Question

This may be beyond the capabilities of the Java VM due to the size of the files being dealt with (50-100MB xml files)

Right now I have a set of xml files sent as zips, which are in turn all decompressed and then all XML in the directory are processed one at a time using SAX.

To save time and space (since the compression is about 1:10) I was wondering if there is a way to pass a ZipFileEntry that is an xml file to a SAX handler.

I’ve seen it done using DocumentBuilder and other xml parsing methods, but for peformance (and especially memory) I’m sticking with SAX.

Currently I am using SAX in the following way

        SAXParserFactory factory = SAXParserFactory.newInstance();
        SAXParser saxParser = factory.newSAXParser();

        MyHandler handler = new MyHandler();

        for( String curFile : xmlFiles )
        {
            System.out.println( "\n\n\t>>>>> open " + curFile + " <<<<<\n");
            saxParser.parse( "file://" + new File( dirToProcess + curFile ).getAbsolutePath(), handler );
        }

Answer 1

You can parse a XML using an InputStream as a source. So you can open a ZipFile, get the InputStream of the entry you want, and then parse it. See the getInputStream method.

—- Edit —-

Here is some code to guide you:

for( String curFile : xmlFiles )
{
        ZipFile zip = new ZipFile(new File( dirToProcess + curFile));
        Enumeration<? extends ZipEntry> entries = zip.entries();
        while (entries.hasMoreElements()){
            ZipEntry entry = entries.nextElement();
            InputStream xmlStream = zip.getInputStream(entry);
            saxParser.parse( xmlStream, handler );
            xmlStream.close();
        }
}