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
".
Author by
Mg.
Updated on August 01, 2022Comments
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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?
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Mg. over 15 yearsyes, but this returns me a list of all the nodes that matches that expression, so all the childs are included ... thanks
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jbwharris over 15 yearsWhat do you mean all the children are included? I think you're leaving out some detail.