how to view and download particular file from db in asp.net

Question

i want to view and download particular file whish is stored in database in asp.net

Arul R · Answer

Paul Raj, Kindly refer the below links, Hope this will solve your problem. http://www.aspsnippets.com/Articles/Upload-and-Download-files-from-SQL-Server-Database-in-ASPNet.aspx http://www.codeproject.com/Articles/308552/Upload-and-Download-Files-to-SQL-Servers-in-ASP-Ne http://www.codeproject.com/Questions/380226/How-to-download-upload-particular-file-in-ASP-NET Let me know for any thing more.... If your problem solved means kindly Accept my answer

Paul Rajs · Answer

Thanks a lot for all of your reply, i can able to accept only one answer, So i can't able accept all of your answers. Paul.S

Ankur Mistry · Answer

Hi, Check : http://www.aspsnippets.com/Articles/Upload-and-Download-files-from-SQL-Server-Database-in-ASPNet.aspx

Rajeesh Menoth · Answer

Hi, Check the following reference : Video Tutorial : https://www.youtube.com/watch?v=-fkdrgFiaeI Other Reference : http://www.aspsnippets.com/Articles/Upload-and-Download-files-from-SQL-Server-Database-in-ASPNet.aspx http://www.aspdotnet-suresh.com/2013/01/aspnet-download-and-upload-files-to.html http://www.aspsnippets.com/Articles/Retrieve-and-display-PDF-Files-from-database-in-browser-in-ASPNet.aspx

Sandy Surwase · Answer

string binaryImageQuery = String.Format(@" ",_customEntityName.ToLower()); EntityCollection binaryImageResults = _serviceProxy.RetrieveMultiple(new FetchExpression(binaryImageQuery)); Console.WriteLine("Records retrieved and image files saved to: {0}", Directory.GetCurrentDirectory()); foreach (Entity record in binaryImageResults.Entities) { String recordName = record["sample_name"] as String; String downloadedFileName = String.Format("Downloaded_{0}", recordName); byte[] imageBytes = record["entityimage"] as byte[]; var fs = new BinaryWriter(new FileStream(downloadedFileName, FileMode.Append, FileAccess.Write)); fs.Write(imageBytes); fs.Close(); Console.WriteLine(downloadedFileName); a ccept the answer if its helpful for you