Convert HttpPostedFileBase to byte[]

In my MVC application, I am using following code to upload a file.

MODEL

 public HttpPostedFileBase File { get; set; }

VIEW

@Html.TextBoxFor(m => m.File, new { type = "file" })

Everything working fine .. But I am trying to convert the result fiel to byte[] .How can i do this

CONTROLLER

 public ActionResult ManagePhotos(ManagePhotos model)
    {
        if (ModelState.IsValid)
        {
            byte[] image = model.File; //Its not working .How can convert this to byte array
        }
     }

As Darin says, you can read from the input stream - but I'd avoid relying on all the data being available in a single go. If you're using .NET 4 this is simple:

MemoryStream target = new MemoryStream();
model.File.InputStream.CopyTo(target);
byte[] data = target.ToArray();

It's easy enough to write the equivalent of CopyTo in .NET 3.5 if you want. The important part is that you read from HttpPostedFileBase.InputStream.

For efficient purposes you could check whether the stream returned is already a MemoryStream:

byte[] data;
using (Stream inputStream = model.File.InputStream)
{
    MemoryStream memoryStream = inputStream as MemoryStream;
    if (memoryStream == null)
    {
        memoryStream = new MemoryStream();
        inputStream.CopyTo(memoryStream);
    }
    data = memoryStream.ToArray();
}

You can read it from the input stream:

public ActionResult ManagePhotos(ManagePhotos model)
{
    if (ModelState.IsValid)
    {
        byte[] image = new byte[model.File.ContentLength];
        model.File.InputStream.Read(image, 0, image.Length); 

        // TODO: Do something with the byte array here
    }
    ...
}

And if you intend to directly save the file to the disk you could use the model.File.SaveAs method. You might find the following blog post useful.

byte[] file = new byte[excelFile.ContentLength];
excelFile.InputStream.Read(file, 0, file.Length);

//Create memory stream object from your bytes
MemoryStream ms = new MemoryStream(file);

// Set WorkbookPart , Sheet
using (var myDoc = DocumentFormat.OpenXml.Packaging.SpreadsheetDocument.Open(ms, true))