parsing XML using XPath in Java

Question

Say I have the following XML file:

<?xml version="1.0" encoding="utf-8"?> <venues> <group type="Nearby"> <venue> <id>222307</id> <name>Union Chapel</name> <primarycategory> <id>78967</id> <fullpathname>Arts &amp; Entertainment:Music Venue</fullpathname> <nodename>Music Venue</nodename> <iconurl>http://foursquare.com/img/categories/arts_entertainment/musicvenue.png</iconurl> </primarycategory> <address>Compton Ave</address> <city>Islington</city> <state>Greater London</state> <zip>N1 2XD</zip> <verified>false</verified> <geolat>51.5439732</geolat> <geolong>-0.1020908</geolong> <stats> <herenow>0</herenow> </stats> <phone>02073594019</phone> <distance>33</distance> </venue>

.............

and my code is the following:

 XPathFactory factory = XPathFactory.newInstance(); XPath xpath = factory.newXPath(); XPathExpression expr = xpath.compile("//venue/*"); Object result = expr.evaluate(document, XPathConstants.NODESET); NodeList nodes = (NodeList) result; //System.out.println(nodes.getLength()); Venue ven = new Venue(); for (int i = 0; i < nodes.getLength(); i++) { String nodeName = nodes.item(i).getNodeName(); String nodeValue = nodes.item(i).getNodeValue(); if (nodeName.equals("id")){ ven = new Venue(); if (nodeValue != null) ven.id = Integer.parseInt(nodeValue); System.out.println(ven.id); } if (nodeName.equals("name")){ ven.name = nodeValue; System.out.println(ven.name); } if (nodeName.equals("address")){ ven.address = nodeValue; System.out.println(ven.address); }

How can I do all of this in one for loop for efficiency? Otherwise for every attribute in the xml that I want to extract I need to create a for loop for each one of them

javamonkey79 · Accepted Answer · 2010-11-03 17:15:54Z

If you use this as your xpath:

//venue/*

You'll get all the child nodes of venue. You can then iterate over this and do a big if else on the node name's and assign them as needed.

Like this:

XPathFactory factory = XPathFactory.newInstance(); XPath xpath = factory.newXPath(); XPathExpression expr = xpath.compile("//venue/*"); Object result = expr.evaluate(document, XPathConstants.NODESET); NodeList nodes = (NodeList) result; for (int i = 0; i < nodes.getLength(); i++) { Node node = nodes.item( i ); String nodeName = node.getNodeName(); String nodeValue = node.getChildNodes().item( 0 ).getNodeValue(); if( nodeName.equals( "name" ) ) { name = nodeValue; } else if( nodeName.equals( "address" ) ) { address = nodeValue; } // ... the rest goes here }

If you don't want to iterate over all child elements you could do something like this:

 XPathExpression expr = xpath.compile( "//venue" ); Object result = expr.evaluate( document, XPathConstants.NODESET ); NodeList nodes = (NodeList)result; for( int i = 0; i < nodes.getLength(); i++ ) { Node node = nodes.item( i ); NodeList venueChildNodes = node.getChildNodes(); String id = venueChildNodes.item( 1 ).getChildNodes().item( 0 ).getNodeValue(); System.out.println( "id: " + id ); String name = venueChildNodes.item( 3 ).getChildNodes().item( 0 ).getNodeValue(); System.out.println( "name: " + name ); String address = venueChildNodes.item( 7 ).getChildNodes().item( 0 ).getNodeValue(); System.out.println( "address: " + address ); }

Where you get all venue nodes and then map it's children. Though, this approach would require a fairly consistent xml structure. Though, something like this seems safest to me:

 XPathExpression expr = xpath.compile( "//venue" ); Object result = expr.evaluate( document, XPathConstants.NODESET ); NodeList nodes = (NodeList)result; for( int i = 0; i < nodes.getLength(); i++ ) { Node node = nodes.item( i ); NodeList venueChildNodes = node.getChildNodes(); String address = null; String name = null; for( int j = 0; j < venueChildNodes.getLength(); j++ ) { Node item = venueChildNodes.item( j ); String nodeName = item.getNodeName(); if ( nodeName.equals( "address" ) ) { address = item.getChildNodes().item( 0 ).getNodeValue(); } if ( nodeName.equals( "name" ) ) { name = item.getChildNodes().item( 0 ).getNodeValue(); } } System.out.println( "address: " + address ); System.out.println( "name: " + name ); }

If you do much xpath, I'd highly recommend getting sketchpath. It's a free tool that is well worth using IMHO (there is a free version out there, but it looks like they're now pushing a pro version). — javamonkey79, CommentedNov 2, 2010 at 23:11
why is it that I get 120 as the nodesLength, while there are only 10 venues... ? I have a class called venue and I want to store this information in those class — aherlambang, CommentedNov 2, 2010 at 23:36
Because //venue means "venue, anywhere in the dom tree". So, //venue/* means "venue, anywhere in the dom tree all of it's children". My guess is that when you're checking the node name you'll need to do a "reset" as you've entered into another instance of venue. — javamonkey79, CommentedNov 2, 2010 at 23:41
I would take a look at XStream - it's a pretty handy little library for XML <-> Object mapping. — javamonkey79, CommentedNov 2, 2010 at 23:42
what do you mean by reseting it again? can you show how to do that in your code example above? — aherlambang, CommentedNov 2, 2010 at 23:46

peter.murray.rust · Accepted Answer · 2010-11-02 23:11:53Z

You can combine them like:

//venue/address/text()|//venue/name/text()

This will return the nodes in document order and you can simply iterate over them.

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