Parsing XML with XPath in Java

35,664

Solution 1

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()); 
    }
  }

Solution 2

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".

Share:
35,664
Mg.
Author by

Mg.

Updated on August 01, 2022

Comments

  • Mg.
    Mg. almost 2 years

    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?

  • Mg.
    Mg. over 15 years
    yes, but this returns me a list of all the nodes that matches that expression, so all the childs are included ... thanks
  • jbwharris
    jbwharris over 15 years
    What do you mean all the children are included? I think you're leaving out some detail.