Эй, попытка чтения потока в управление изображением может помочь, у меня есть некоторые ошибки на моей странице обработчика.
protected void GridView1_SelectedIndexChanged(object sender, EventArgs e)
{
string PhotoPath;
GridViewRow row = GridView1.Rows[GridView1.SelectedIndex];
PhotoPath = row.Cells[5].Text;
PhotoPath = HttpUtility.UrlEncode(PhotoPath);
HttpWebRequest request = (HttpWebRequest)
WebRequest.Create(PhotoPath);
HttpWebResponse response = (HttpWebResponse)
request.GetResponse();
Stream resStream = response.GetResponseStream();
// want to output to image control upon selection of gridview
//using (System.Drawing.Image img = System.Drawing.Image.FromStream(resStream))
//{
// img.Save("temp.jpg", ImageFormat.Jpeg);
//}
}
}
Я получаю сообщение об ошибке.Имя параметра: uriString
public class GetImage : IHttpHandler
{
public void ProcessRequest(HttpContext context)
{
{
string PhotoPath = System.Web.HttpContext.Current.Request.QueryString["PhotoPath"];
PhotoPath = HttpUtility.UrlDecode(PhotoPath);
FtpWebRequest request = (FtpWebRequest)FtpWebRequest.Create(new Uri(PhotoPath));
// error is here -------------------
request.Method = WebRequestMethods.Ftp.DownloadFile;
request.Credentials = new NetworkCredential("Administrator", "commando");
try
{
FtpWebResponse response = (FtpWebResponse)request.GetResponse();
Stream stream = response.GetResponseStream();
byte[] bytes = new byte[2048];
int i = 0;
MemoryStream mStream = new MemoryStream();
do
{
i = stream.Read(bytes, 0, bytes.Length);
mStream.Write(bytes, 0, i);
} while (i != 0);
context.Response.Clear();
context.Response.ClearHeaders();
context.Response.ClearContent();
context.Response.ContentType = "image/jpeg";
context.Response.BinaryWrite(mStream.GetBuffer());
}
catch (WebException wex)
{
//throw new Exception("Unable to locate or access your file.\\nPlease try a different file.");
}
catch (Exception ex)
{
throw new Exception("An error occurred: " + ex);
}
}
}
public bool IsReusable
{
get
{
return false;
}
}
}