TeamSuperGood

This is how you can validate an xml file with a xsd schema using the Java built-in libraries. Look for the sample code at the bottom of the page. After some painful hours, I found out that it doesn’t work with DTDs (although it says it does…). If you have a DTD, then refer to earlier post and look for a conversion tool (DTD to XSD).

Similarly, this is how you would create an xml document with the built-in java library:

try {
Document doc =
DocumentBuilderFactory.newInstance()
.newDocumentBuilder().newDocument();

Element root = doc.createElement(”book”);
Element child = doc.createElement(”author”);
child.setAttribute(”id”, “Eric The Great”);
root.appendChild(child);
doc.appendChild(root);

} catch (ParserConfigurationException e) { … }

The previous code should generate this xml document: (please note that I changed pointy brackets to square ones on purpose)

[book]
[author id=”Eric The Great”/]
[/book]

For XPath to find 1 node:

XPath xpath = XPathFactory
.newInstance().newXPath();

String expression = “book/author”;

Node node = (Node)
xpath.evaluate(expression,
document, XPathConstants.NODE);

String value =
node.getAttributes()
.getNamedItem(ATTRIBUTE_NAME)
.getNodeValue();

For XPath to find a list of nodes:

NodeList nodeList =
(NodeList) xpath.evaluate(expression,
document, XPathConstants.NODESET);

Note: Be careful which library you import because many classnames are very common, like Document, Node, Element, etc. The classes shown here are either from org.w3c.dom or from javax.xml.

For more complicated things……….. LOOK IT UP YOURSELF!!

Explore posts in the same categories: Uncategorized, Java, XML

This entry was posted on Thursday, January 5th, 2006 at 9:32 pm and is filed under Uncategorized, Java, XML. You can subscribe via RSS 2.0 feed to this post's comments. You can comment below, or link to this permanent URL from your own site. Your comments will appear immediately, but I reserve the right to delete innapropriate comments. Edit this entry.