How to save FormData into sql - sql-server

Question

I need to store the path of the file along with its properties into a sql database. Currently all I can do is store the file on the server. I can see the file properties in the controller but I do not know how to access them.
public class File
{
public int FileId { get; set; }
public string FileType { get; set; }
public string FileDate { get; set; }
public string FilePdf { get; set; }
public string FileLocation { get; set; }
public string FilePlant { get; set; }
public string FileTerm { get; set; }
public DateTime? FileUploadDate { get; set; }
public string FileUploadedBy { get; set; }
public string CompanyName { get; set; }
public virtual ApplicationUser User { get; set; }
}
Controller
[HttpPost]
public async Task<HttpResponseMessage> PostFile()
{
if (!Request.Content.IsMimeMultipartContent())
{
this.Request.CreateResponse(HttpStatusCode.UnsupportedMediaType);
}
string root = HttpContext.Current.Server.MapPath("~/App_Data");
var provider = new MultipartFormDataStreamProvider(root);
var result = await Request.Content.ReadAsMultipartAsync(provider);
foreach (var key in provider.FormData.AllKeys)
{
foreach (var val in result.FormData.GetValues("companyname")
.FirstOrDefault())
{
if (key == "companyName")
{
var companyName = val;
var fileDate = val;
var fileLocation = val;
var filePlant = val;
var fileTerm = val;
var fileType = val;
var fileUploadDate = val;
var fileUploadedBy = val;
}
}
}
// On upload, files are given a generic name like "BodyPart_26d6abe1-3ae1-416a-9429-b35f15e6e5d5"
// so this is how you can get the original file name
var originalFileName = GetDeserializedFileName(result.FileData.First());
var uploadedFileInfo = new FileInfo(result.FileData.First().LocalFileName);
string path = result.FileData.First().LocalFileName;
//Do whatever you want to do with your file here
//db.Files.Add();
db.SaveChanges();
return this.Request.CreateResponse(HttpStatusCode.OK, originalFileName);
}
private string GetDeserializedFileName(MultipartFileData fileData)
{
var fileName = GetFileName(fileData);
return JsonConvert.DeserializeObject(fileName).ToString();
}
public string GetFileName(MultipartFileData fileData)
{
return fileData.Headers.ContentDisposition.FileName;
}

Answer 1

Try using a function like this. You can replace with
private object GetFormData<T>(MultipartFormDataStreamProvider result)
{
if (result.FormData.HasKeys())
{
var unescapedFormData = Uri.UnescapeDataString(result.FormData
.GetValues(0).FirstOrDefault() ?? String.Empty);
if (!String.IsNullOrEmpty(unescapedFormData))
return JsonConvert.DeserializeObject<T>(unescapedFormData);
}
return null;
}
Use it like this
File file = GetFormData(result);
The main line of code you want is:
JsonConvert.DeserializeObject<File>(result.FormData.GetValues(0).FirstOrDefault());