Parsing XML with XPath in Java

Question

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?

What's the "and so on" ...that's important – tcurdt Dec 05 '08 at 01:50 — tcurdt, Dec 05 '08 at 01:50
Also the question is how big will your file be. – tcurdt Dec 05 '08 at 01:51 — tcurdt, Dec 05 '08 at 01:51

score 19 · Accepted Answer · answered May 12 '10 at 11:18

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()); 
    }
  }

sblundy · Answer 2 · 2008-12-04T15:12:16.077

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".

edited Dec 04 '08 at 15:12

answered Dec 04 '08 at 14:52

sblundy

60,628
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yes, but this returns me a list of all the nodes that matches that expression, so all the childs are included ... thanks – Mg. Dec 04 '08 at 14:53
What do you mean all the children are included? I think you're leaving out some detail. – sblundy Dec 04 '08 at 14:57

score 0 · Answer 3 · answered Dec 04 '08 at 14:52

0

This will work for you:

NodeList nodes = (NodeList) xpath.evaluate("//category//subCategoryList/category",
inputSource, XPathConstants.NODESET);

Then you can parse the children of category as you wish.

answered Dec 04 '08 at 14:52

kgiannakakis

103,016
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Parsing XML with XPath in Java

3 Answers3

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