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.
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