How to divide the request.getInputStream()?

Question

I use a form in a web page to update files, the form code is
<form action="/Mywork/servlet/fileupdate1" method="post" enctype="multipart/form-data">
<p>
Please select a file1 to upload: <input type="file" name="foo">
</p>
<p>
Please select a file2 to upload:<input type="file" name="foo1">
</p>
<input type="submit" value="Upload File!">
</form>
and the "fileupdate1" servlet which the "action" in the form tag refers to will get the inputstream by
"ServletInputStream servIn = request.getInputStream();", and create new OutputFileStream in the disk. But
new files which the servlet stored always have some additional codes "
-----------------------------7d419320202be
Content-Disposition: form-data; name="foo"; filename="H:\Documents and
Settings\administrator\desktop\ee.java"
Content-Type: application/octet-stream
If I updating two files, each file will has these kinds of codes and the two files only have one inputstream. I don't
how to remove these additional codes and how to divide the stream into pieces according to the updated-file-size.
I don't know wether I say it clearly?And
does anyone know the reasons and can help me?
Thank you very much.

Answer 1

Well, if you know the format, you'd know that the file data starts after the first blank line (\r\n\r\n).
But why don't you just use Jakarta Commons File Upload or some other existing free package to handle multipart forms and save yourself the trouble.