Generate/get Xpath from XML in Java

Here is how this can be done with SAX:

import java.util.HashMap;
import java.util.Map;

import org.xml.sax.Attributes;
import org.xml.sax.SAXException;
import org.xml.sax.XMLReader;
import org.xml.sax.helpers.DefaultHandler;

public class FragmentContentHandler extends DefaultHandler {

    private String xPath = "/";
    private XMLReader xmlReader;
    private FragmentContentHandler parent;
    private StringBuilder characters = new StringBuilder();
    private Map<String, Integer> elementNameCount = new HashMap<String, Integer>();

    public FragmentContentHandler(XMLReader xmlReader) {
        this.xmlReader = xmlReader;
    }

    private FragmentContentHandler(String xPath, XMLReader xmlReader, FragmentContentHandler parent) {
        this(xmlReader);
        this.xPath = xPath;
        this.parent = parent;
    }

    @Override
    public void startElement(String uri, String localName, String qName, Attributes atts) throws SAXException {
        Integer count = elementNameCount.get(qName);
        if(null == count) {
            count = 1;
        } else {
            count++;
        }
        elementNameCount.put(qName, count);
        String childXPath = xPath + "/" + qName + "[" + count + "]";

        int attsLength = atts.getLength();
        for(int x=0; x<attsLength; x++) {
            System.out.println(childXPath + "[@" + atts.getQName(x) + "='" + atts.getValue(x) + ']');
        }

        FragmentContentHandler child = new FragmentContentHandler(childXPath, xmlReader, this);
        xmlReader.setContentHandler(child);
    }

    @Override
    public void endElement(String uri, String localName, String qName) throws SAXException {
        String value = characters.toString().trim();
        if(value.length() > 0) {
            System.out.println(xPath + "='" + characters.toString() + "'");
        }
        xmlReader.setContentHandler(parent);
    }

    @Override
    public void characters(char[] ch, int start, int length) throws SAXException {
        characters.append(ch, start, length);
    }

}

It can be tested with:

import java.io.FileInputStream;

import javax.xml.parsers.SAXParser;
import javax.xml.parsers.SAXParserFactory;

import org.xml.sax.InputSource;
import org.xml.sax.XMLReader;

public class Demo {

    public static void main(String[] args) throws Exception {
        SAXParserFactory spf = SAXParserFactory.newInstance();
        SAXParser sp = spf.newSAXParser();
        XMLReader xr = sp.getXMLReader();

        xr.setContentHandler(new FragmentContentHandler(xr));
        xr.parse(new InputSource(new FileInputStream("input.xml")));
    }
}

This will produce the desired output:

//root[1]/elemA[1]='one'
//root[1]/elemA[2][@attribute1='first]
//root[1]/elemA[2][@attribute2='second]
//root[1]/elemA[2]='two'
//root[1]/elemB[1]='three'
//root[1]/elemA[3]='four'
//root[1]/elemC[1]/elemB[1]='five'

With jOOX (a jquery API port to Java, disclaimer - I work for the company behind the library), you can almost achieve what you want in a single statement:

// I'm assuming this:
import static org.joox.JOOX.$;

// And then...
List<String> coolList = $(document).xpath("//*[not(*)]").map(
    context -> $(context).xpath() + "='" + $(context).text() + "'"
);

If document is your sample document:

<root>
    <elemA>one</elemA>
    <elemA attribute1='first' attribute2='second'>two</elemA>
    <elemB>three</elemB>
    <elemA>four</elemA>
    <elemC>
        <elemB>five</elemB>
    </elemC>
</root>

This will produce

/root[1]/elemA[1]='one'
/root[1]/elemA[2]='two'
/root[1]/elemB[1]='three'
/root[1]/elemA[3]='four'
/root[1]/elemC[1]/elemB[1]='five'

By "almost", I mean that jOOX does not (yet) support matching/mapping attributes. Hence, your attributes will not produce any output. This will be implemented in the near future, though.

