Cannot open zip file generated with java

Question

Hello everybody !
I work on a web application ( servlet, JSP... ).
I generate a zip file containing only one file corresponding to a JSP page but when I try to open it thanks to Winzip 8.0, I have the following message :
Error : invalid compressed data to inflate.
The zipped file is not empty, does anybody know what is the problem, here is the source code of my servlet :
public void trt (HttpServletRequest req, HttpServletResponse res)     throws ServletException, IOException
URL l_JspPage;
String temp;
ByteArrayOutputStream l_ByteArrayOutputStream = new ByteArrayOutputStream();
ZipOutputStream l_ZipOutputStream = new ZipOutputStream(l_ByteArrayOutputStream);
l_ZipOutputStream.setMethod(ZipOutputStream.DEFLATED);
ServletOutputStream out = res.getOutputStream();
ServletContext servletContext = getServletContext();
if ((temp=req.getParameter("url"))!=null)
try
l_JspPage = new URL(temp);
DataInputStream l_DataInputStream = new DataInputStream(l_JspPage.openStream());
int l_iBufferSize = l_DataInputStream.available();
byte l_byJspPageData[] = new byte[l_iBufferSize];
l_DataInputStream.read(l_byJspPageData, 0, l_iBufferSize);
l_ZipOutputStream.putNextEntry(new ZipEntry("file.html"));
l_ZipOutputStream.write(l_byJspPageData,0,l_iBufferSize);
l_ZipOutputStream.closeEntry();
l_ZipOutputStream.finish();
String zip=l_ByteArrayOutputStream.toString();
res.setContentType("application/zip");
res.setHeader("Content-Disposition","inline; filename=output.zip;");
out.println(zip);
out.flush();
Thanks for your help.

Answer 1

Your code is somewhat obscure, but it appears that you are writing a zip file to a byte array, converting the byte array to a string, and writing the string to "out" - whatever that is.
I'm not sure why you are going through all this, but in any case the string conversion is probably a wrong step. Zipping results in binary data, and string conversion will probably result in changes to the binary values.