下面给定的代码我想在xml文件中打开 [英] below given code i want to open in xml file

问题描述

我要在上传文件时在xml文件中打开
请任何人发送代码

i want open in xml file when i upload the file
plz any one send the code

using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
using System.Web.UI;
using System.Web.UI.WebControls;
using System.IO;
using System.Xml;

public partial class Default2 : System.Web.UI.Page
{
    protected void Page_Load(object sender, EventArgs e)
    {

    }
    protected void Button1_Click(object sender, EventArgs e)
    {
        if (FileUploadControl.HasFile)
        {
            try
            {
                if (FileUploadControl.PostedFile.ContentType == "My Documents")
                {
                    if (FileUploadControl.PostedFile.ContentLength < 102400)
                    {
                        string filename = Path.GetFileName(FileUploadControl.FileName);
                        FileUploadControl.SaveAs(Server.MapPath("C:\\Documents and Settings\\azhar\\My Documents\\flashrecorder.xml") + filename);
                        Label1.Text = "Upload status: File uploaded!";
                    }
                    else
                        Label1.Text = "Upload status: The file has to be less than 100 kb!";
                }
                else
                    Label1.Text = "Upload status: Files are accepted!";
            }
            catch (Exception ex)
            {
                Label1.Text = "Upload status: The file could not be uploaded. The following error occured: " + ex.Message;
            }
        }
    }
}

推荐答案

if (FileUploadControl.HasFile)
       {
           try
           {
               if (FileUploadControl.PostedFile.ContentType == "My Documents")
               {
                   if (FileUploadControl.PostedFile.ContentLength < 102400)
                   {
                       string filename = Path.GetFileName(FileUploadControl.FileName);
                       FileUploadControl.SaveAs(Server.MapPath("C:\\Documents and Settings\\azhar\\My Documents\\flashrecorder.xml") + filename);
                       Label1.Text = "Upload status: File uploaded!";

Response.Redirect(Server.MapPath("C:\\Documents and Settings\\azhar\\My Documents\\flashrecorder.xml
           "        }
                   else
                       Label1.Text = "Upload status: The file has to be less than 100 kb!";
               }
               else
                   Label1.Text = "Upload status: Files are accepted!";
           }
           catch (Exception ex)
           {
               Label1.Text = "Upload status: The file could not be uploaded. The following error occured: " + ex.Message;
           }
       }

这篇关于下面给定的代码我想在xml文件中打开的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！