private static void buildEntryList( List<String> entries, String parentXPath, Element parent ) {
    NamedNodeMap attrs = parent.getAttributes();
    for( int i = 0; i < attrs.getLength(); i++ ) {
        Attr attr = (Attr)attrs.item( i );
        //TODO: escape attr value
        entries.add( parentXPath+"[@"+attr.getName()+"='"+attr.getValue()+"']"); 
    }
    HashMap<String, Integer> nameMap = new HashMap<String, Integer>();
    NodeList children = parent.getChildNodes();
    for( int i = 0; i < children.getLength(); i++ ) {
        Node child = children.item( i );
        if( child instanceof Text ) {
            //TODO: escape child value
            entries.add( parentXPath+"='"+((Text)child).getData()+"'" );
        } else if( child instanceof Element ) {
            String childName = child.getNodeName();
            Integer nameCount = nameMap.get( childName );
            nameCount = nameCount == null ? 1 : nameCount + 1;
            nameMap.put( child.getNodeName(), nameCount );
            buildEntryList( entries, parentXPath+"/"+childName+"["+nameCount+"]", (Element)child);
        }
    }
}

public static List<String> getEntryList( Document doc ) {
    ArrayList<String> entries = new ArrayList<String>();
    Element root = doc.getDocumentElement();
    buildEntryList(entries, "/"+root.getNodeName()+"[1]", root );
    return entries;
}

This code works with two assumptions: you aren't using namespaces and there are no mixed content elements. The namespace limitation isn't a serious one, but it'd make your XPath expression much harder to read, as every element would be something like *:<name>[namespace-uri()='<nsuri>'][<index>], but otherwise it's easy to implement. Mixed content on the other hand would make the use of xpath very tedious, as you'd have to be able to individually address the second, third and so on text node within an element.

1. use w3c.dom
2. go recursively down
3. for each node there is easy way to get it's xpath: either by storing it as array/list while #2, or via function which goes recursively up until parent is null, then reverses array/list of encountered nodes.

something like that.

UPD: and concatenate final list in order to get final xpath. don't think attributes will be a problem.

I've done a similar task once. The main idea used was that you can use indexes of the element in the xpath. For example in the following xml

<root>
    <el />
    <something />
    <el />
</root>

xpath to the second <el/> will be /root[1]/el[2] (xpath indexes are 1-based). This reads as "take the first root, then take the second one from all elements with the name el". So element something does not affect indexing of elements el. So you can in theory create an xpath for each specific element in your xml. In practice I've accomplished this by walking the tree recursevely and remembering information about elements and their indexes along the way.
Creating xpath referencing specific attribute of the element then was just adding '/@attrName' to element's xpath.

I have written a method to return the absolute path of an element in the Practical XML library. To give you an idea of how it works, here's an extract form one of the unit tests:

assertEquals("/root/wargle[2]/zargle",
             DomUtil.getAbsolutePath(child3a)); 

So, you could recurse through the document, apply your tests, and use this to return the XPath. Or, what is probably better, is that you could use the XPath-based assertions from that same library.

I did the exact same thing last week for processing my xml to solr compliant format.

Since you wanted a pseudo code: This is how I accomplished that.

// You can skip the reference to parent and child.

1_ Initialize a custom node object: NodeObjectVO {String nodeName, String path, List attr, NodeObjectVO parent, List child}

2_ Create an empty list

3_ Create a dom representation of xml and iterate thro the node. For each node, get the corresponding information. All the information like Node name,attribute names and value should be readily available from dom object. ( You need to check the dom NodeType, code should ignore processing instruction and plain text nodes.)

// Code Bloat warning. 4_ The only tricky part is get path. I created an iterative utility method to get the xpath string from NodeElement. (While(node.Parent != null ) { path+=node.parent.nodeName}.

(You can also achieve this by maintaining a global path variable, that keeps track of the parent path for each iteration.)

5_ In the setter method of setAttributes (List), I will append the object's path with all the available attributes. (one path with all available attributes. Not a list of path with each possible combination of attributes. You might want to do someother way. )

6_ Add the NodeObjectVO to the list.

7_ Now we have a flat (not hierrarchial) list of custom Node Objects, that have all the information I need.

(Note: Like I mentioned, I maintain parent child relationship, you should probably skip that part. There is a possibility of code bloating, especially while getparentpath. For small xml this was not a problem, but this is a concern for large xml).