environment

org.dom4j.io.SAXReader

`pom.xml`


<dependency>
    <groupId>dom4j</groupId>
    <artifactId>dom4j</artifactId>
    <version>1.6.1</version>
</dependency>

<dependency>
    <groupId>jaxen</groupId>
    <artifactId>jaxen</artifactId>
    <version>1.1.1</version>
</dependency>

Thing you want to do

I want to get the XML with the default namespace set as below in XPath format.

`web.xml`


<?xml version="1.0" encoding="UTF-8" ?>
<web-app xmlns="http://java.sun.com/xml/ns/j2ee"
         xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
         xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd"
         version="2.4">
  <context-param>
    <param-name>version</param-name>
    <param-value>1.2.3</param-value>
  </context-param>
 </web-apps>

Now I want to get the value 1.2.3 in the<param-value>element, which corresponds to the key version in the <param-name> element.

What I tried

The code below returned an empty string and couldn't get the value 1.2.3.

`Bad.java`


 BaseXPath xpath = new Dom4jXPath("//context-param/param-name[text()='version']/following-sibling::param-value");
String version = xpath.stringValueOf(doc); //⇒ Empty string

The method that was realized

You need to set the default namespace yourself.

`Good.java`


 //Default namespace"p"Prefix with.
BaseXPath xpath = new Dom4jXPath(//p:context-param/p:param-name[text()='version']/following-sibling::p:param-value");
xpath.addNamespace("p", "http://java.sun.com/xml/ns/j2ee");
String version = xpath.stringValueOf(doc); //⇒1.2.3

Reference site

http://waman.hatenablog.com/entry/20091030/1256858160 http://www.edankert.com/defaultnamespaces.html#What_s_the_Problem_

`SAXReader`: Get the xml file with the default namespace set in XPath format

environment

pom.xml