP.NET中的Httphandler与JSP中的Serverlet到有一些相似之处,可是在JSP中用Serverlet来实现一个AJAX无刷新上传很容易，现在真的有点困惑了，难道ASP.NET要实现无刷新的上传一定得要用UpdatePanel。都不知道.NET到什么都封装到控件，我们只能无脑的拖出来使用，就不能像JSP一样，多让我们程序员按照自己的思路去实现一个东西。

 

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private void Uploadfile(string filepath)
        {
           //string filepath = Request["filepath"].ToString();
            Response.ContentType = "text/plain";
            if (filepath == "" || filepath == null)
            {
                return;
            }
            else
            {
                FileStream fs=null;
                BinaryReader br=null;
                Stream postStream=null;
                //use a try-catch to catch an exception if happened
                try
                {
                    //Get a filestream form the file
                    fs = new FileStream(filepath, FileMode.Open, FileAccess.Read);
                    //Create a binaryreader object for read the filestream which you can deliver an parmetter filestream
                    br = new BinaryReader(fs);
                    //Create a webclient for send data or receive data from uri
                    WebClient myWebClient = new WebClient();
                    //Get the file name from parm filepath
                    string filename = filepath.Substring(filepath.LastIndexOf("//") + 1);
                    //Get server path
                    string tofilepath = Server.MapPath("~/uploadFiles/" + filename);
                    //Get the windows safety credential
                    myWebClient.Credentials = CredentialCache.DefaultCredentials;

                    byte[] postArr = br.ReadBytes((int)fs.Length);
                    postStream = myWebClient.OpenWrite(tofilepath, "PUT");
                    if (postStream.CanWrite)
                    {
                        postStream.Write(postArr, 0, postArr.Length);
                        Response.Write("File upload successfully !");
                    }
                    else
                    {
                        Response.Write("File can not be writed !");
                    } 
                }
                catch (Exception ex)
                {
                    Response.Write("File upload failed the reason is :" + ex.Message);
                }
                finally
                {
                    if(postStream!=null)
                        postStream.Close();
                    if(br!=null)
                        br.Close();
                    if(fs!=null)
                        fs.Close();
                    Response.End();
                }
            }
        }