在没有自定义内容处理程序的情况下转换流式 XSLT

Question

Take a look at this website:

http://xmpp.wordpress.com:8008/firehose.xml?type=text/plain

It constantly streams data. You can transform this content using the newest version of XSLT (v3), with a command like this:

<xsl:stream href="http://xmpp.wordpress.com:8008/firehose.xml?type=text/plain">

If I want to write some Java code to initiate the transformation (using Saxon, which has implemented xsl:stream), I can do this:

// XSL
TransformerFactory transformerFactory = TransformerFactory.newInstance();
Transformer transformer = transformerFactory.newTransformer(new StreamSource(new FileInputStream(xslFile)));
// XML
StreamSource xmlSource = new StreamSource(new FileInputStream(xmlFile));
// Output
MyCustomContentHandler handler = new MyCustomContentHandler();  
PrintStream outputPrintStream = new PrintStream(new BufferedOutputStream(new FileOutputStream(outputFile)), true);
handler.setPrintStream(outputPrintStream);
Result result = new SAXResult(handler);
// Transform
transformer.transform(xmlSource, result);

This works. If you let it run for a bit, then open the output file, you'll see data in it. If you re-open it a bit later, you'll see even more data. The key to this is the custom content handler that processes the various SAX events.

But suppose that I don't really want a custom content handler. Suppose I just want to keep the output of the XSLT as is. I can modify my code like this:

// XSL
TransformerFactory transformerFactory = TransformerFactory.newInstance();
Transformer transformer = transformerFactory.newTransformer(new StreamSource(new FileInputStream(xslFile)));
// XML
StreamSource xmlSource = new StreamSource(new FileInputStream(xmlFile));
// Output
TransformerHandler transformerHandler = ((SAXTransformerFactory) SAXTransformerFactory.newInstance()).newTransformerHandler();
transformerHandler.setResult(new StreamResult(new PrintWriter(new FileOutputStream(outputFile, true), true)));
// or this…
//transformerHandler.setResult(new StreamResult(new FileOutputStream(outputFile)));
// or this…
//transformerHandler.setResult(new StreamResult(new FileWriter(outputFile)));
ContentHandler contentHandler = (ContentHandler) transformerHandler;
SAXResult result = new SAXResult(transformerHandler);
// Transform
transformer.transform(xmlSource, result);

The good news is that I no longer need a custom content handler, and my output now matches the output of the XSLT exactly. The bad news is that although this code works with non-streaming XSLT, it does not work with streaming XSLT. Despite my various attempts at setting the result (see the “or this…” statements above), nothing is written to the file. I suspect there's a buffering problem of some sort.

Question: How can I combine the best of these two together? How can I transform a streaming XSLT without having to use a custom content handler?

Answer 1

This seems to be a rerun of a thread on the saxon-help list in June:

http://sourceforge.net/p/saxon/mailman/message/32472658/

The conclusion there was that the output was somehow being buffered in the output stream pipeline. Saxon is emitting events representing the transformation result, as you see by supplying a ContentHandler, but the serialization of these events is being buffered in the I/O system.

Answer 2

At this time, it does not appear to be possible to do what I want to do. My current solution is to use a custom content handler (per my question above) and run its results through a standard XSLT identity transformation. A bit ugly and not very efficient, but it works.