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    <p>So basically using the SAX parser is a good choice for you (it is fast, allows you to filter out all the unnecessary data, consumes low memory). When working with SAX for the first time, you may find the following example useful. I do not say the code is perfect (it misses e.g. exception handling, safe stream closing, etc.), but it could be a good start point for you.</p> <pre> <code> import java.io.FileInputStream; import java.io.InputStream; import java.util.ArrayList; import java.util.List; import javax.xml.parsers.SAXParser; import javax.xml.parsers.SAXParserFactory; import org.xml.sax.Attributes; import org.xml.sax.SAXException; import org.xml.sax.helpers.DefaultHandler; public class Test { private static final String HTML_INSTRUCTIONS = "html_instructions"; public static void main(String[] args) throws Exception { final List htmlInstructions = new ArrayList(); SAXParserFactory spf = SAXParserFactory.newInstance(); SAXParser sp = spf.newSAXParser(); DefaultHandler dh = new DefaultHandler() { private boolean isHtmlInstructions = false; private StringBuilder sb = new StringBuilder(); @Override public void startElement(String uri, String localName, String name, Attributes attributes) throws SAXException { super.startElement(uri, localName, name, attributes); if (HTML_INSTRUCTIONS.equals(name)) { isHtmlInstructions = true; } } @Override public void characters(char ch[], int start, int length) throws SAXException { if (isHtmlInstructions) { sb.append(ch, start, length); } } @Override public void endElement(String uri, String localName, String name) throws SAXException { super.endElement(uri, localName, name); if (HTML_INSTRUCTIONS.equals(name)) { htmlInstructions.add(sb.toString()); sb.delete(0, sb.length()); isHtmlInstructions = false; } } }; InputStream is = new FileInputStream("test.xml"); sp.parse(is, dh); for (String htmlInstruction : htmlInstructions) { System.out.println(htmlInstruction); } } } </code> </pre> <p>The output should look like this:</p> <pre> <code> Head &lt;b>east on &lt;b>S Street Viaduct&lt;/b> </code> </pre>
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    1. COThanks so much for this response! It's really helped me understand what's going on. I didn't mention that I am trying to parse a generated xml file from the web, which seems to complicate things, then display the content as a web page. From the ex from my prof I see that I can create an HttpRequest instance and then HttpGet getMethod=new HttpGet(ADDRESS); Then use a responseHandler to get the file as a String. But then I am not sure how to bring the SAX code and the HttpRequest together. I hope that's clear enough. I'll be working on this all weekend so any insight is greatly appreciated.
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    2. COIt depends on what "a file from the web" and "display a content as a web page" mean. If it is a file whose URL is always the same, you can easily use the `java.net.URLConnection` class and get the file as an `java.io.InputStream`. I recommend working with files as with streams and that holds especially for generated files that can easily get very large. Then you can generate an HTML file as an output. Or are you working on a JavaServlet and the XML file will be uploaded by a user and the output will be generated as a direct response to the web browser?
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