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Parsing XML with XPath in Java [duplicate]

Tags:

java

xml

xpath

I have a XML with a structure similar to this:

<category>
   <subCategoryList>
      <category>

      </category>
      <category>
         <!--and so on -->
      </category>
   </subCategoryList>
</category>

I have a Category class that has a subcategory list (List<Category>). I'm trying to parse this XML file with XPath, but I can't get the child categories of a category.

How can I do this with XPath? Is there a better way to do this?

like image 737
Mg. Avatar asked Dec 04 '08 14:12

Mg.


2 Answers

This link has everything you need. In shorts:

 public static void main(String[] args) 
   throws ParserConfigurationException, SAXException, 
          IOException, XPathExpressionException {

    DocumentBuilderFactory domFactory = 
    DocumentBuilderFactory.newInstance();
          domFactory.setNamespaceAware(true); 
    DocumentBuilder builder = domFactory.newDocumentBuilder();
    Document doc = builder.parse("persons.xml");
    XPath xpath = XPathFactory.newInstance().newXPath();
       // XPath Query for showing all nodes value
    XPathExpression expr = xpath.compile("//person/*/text()");

    Object result = expr.evaluate(doc, XPathConstants.NODESET);
    NodeList nodes = (NodeList) result;
    for (int i = 0; i < nodes.getLength(); i++) {
     System.out.println(nodes.item(i).getNodeValue()); 
    }
  }
like image 89
ripper234 Avatar answered Nov 07 '22 13:11

ripper234


I believe the XPath expression for this would be "//category/subCategoryList/category". If you just want the children of the root category node (assuming it is the document root node), try "/category/subCategoryList/category".

like image 26
sblundy Avatar answered Nov 07 '22 12:11

sblundy