How to get the value of a tag in xml using java

[Raaja Gotluru]

Hi,

I am having an xml like below
<?xml version="1.0" encoding="utf-8"?> <Movies> <servername> raaja </servername> <moviename> xyz </moviename> <city> Hyd </city> <theatername> abc </theatername> <noofreels> 16 </noofreels> <aspectratio> 216 </aspectratio> </Movies>
I want the values of tags servername and theatername. Rest I dont want. How to get these using java. Is it possible to get the value using tagnames.

Thanks.

--raaja.g

[Rob Spoor]

You can use XPath for that; both the interface and more generally the technique.

[Raaja Gotluru]

Thanks for the reply. The problem is solved. I used dom parser to acieve this.

                        DocumentBuilderFactory domFactory = DocumentBuilderFactory.newInstance();
		 domFactory.setNamespaceAware(true);
		 DocumentBuilder builder;
		try {
			new File(destinationFolder).mkdir();
			builder = domFactory.newDocumentBuilder();
			File fileDirectory = new File(fileDir);
			//System.out.println("filedir= "+fileDir);
			String files[] = fileDirectory.list();
			
			for (String fileName:files) {
				String xmlData="";
				Document doc;
				doc = builder.parse(fileDir+fileName);
				doc.getDocumentElement ().normalize ();
				String rootNode =  doc.getDocumentElement().getNodeName();
				NodeList bookslist = doc.getElementsByTagName(rootNode);
				String servername = (String) ((Element) bookslist.item(0)).getElementsByTagName("servername").
								item(0).getChildNodes().item(0).getNodeValue();
				xmlData+=servername.trim();
				String theatername = (String) ((Element) bookslist.item(0)).getElementsByTagName("theatername").
								item(0).getChildNodes().item(0).getNodeValue();
				xmlData+=theatername.trim();
			}
		}catch (ParserConfigurationException e) {
				e.printStackTrace();
		}catch (SAXException e) {
				e.printStackTrace();
		} catch (IOException e) {
					e.printStackTrace();
		}  

[Rob Spoor]

Why do you use a NodeList bookslist and not just doc.getDocumentElement()? Other than that, and the lack of error checking, that code looks decent.

[John Jai]

I dont know if XPath is more proficient but looks more readable. please use code tag henceforth.

XPathFactory factory = new net.sf.saxon.xpath.XPathFactoryImpl();
XPath xpath = factory.newXPath();
xmlData += (xpath.evaluate("/Movies/servername", doc.getDocumentElement())).trim();
xmlData += (xpath.evaluate("/Movies/theatername", doc.getDocumentElement())).trim();

[Raaja Gotluru]

Hi,

Thanks for the reply. But the problem is how to get the root node at run time. Because the root node differes from xml to xml. Only servername and theatername are common.
If I use

xmlData += (xpath.evaluate("/Movies/servername", doc.getDocumentElement())).trim();

it will give problem as the root node differs.

Can we get the root node at run time. Please guide me.

Thanking you.

--raaja.g

[Rob Spoor]

That's what's being returned by doc.getDocumentElement(). Each XML document has only one single root element. Here it's called the document element.