CODEWITHSUNDEEP
javaxml
Sameek Mishra
Sameek Mishra

Reputation: 9384

How to retrieve element value of XML using Java?

I am new to XML. I want to read the following XML on the basis of request name. Please help me on how to read the below XML in Java -

<?xml version="1.0"?>
    <config>
        <Request name="ValidateEmailRequest">
            <requestqueue>emailrequest</requestqueue>
            <responsequeue>emailresponse</responsequeue>
        </Request>
        <Request name="CleanEmail">
            <requestqueue>Cleanrequest</requestqueue>
            <responsequeue>Cleanresponse</responsequeue>
        </Request>
    </config>

Upvotes: 56

Views: 367655

Answers (10)

Yash
Yash

Reputation: 9568

If the XML is well formed then you can convert it to Document. By using the XPath you can get the XML Elements.

String xml = "<stackusers><name>Yash</name><age>30</age></stackusers>";

Form XML-String Create Document and find the elements using its XML-Path.

Document doc = getDocument(xml, true);

    public static Document getDocument(String xmlData, boolean isXMLData) throws Exception {
        DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
        dbFactory.setNamespaceAware(true);
        dbFactory.setIgnoringComments(true);
        DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
        Document doc;
        if (isXMLData) {
            InputSource ips = new org.xml.sax.InputSource(new StringReader(xmlData));
            doc = dBuilder.parse(ips);
        } else {
            doc = dBuilder.parse( new File(xmlData) );
        }
        return doc;
    }

Use org.apache.xpath.XPathAPI to get Node or NodeList.

System.out.println("XPathAPI:"+getNodeValue(doc, "/stackusers/age/text()"));

NodeList nodeList = getNodeList(doc, "/stackusers");
System.out.println("XPathAPI NodeList:"+ getXmlContentAsString(nodeList));
System.out.println("XPathAPI NodeList:"+ getXmlContentAsString(nodeList.item(0)));

    public static String getNodeValue(Document doc, String xpathExpression) throws Exception {
        Node node = org.apache.xpath.XPathAPI.selectSingleNode(doc, xpathExpression);
        String nodeValue = node.getNodeValue();
        return nodeValue;
    }
    public static NodeList getNodeList(Document doc, String xpathExpression) throws Exception {
        NodeList result = org.apache.xpath.XPathAPI.selectNodeList(doc, xpathExpression);
        return result;
    }

Using javax.xml.xpath.XPathFactory

System.out.println("javax.xml.xpath.XPathFactory:"+getXPathFactoryValue(doc, "/stackusers/age"));

    static XPath xpath = javax.xml.xpath.XPathFactory.newInstance().newXPath();
    public static String getXPathFactoryValue(Document doc, String xpathExpression) throws XPathExpressionException, TransformerException, IOException {
        Node node = (Node) xpath.evaluate(xpathExpression, doc, XPathConstants.NODE);
        String nodeStr = getXmlContentAsString(node);
        return nodeStr;
    }

Using Document Element.

System.out.println("DocumentElementText:"+getDocumentElementText(doc, "age"));

    public static String getDocumentElementText(Document doc, String elementName) {
        return doc.getElementsByTagName(elementName).item(0).getTextContent();
    }

Get value in between two strings.

String nodeVlaue = org.apache.commons.lang.StringUtils.substringBetween(xml, "<age>", "</age>");
System.out.println("StringUtils.substringBetween():"+nodeVlaue);

Full Example:

public static void main(String[] args) throws Exception {
    String xml = "<stackusers><name>Yash</name><age>30</age></stackusers>";
    Document doc = getDocument(xml, true);
    
    String nodeVlaue = org.apache.commons.lang.StringUtils.substringBetween(xml, "<age>", "</age>");
    System.out.println("StringUtils.substringBetween():"+nodeVlaue);
    
    System.out.println("DocumentElementText:"+getDocumentElementText(doc, "age"));
    System.out.println("javax.xml.xpath.XPathFactory:"+getXPathFactoryValue(doc, "/stackusers/age"));
    
    System.out.println("XPathAPI:"+getNodeValue(doc, "/stackusers/age/text()"));
    NodeList nodeList = getNodeList(doc, "/stackusers");
    System.out.println("XPathAPI NodeList:"+ getXmlContentAsString(nodeList));
    System.out.println("XPathAPI NodeList:"+ getXmlContentAsString(nodeList.item(0)));
}
public static String getXmlContentAsString(Node node) throws TransformerException, IOException {
    StringBuilder stringBuilder = new StringBuilder();
    NodeList childNodes = node.getChildNodes();
    int length = childNodes.getLength();
    for (int i = 0; i < length; i++) {
        stringBuilder.append( toString(childNodes.item(i), true) );
    }
    return stringBuilder.toString();
}

OutPut:

StringUtils.substringBetween():30
DocumentElementText:30
javax.xml.xpath.XPathFactory:30
XPathAPI:30
XPathAPI NodeList:<stackusers>
   <name>Yash</name>
   <age>30</age>
</stackusers>
XPathAPI NodeList:<name>Yash</name><age>30</age>

Upvotes: 4

Yuriy N.
Yuriy N.

