RadGridAttachmentColumn에서 바이너리 데이터를 다운로드하는 방법은 무엇입니까?

Question

내 코드

protected void grdFiles_ItemCommand(object source, GridCommandEventArgs e) 
     { 
      if (e.CommandName == RadGrid.DownloadAttachmentCommandName) 
      { 

       GridDownloadAttachmentCommandEventArgs args = e as GridDownloadAttachmentCommandEventArgs; 
       string fileName = args.FileName; 
       int attachmentId = (int)args.AttachmentKeyValues["ProjectFileId"]; 

       ProTrakEntities objEntity = new ProTrakEntities(); 
       ProjectFile objFile = (from type in objEntity.ProjectFiles where type.ProjectFileId == attachmentId select type).First(); 
       string filename = objFile.FileName; 
       string Filetype = objFile.FileType; 
       byte[] binaryData = (byte[])objFile.FileData; 
       //byte[] binaryData = (byte[])data.Tables[0].Rows[0]["BinaryData"]; 
       // Response.AppendHeader("Content-Disposition", "attachment; filename = "+filename); 
       // Response.AppendHeader("Content-Length", filename.Length.ToString()); 
       //Response.ContentType = Filetype; 
       //Response.WriteFile(Server.MapPath("Uploads/"+filename)); 
       //Response.Flush(); 
       //Response.Close(); 
       //Response.End(); 


       // grdFiles.Items[0].FireCommandEvent(RadGrid.DownloadAttachmentCommandName, parameters); 
       Response.Clear(); 
       Response.ContentType = Filetype; 
       Response.AddHeader("content-disposition", "attachment; filename=" + fileName); 
       Response.BinaryWrite(binaryData); 
       Response.End(); 

     } 
    } 

내가 서버 측에서 다운로드 할 때 작동하지 않는 지역 고객과의 면담에서 다운로드 할 수 Database.I에 바이너리 데이터로 변환 및 저장 업로드 된 데이터를 디버깅.

Answer 1

protected void grdFiles_ItemCommand(object source, GridCommandEventArgs e) 
{ 

    if (e.CommandName == RadGrid.DownloadAttachmentCommandName) 
    { 
     RadAjaxManager Manager = new RadAjaxManager(); 
     Manager.EnableAJAX = false; 
     GridDownloadAttachmentCommandEventArgs args = e as GridDownloadAttachmentCommandEventArgs; 
     string fileName = args.FileName; 
     int attachmentId = (int)args.AttachmentKeyValues["ProjectFileId"]; 

     ProTrakEntities objEntity = new ProTrakEntities(); 
     ProjectFile objFile = (from type in objEntity.ProjectFiles where type.ProjectFileId == attachmentId select type).First(); 
     byte[] binaryData = (byte[])objFile.FileData; 
     //grdFiles.DataSource = objFile; 
     Response.Clear(); 
     Response.ContentType = "application/octet-stream"; 
     Response.AddHeader("content-disposition", "attachment; filename=" + fileName); 
     Response.BinaryWrite(binaryData); 
     Response.Flush(); 
     Response.Close(); 
     Response.End(); 



    } 
} 

고유 이름이 고유해야합니다.