Create ZIP file from servlet

[Jeppe Sommer]

Hello,

I want to be able to add some files (java.sql.Blob) to a ZIP file in a servlet. At this moment I am able to create the ZIP file, but when opening the ZIP file it´s empty?

Below you see some of my code. I get the files from Blob values:

ServletOutputStream sout = response.getOutputStream(); 

String outFilename = "invoice.zip";
ZipOutputStream out = new ZipOutputStream( new FileOutputStream( outFilename ) );

//Set header to HttpServletResponse response
response.reset();
response.setContentType("application/zip");
response.addHeader("Content-Disposition","inline; filename=oioxml.zip");

for ( int x = 0 ; x < a_invoice.length ; x++ ){
    
  ...
  //Get fileName to be Zipped
  String fileName = obj_invoice.getFileName();

  //Get the blob
  java.sql.Blob blob = obj_invoice.getBlob();

  //Converting the blob to InputStream 
  java.io.InputStream in = blob.getBinaryStream();

  // Add ZIP entry to output stream.
  out.putNextEntry( new ZipEntry( fileName ) );

  byte[] bytearray = new byte[4096];
  int size = 0;

  while ( ( size = in.read( bytearray ) ) != -1 ) {
        		  		
     response.getOutputStream().write( bytearray, 0 , size );
  }

}

out.closeEntry(); 
out.finish(); 
out.close();
in.close();

I guess the problem is my for loop?! I appreciate any help...

[David Newton]

Are you trying to send a dynamically-created zip file to the browser?

[Jeppe Sommer]

David Newton wrote:Are you trying to send a dynamically-created zip file to the browser?

Yes, I am trying to send a dynamically-created zip file to the browser.

When I call the servlet a window is shown, but when opening the zip file there is no files added to the zip file!

[Andrew Monkhouse]

Without looking into this in too much depth, it looks like you are writing the contents of the bytearray to the response in line 31. And the bytearray is being filled from in at line 29. in is defined in line 21 as the binary stream from the Blob itself. So it looks to me like your code should be sending the raw blob data to the client - not a zipfile.

Of course there is some other problem with the code if you are getting a zero byte zipfile - I have not looked at what that might be. If the file the client receives does contain data, just is not a zipfile, try changing the extension so you can see if it contains the data you think should be in the zipfile.

[Andrew Monkhouse]

Just tried a simple test case for myself, and tested it on my computer:

    protected void doGet(HttpServletRequest req, HttpServletResponse resp)
            throws ServletException, IOException {
        String webXMLFile = getServletContext().getRealPath("/WEB-INF/web.xml");
        
        resp.setContentType("application/zip");
        resp.setHeader("Content-Disposition","inline; filename=output.zip;");
        
        ServletOutputStream outputStream = resp.getOutputStream();
        ZipOutputStream zip = new ZipOutputStream(outputStream);
        zip.putNextEntry(new ZipEntry("web.xml"));
        
        byte[] b = new byte[1024];
        int len;
        FileInputStream fis = new FileInputStream(webXMLFile);
        while ((len = fis.read(b)) != -1) {
            zip.write(b, 0, len);
        }
        fis.close();
        zip.flush();
        zip.close();
        outputStream.flush();
    }

Although I am just using a file as my input stream in this case, the theory should be the same. I am leaving the writing of the servlet response up to the ZipOutputStream. All I do is add data to the ZipOutputStream.

Can you see how to modify your code now?

[Jeppe Sommer]

Thanks Andrew,

Your code example worked at first try :-)