When to Use XPath on XML Documents with a Default Namespace?
Using XPath on XML Documents with a Default Namespace
When manipulating XML documents with a default namespace and no prefix, it's possible to use XPath without the namespace URI, as if no namespace exists. However, setting the namespaceAware property of DocumentBuilderFactory to false may not always suffice.
To successfully use XPath on such documents, you can employ a NamespaceContext when executing the XPath. This involves prefixing the XPath fragments to match the NamespaceContext. The prefixes used do not need to align with those in the document.
Here's an updated version of your code that incorporates this approach:
<code class="java">import java.util.Iterator; import javax.xml.namespace.NamespaceContext; import javax.xml.parsers.DocumentBuilder; import javax.xml.parsers.DocumentBuilderFactory; import javax.xml.xpath.XPath; import javax.xml.xpath.XPathConstants; import javax.xml.xpath.XPathFactory; import org.w3c.dom.Document; import org.w3c.dom.NodeList; public class Demo { public static void main(String[] args) { DocumentBuilderFactory domFactory = DocumentBuilderFactory.newInstance(); domFactory.setNamespaceAware(true); try { DocumentBuilder builder = domFactory.newDocumentBuilder(); Document dDoc = builder.parse("E:/test.xml"); XPath xPath = XPathFactory.newInstance().newXPath(); xPath.setNamespaceContext(new MyNamespaceContext()); NodeList nl = (NodeList) xPath.evaluate("/ns:root/ns:author", dDoc, XPathConstants.NODESET); System.out.println(nl.getLength()); } catch (Exception e) { e.printStackTrace(); } } private static class MyNamespaceContext implements NamespaceContext { public String getNamespaceURI(String prefix) { if("ns".equals(prefix)) { return "http://www.mydomain.com/schema"; } return null; } public String getPrefix(String namespaceURI) { return null; } public Iterator getPrefixes(String namespaceURI) { return null; } } }</code>
Note that the XPath expression has also been corrected, as per Dennis' suggestion.
Alternatively, the following simplified code snippet may also work:
<code class="java">import javax.xml.parsers.DocumentBuilder; import javax.xml.parsers.DocumentBuilderFactory; import javax.xml.xpath.XPath; import javax.xml.xpath.XPathConstants; import javax.xml.xpath.XPathFactory; import org.w3c.dom.Document; import org.w3c.dom.NodeList; public class Demo { public static void main(String[] args) { DocumentBuilderFactory domFactory = DocumentBuilderFactory.newInstance(); try { DocumentBuilder builder = domFactory.newDocumentBuilder(); Document dDoc = builder.parse("E:/test.xml"); XPath xPath = XPathFactory.newInstance().newXPath(); NodeList nl = (NodeList) xPath.evaluate("/root/author", dDoc, XPathConstants.NODESET); System.out.println(nl.getLength()); } catch (Exception e) { e.printStackTrace(); } } }</code>
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