Who knows: 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

Use something like Jakarta Commons FileUpload to process the multipart/form-data request:
http://jakarta.apache.org/commons/fileupload/