怎么将图片上传到服务器上

如何将图片上传到服务器上？
我想要一个能将图片上传到服务器上的一个demo 那位达人能发一个 谢了！！！

[解决办法]
用html的file控件
[解决办法]

HTML code

 <asp:FileUpload ID="FileUpload1" runat="server" /><asp:Button ID="Button1" runat="server" OnClick="Button1_Click" Text="Button" />[code=C#]
[解决办法]
我用的是上传 组件
[解决办法]
参考下面两种情况,一是存入Database,另一个是存Folder:

http://blog.csdn.net/insus/archive/2008/01/20/2055033.aspx

http://blog.csdn.net/insus/archive/2008/01/20/2055030.aspx
[解决办法]
http://www.svnhost.cn/Article/Detail-45.shtml
http://www.svnhost.cn/Article/Detail-62.shtml
这些都是例子
[解决办法]
C# code
 protected void fUpload_Click(object sender, EventArgs e)    {        Boolean fileOk = false;        String path = Server.MapPath("~/");        if (FileUpload.HasFile)        {            String fileExtension =                 System.IO.Path.GetExtension(FileUpload.FileName).ToLower();            String[] allowExtension ={ ".gif", ".jpg", ".jpeg", "png" };            for (int i = 0; i < allowExtension.Length; i++)            {                if (fileExtension==allowExtension[i])                {                    fileOk = true;                }            }        }        if (fileOk)        {            try            {                FileUpload.PostedFile.SaveAs(path + FileUpload.FileName);                Label6.Text = "文件上传成功";            }            catch (System.Exception ex)            {                Label6.Text = "文件上传失败";            }        }        else        {            Label6.Text = "只能上传jpg,png图片";            Label6.BackColor = Color.Red;        }    }
[解决办法]
C# code
  /// <summary>    /// 上传图片    /// </summary>    /// <param name="sender"></param>    /// <param name="e"></param>    protected void btnUpLoad_Click(object sender, EventArgs e)    {        try        {            string test = Server.MapPath("ImageFiles/" + DateTime.Now.ToString("yyyy-MM-dd"));  //用来生成文件夹            if (!Directory.Exists(test))            {                Directory.CreateDirectory(test);            }            if (FileUpload1.PostedFile.FileName != "")            {               string  imgname = FileUpload1.PostedFile.FileName;               string imgType = imgname.Substring(imgname.LastIndexOf(".") + 1);               string quanname =DateTime.Now.ToString("yyyyMMddHHmmss")+imgname.LastIndexOf("\\")+"."+imgType ;               if ("gif" != imgType && "jpg" != imgType && "GIF" != imgType && "JPG" != imgType)                {                    Response.Write("<script>alert('请选择gif,jpg格式的文件!');</script>");                    return;                }                         string imgurl = "ImageFiles/" + DateTime.Now.ToString("yyyy-MM-dd") + "/" + quanname;                 FileUpload1.PostedFile.SaveAs(Server.MapPath(imgurl));}}//以下就执行更新                     sqlt.Operate("update dbo.NewsAdd set imgurl='" + [color=#FF0000]imgurl[/color] + "' where  newsid ='" + dtNewsID.Rows[0]["newsid"].ToString() + "'"); 
 
[解决办法]
嗯嗯,代码够多的了
呵呵,感谢CSDn的热心人士
filename = filename.Substring(filename.LastIndexOf('.'), filename.Length - filename.LastIndexOf('.'));//文件名后缀  
           filename = DateTime.Now.ToString("yyyyMMddmmss") + ran.Next(100, 999) + filename;//重命名  
我觉得这个方法够好了
解决了获取文件名的又一个方法
还生成随机数呢