Is it possible to convert XSD to XPath in Java?

Question

I need to represent all the elements from an XSD Schema as XPath. Is there any way for it? Like consider there are five elements in XSD Schema I need to display XPath of all the five elements separately. My suggestion is at the background XML corresponding to XSD has to be created and XPath has to be generated. Please suggest solution for the same if the approach is correct or suggest other approaches..

Thanks. M.Sasi kumar

It would be great if some one shares the java program to convert xpath to xsd if possible. I am trying for the same. If I get solution I will post the same. — Sasi Kumar M, Aug 02 '13 at 02:58
Your question asks for XSD to XPath, but your above comment asks for XPath to XSD. Please clarify. If you need both, please start a separate question. — kjhughes, Oct 04 '13 at 13:28

score 0 · Answer 1 · answered Apr 28 '15 at 05:55

import java.io.File;
import java.util.HashMap;
import java.util.Map;
import java.util.Stack;
import javax.xml.parsers.*;

import org.xml.sax.*;
import org.xml.sax.helpers.DefaultHandler;

/**
 * SAX handler that creates and prints XPath expressions for each element encountered.
 *
 * The algorithm is not infallible, if elements appear on different levels in the hierarchy.
 * Something like the following is an example:
 * - <elemA/>
 * - <elemA/>
 * - <elemB/>
 * - <elemA/>
 * - <elemC>
 * -     <elemB/>
 * - </elemC>
 *
 * will report
 *
 * //elemA[0]
 * //elemA[1]
 * //elemB[0]
 * //elemA[2]
 * //elemC[0]
 * //elemC[0]/elemB[1]       (this is wrong: should be //elemC[0]/elemB[0] )
 *
 * It also ignores namespaces, and thus treats <foo:elemA> the same as <bar:elemA>.
 */

public class SAXCreateXPath extends DefaultHandler {

    // map of all encountered tags and their running count
    private Map<String, Integer> tagCount;
    // keep track of the succession of elements
    private Stack<String> tags;

    // set to the tag name of the recently closed tag
    String lastClosedTag;

    /**
     * Construct the XPath expression
     */
    private String getCurrentXPath() {
        String str = "//";
        boolean first = true;
        for (String tag : tags) {
            if (first)
                str = str + tag;
            else
                str = str + "/" + tag;
            str += "["+tagCount.get(tag)+"]";
            first = false;
        }
        return str;
    }

    @Override
    public void startDocument() throws SAXException {
        tags = new Stack();
        tagCount = new HashMap<String, Integer>();
    }

    @Override
    public void startElement (String namespaceURI, String localName, String qName, Attributes atts)
        throws SAXException
    {
        boolean isRepeatElement = false;

        if (tagCount.get(localName) == null) {
            tagCount.put(localName, 0);
        } else {
            tagCount.put(localName, 1 + tagCount.get(localName));
        }

        if (lastClosedTag != null) {
            // an element was recently closed ...
            if (lastClosedTag.equals(localName)) {
                // ... and it's the same as the current one
                isRepeatElement = true;
            } else {
                // ... but it's different from the current one, so discard it
                tags.pop();
            }
        }

        // if it's not the same element, add the new element and zero count to list
        if (! isRepeatElement) {
            tags.push(localName);
        }

        System.out.println(getCurrentXPath());
        lastClosedTag = null;
    }

    @Override
    public void endElement (String uri, String localName, String qName) throws SAXException {
        // if two tags are closed in succession (without an intermediate opening tag),
        // then the information about the deeper nested one is discarded
        if (lastClosedTag != null) {
            tags.pop();
        }
        lastClosedTag = localName;
    }

    public static void main (String[] args) throws Exception {
        if (args.length < 1) {
            System.err.println("Usage: SAXCreateXPath <file.xml>");
            System.exit(1);
        }

        // Create a JAXP SAXParserFactory and configure it
        SAXParserFactory spf = SAXParserFactory.newInstance();
        spf.setNamespaceAware(true);
        spf.setValidating(false);

        // Create a JAXP SAXParser
        SAXParser saxParser = spf.newSAXParser();

        // Get the encapsulated SAX XMLReader
        XMLReader xmlReader = saxParser.getXMLReader();

        // Set the ContentHandler of the XMLReader
        xmlReader.setContentHandler(new SAXCreateXPath());

        String filename = args[0];
        String path = new File(filename).getAbsolutePath();
        if (File.separatorChar != '/') {
            path = path.replace(File.separatorChar, '/');
        }
        if (!path.startsWith("/")) {
            path = "/" + path;
        }

        // Tell the XMLReader to parse the XML document
        xmlReader.parse("file:"+path);
    }
}

Is it possible to convert XSD to XPath in Java?

1 Answers1