Reputation: 6097

In case you just need one (first) value to retrieve from xml: 

public static String getTagValue(String xml, String tagName){
    return xml.split("<"+tagName+">")[1].split("</"+tagName+">")[0];
}

In case you want to parse whole xml document use JSoup:

Document doc = Jsoup.parse(xml, "", Parser.xmlParser());
for (Element e : doc.select("Request")) {
    System.out.println(e);
}

Upvotes: 38

Buhake Sindi
Buhake Sindi

Reputation: 89169

If your XML is a String, Then you can do the following:

String xml = ""; //Populated XML String....

DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
DocumentBuilder builder = factory.newDocumentBuilder();
Document document = builder.parse(new InputSource(new StringReader(xml)));
Element rootElement = document.getDocumentElement();

If your XML is in a file, then Document document will be instantiated like this:

Document document = builder.parse(new File("file.xml"));

The document.getDocumentElement() returns you the node that is the document element of the document (in your case <config>).

Once you have a rootElement, you can access the element's attribute (by calling rootElement.getAttribute() method), etc. For more methods on java's org.w3c.dom.Element

More info on java DocumentBuilder & DocumentBuilderFactory. Bear in mind, the example provided creates a XML DOM tree so if you have a huge XML data, the tree can be huge.

Update Here's an example to get "value" of element <requestqueue>

protected String getString(String tagName, Element element) {
        NodeList list = element.getElementsByTagName(tagName);
        if (list != null && list.getLength() > 0) {
            NodeList subList = list.item(0).getChildNodes();

            if (subList != null && subList.getLength() > 0) {
                return subList.item(0).getNodeValue();
            }
        }

        return null;
    }

You can effectively call it as, 

String requestQueueName = getString("requestqueue", element);

Upvotes: 76

bdoughan
bdoughan

Reputation: 149017

If you are just looking to get a single value from the XML you may want to use Java's XPath library. For an example see my answer to a previous question:

It would look something like:

import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.xpath.XPath;
import javax.xml.xpath.XPathConstants;
import javax.xml.xpath.XPathFactory;

import org.w3c.dom.Document;
import org.w3c.dom.NodeList;

public class Demo {

    public static void main(String[] args) {
        DocumentBuilderFactory domFactory = DocumentBuilderFactory.newInstance();
        try {
            DocumentBuilder builder = domFactory.newDocumentBuilder();
            Document dDoc = builder.parse("E:/test.xml");

            XPath xPath = XPathFactory.newInstance().newXPath();
            Node node = (Node) xPath.evaluate("/Request/@name", dDoc, XPathConstants.NODE);
            System.out.println(node.getNodeValue());
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

}

Upvotes: 29

Bozho
Bozho

Reputation: 597106

Since you are using this for configuration, your best bet is apache commons-configuration. For simple files it's way easier to use than "raw" XML parsers.

See the XML how-to

Upvotes: 1

khachik
khachik

Reputation: 28693

You can make a class which extends org.xml.sax.helpers.DefaultHandler and call

start_<tag_name>(Attributes attrs);

and

end_<tag_name>();

For it is:

start_request_queue(attrs);

etc.

And then extends that class and implement xml configuration file parsers you want. Example:

  ...
  public void startElement(String uri, String name, String qname,
                           org.xml.sax.Attributes attrs) 
                  throws org.xml.sax.SAXException {
    Class[] args = new Class[2];
    args[0] = uri.getClass();
    args[1] = org.xml.sax.Attributes.class;
    try {
      String mname = name.replace("-", "");
      java.lang.reflect.Method m = 
            getClass().getDeclaredMethod("start" + mname, args);
      m.invoke(this, new Object[] { uri, (org.xml.sax.Attributes)attrs }); 
    }

    catch (IllegalAccessException e) {
      throw new RuntimeException(e);
    }

    catch (NoSuchMethodException e) {
      throw new RuntimeException(e);    }

    catch (java.lang.reflect.InvocationTargetException e) {
      org.xml.sax.SAXException se = 
            new org.xml.sax.SAXException(e.getTargetException());
      se.setStackTrace(e.getTargetException().getStackTrace());
    }

and in a particular configuration parser:

   public void start_Request(String uri, org.xml.sax.Attributes attrs) {
     // make sure to read attributes correctly
     System.err.println("Request, name="+ attrs.getValue(0);
   }

Upvotes: 1

Vishal
Vishal

Reputation: 2752

There are various APIs available to read/write XML files through Java. I would refer using StaX

Also This can be useful - Java XML APIs

Upvotes: 1

lisak
lisak

Reputation: 21971

There are two general ways of doing that. You will either create a Domain Object Model of that XML file, take a look at this

and the second choice is using event driven parsing, which is an alternative to DOM xml representation. Imho you can find the best overall comparison of these two basic techniques here. Of course there are much more to know about processing xml, for instance if you are given XML schema definition (XSD), you could use JAXB.

Upvotes: 1

Abdul Khaliq
Abdul Khaliq

Reputation: 2453

following links might help

http://labe.felk.cvut.cz/~xfaigl/mep/xml/java-xml.htm

http://developerlife.com/tutorials/?p=25

http://www.java-samples.com/showtutorial.php?tutorialid=152

Upvotes: 1

posdef
posdef

Reputation: 6532

There are a number of different ways to do this. You might want to check out XStream or JAXB. There are tutorials and the examples.

Upvotes: 5

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