FileUpload to FileStream

问题:

I am in process of sending the file along with HttpWebRequest. My file will be from FileUpload UI. Here I need to convert the File Upload to filestream to send the stream along with HttpWebRequest. How do I convert the FileUpload to a filestream?

回答1:

Since FileUpload.PostedFile.InputStream gives me Stream, I used the following code to convert it to byte array

public static byte[] ReadFully(Stream input)
{
    byte[] buffer = new byte[input.Length];
    //byte[] buffer = new byte[16 * 1024];
    using (MemoryStream ms = new MemoryStream())
    {
        int read;
        while ((read = input.Read(buffer, 0, buffer.Length)) > 0)
        {
            ms.Write(buffer, 0, read);
        }
        return ms.ToArray();
    }
}

回答2:

Might be better to pipe the input stream directly to the output stream:

inputStream.CopyTo(outputStream);

This way, you are not caching the entire file in memory before re-transmission. For example, here is how you would write it to a FileStream:

FileUpload fu;  // Get the FileUpload object.
using (FileStream fs = File.OpenWrite("file.dat"))
{
    fu.PostedFile.InputStream.CopyTo(fs);
    fs.Flush();
}

If you wanted to write it directly to another web request, you could do the following:

FileUpload fu; // Get the FileUpload object for the current connection here.
HttpWebRequest hr;  // Set up your outgoing connection here.
using (Stream s = hr.GetRequestStream())
{
    fu.PostedFile.InputStream.CopyTo(s);
    s.Flush();
}

That will be more efficient, as you will be directly streaming the input file to the destination host, without first caching in memory or on disk.

回答3:

You can't convert a FileUpload into a FileStream. You can, however, get a MemoryStream from that FileUpload's PostedFile property. You can then use that MemoryStream to fill your HttpWebRequest.

回答4:

You can put a FileUpload file directly into a MemoryStream by using FileBytes (simplified answer from Tech Jerk)

using (MemoryStream ms = new MemoryStream(FileUpload1.FileBytes))
{
    //do stuff 
}

Or if you do not need a memoryStream

byte[] bin = FileUpload1.FileBytes;