Reputation: 6198
I want to manipulate xml doc having default namespace but no prefix. Is there a way to use xpath without namespace uri just as if there is no namespace?
I believe it should be possible if we set namespaceAware property of documentBuilderFactory to false. But in my case it is not working.
Is my understanding is incorrect or I am doing some mistake in code?
Here is my code:
DocumentBuilderFactory domFactory = DocumentBuilderFactory.newInstance();
domFactory.setNamespaceAware(false);
try {
DocumentBuilder builder = domFactory.newDocumentBuilder();
Document dDoc = builder.parse("E:/test.xml");
XPath xPath = XPathFactory.newInstance().newXPath();
NodeList nl = (NodeList) xPath.evaluate("//author", dDoc, XPathConstants.NODESET);
System.out.println(nl.getLength());
} catch (Exception e) {
e.printStackTrace();
}
Here is my xml:
<?xml version="1.0" encoding="UTF-8"?>
<root xmlns="http://www.mydomain.com/schema">
<author>
<book title="t1"/>
<book title="t2"/>
</author>
</root>
Upvotes: 18
Views: 32791
Reputation: 1590
I've written a simple NamespaceContext
implementation (here), that might be of help. It takes a Map<String, String>
as input, where the key
is a prefix, and the value
is a namespace.
It follows the NamespaceContext spesification, and you can see how it works in the unit tests.
Map<String, String> mappings = new HashMap<>();
mappings.put("foo", "http://foo");
mappings.put("foo2", "http://foo");
mappings.put("bar", "http://bar");
context = new SimpleNamespaceContext(mappings);
context.getNamespaceURI("foo"); // "http://foo"
context.getPrefix("http://foo"); // "foo" or "foo2"
context.getPrefixes("http://foo"); // ["foo", "foo2"]
Note that it has a dependency on Google Guava
Upvotes: 0
Reputation: 6198
Blaise Doughan is right, attached code is correct.
Problem was somewhere elese. I was running all my tests through Application launcher in Eclipse IDE and nothing was working. Then I discovered Eclipse project was cause of all grief. I ran my class from command prompt, it worked. Created a new eclipse project and pasted same code there, it worked there too.
Thank you all guys for your time and efforts.
Upvotes: 1
Reputation: 149047
The XPath processing for a document that uses the default namespace (no prefix) is the same as the XPath processing for a document that uses prefixes:
For namespace qualified documents you can use a NamespaceContext when you execute the XPath. You will need to prefix the fragments in the XPath to match the NamespaceContext. The prefixes you use do not need to match the prefixes used in the document.
Here is how it looks with your code:
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;
}
}
}
Note: I also used the corrected XPath suggested by Dennis.
The following also appears to work, and is closer to your original question:
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();
}
}
}
Upvotes: 23