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Make JXPATH 1.2 namespace aware

Some hours of hard work to find a workaround for this issue…I hope that it will help some of you
as this simple issue should be quite common.

Here is a sample xml that reveal the issue (sample.xml):

   1:  <?xml version="1.0" encoding="UTF-8"?>
   2:  <address xmlns:xsi=""
   3:          xmlns=""
   4:          xsi:schemaLocation=" sample.xsd">
   5:    <name>name</name>
   6:    <street>street</street>
   7:    <city>city</city>
   8:    <country>country</country>
   9:  </address>

A very simple XSD schema (sample.xsd)

   1:  <?xml version="1.0" encoding="UTF-8" standalone="no"?>
   2:  <xs:schema xmlns:xs="" 
   3:             targetNamespace="" 
   4:             xmlns="">
   5:      <xs:element name="address">
   6:          <xs:complexType>
   7:              <xs:sequence>
   8:                  <xs:element name="name" type="xs:string" />
   9:                  <xs:element name="street" type="xs:string" />
  10:                  <xs:element name="city" type="xs:string" />
  11:                  <xs:element name="country" type="xs:string" />
  12:              </xs:sequence>
  13:          </xs:complexType>
  14:      </xs:element>
  15:  </xs:schema>

And a simple java client, using JUNIT4

   1:  import;
   3:  import javax.xml.parsers.DocumentBuilder;
   4:  import javax.xml.parsers.DocumentBuilderFactory;
   6:  import org.apache.commons.jxpath.JXPathContext;
   7:  import org.junit.Assert;
   8:  import org.junit.Test;
   9:  import org.w3c.dom.Document;
  11:  public class JXpath12NameSpaceIssue{
  13:    @Test
  14:    public void testCountNonWorkingXML() {
  15:      InputStream xmlStream = this.getClass().getResourceAsStream("/sample.xml");
  17:      try {
  18:        JXPathContext context = this.getJXPathContext(xmlStream);
  19:        Double value = (Double)context.getValue("count(//name)");
  20:        Assert.assertEquals(1, value, 0.0);
  21:      } catch (Exception e) {
  23:      }
  24:    }
  26:    public JXPathContext getJXPathContext(InputStream inputStream) {
  27:      try {
  28:        DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
  29:        factory.setValidating(false); //This is for xml with DTD only!
  30:        factory.setNamespaceAware(true); //if namespace in xml, make no difference if true or false
  31:        factory.setFeature("", true);
  33:        DocumentBuilder builder = factory.newDocumentBuilder();
  34:        builder.setErrorHandler(new JXPathErrorHandler());
  36:        Document document = builder.parse(inputStream);
  37:        JXPathContext context = JXPathContext.newContext(document);
  39:        context.setLenient(true);
  40:        return context;
  41:      } catch (Throwable throwable) {
  42:        throwable.printStackTrace();
  43:      }
  44:      return null;
  45:    }


This line Double value = (Double)context.getValue("count(//name)"); will always make the
test case fail, as the value of context.getValue("count(//name)");  is 0.0 instead of 1.0

As soon as You remove the namespace from the XML file

   1:  <?xml version="1.0" encoding="UTF-8"?>
   2:  <address xmlns:xsi=""
   3:          xmlns=""
   4:          xsi:schemaLocation=" sample.xsd">


The code will return the correct value aka 1.0. The explanation has been found on internet thanks to Google

From[email protected]/msg07865.html

JXPath 1.2 handles namespaces somewhat differently from JXPath 1.1. It
is following the XPath specification more closely. The specification
describes the procedure of matching a name by comparing so-called
expanded names. An expanded name is a combination of a local name and a
namespace URI. In Quote from the spec: "Two expanded-names are equal if
they have the same local part, and either both have a null namespace
URI or both have non-null namespace URIs that are equal." The notion of
default namespace applies to elements of an XML document, but does not
apply to XPaths. Quote: "if the QName does not have a prefix, then the
namespace URI is null (this is the same way attribute names are
expanded). It is an error if the QName has a prefix for which there is
no namespace declaration in the expression context

To remedy the situation, do the following two things:

1. Register the namespace with the JXPathContext:

Namespaces do not apply to objects, unless, of course, those objects are handled by
custom NodePointers that are made namespace-aware.  The standard distribution of
JXPath does not contain any such NodePointers. As far as the interpretation of XPaths
on XML documents is concerned, we are bound by the XPath 1.0 standard. 
On the other hand, the standard does not say anything about applying XPaths to any
non-XML object models, therefore we were free to make pretty much arbitrary choices. 
One of those choices was to ignore namespaces.

More can also be read here

When using namespaces, it is important to remember that XPath matches qualified
names (QNames) based on the namespace URI, not on the prefix. Therefore the XPath
"//foo:bar" may not find a node named "foo:bar" if the prefix "foo" in the context
of the node and in the execution context of the XPath are mapped to different URIs.
Conversely, "//foo:bar" will find the node named "biz:bar", if "foo" in the
execution context and "biz" in the node context are mapped to the same URI.

In order to use a namespace prefix with JXPath, that prefix should be known to
JXPathContext. JXPathContext knows about namespace prefixes declared on the
document element of the context node (the one passed to
JXPathContext.newContext(node)), as well as the ones explicitly registered using
the JXPathContext.registerNamespace(prefix, namespaceURI) method.

So you should end up with:

   1:  Document document = builder.parse(inputStream);
   2:  JXPathContext context = JXPathContext.newContext(document);
   3:  context.registerNamespace("schema", "");
   4:  Double value = (Double)context.getValue("count(//schema:name)");

You can also go another way, and remove any namespace by forcing JAXB2 to
create XML and DOM without qualified namespace

In go from


   1:  @javax.xml.bind.annotation.XmlSchema
   2:  (namespace = "", 
   3:  elementFormDefault = javax.xml.bind.annotation.XmlNsForm.UNQUALIFIED)
   4:  package com.example.test;

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