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Home/ Questions/Q 135047
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Asked: May 11, 20262026-05-11T06:43:24+00:00 2026-05-11T06:43:24+00:00

What i want to do is that the user selects some fields on a

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What i want to do is that the user selects some fields on a grid and according to these datas i create an xml file on the web server and then i want user to download it like downloading a any file. But the problem is, i cant use this code:

Response.ContentType = 'APPLICATION/OCTET-STREAM';          // initialize the http content-disposition header to         // indicate a file attachment with the default filename         // 'myFile.txt'         System.String disHeader = 'Attachment; Filename=\'' + fileName +            '\'';         Response.AppendHeader('Content-Disposition', disHeader);         FileInfo downloadFile = new FileInfo(fileFullName);         if (downloadFile.Exists)         {             Response.WriteFile(downloadFile.FullName);             HttpContext.Current.ApplicationInstance.CompleteRequest();         } 

Because i need to let the user download 3 files so the header cannot cary it, what i tought was getting the file names and open a popup, list the file names with a linkbutton, and then the user can download it.

For each file i create a linkbutton at runtime and add this code :

lnkProblem.Text = 'Problemler dosyası';     lnkProblem.Visible = true;     lnkProblem.Command += new CommandEventHandler(lnkUser_Command);     lnkProblem.CommandName = Request.QueryString['fileNameProblems'];     lnkProblem.CommandArgument = Request.QueryString['fileNameProblems']; 

Then use this function to make it download by the user :

void lnkUser_Command(object sender, CommandEventArgs e) {     Response.ContentType = 'APPLICATION/XML';      System.String disHeader = 'Attachment; Filename=\'' + e.CommandArgument.ToString() +        '\'';     Response.AppendHeader('Content-Disposition', disHeader);     FileInfo downloadFile = new FileInfo(Server.MapPath('.') + '\\xmls\\' + e.CommandArgument.ToString());     if (downloadFile.Exists)     {         Response.WriteFile(Server.MapPath('.') + '\\xmls\\' + e.CommandArgument.ToString());         HttpContext.Current.ApplicationInstance.CompleteRequest();     }     HttpContext.Current.ApplicationInstance.CompleteRequest(); } 

Application creates the xml file but somewhere of it, app puts the html tags in that xml file so i can’t open the file, is there anyway to do this? Maybe any other example…

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1 Answer

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  1. 2026-05-11T06:43:25+00:00Added an answer on May 11, 2026 at 6:43 am

    The cleaneast way to send a file to the client is to use the TransmitFile method like this:

    FileInfo file = new FileInfo(filePath); // full file path on disk Response.ClearContent(); // neded to clear previous (if any) written content Response.AddHeader('Content-Disposition',      'attachment; filename=' + file.Name); Response.AddHeader('Content-Length', file.Length.ToString()); Response.ContentType = 'text/xml'; //RFC 3023 Response.TransmitFile(file.FullName); Response.End(); 

    For multiple files a common solution is to pack all files in a zip file and send them (the mime type would be application/zip in this case).

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