CODEWITHSUNDEEP
javaxml
phil swenson
phil swenson

Reputation: 8904

How do I extract child element from XML to a string in Java?

If I have an XML document like

<root>   
   <element1>
        <child attr1="blah">
           <child2>blahblah</child2>
        <child>   
   </element1> 
</root>

I want to get an XML string with the first child element. My output string would be 

<element1>
    <child attr1="blah">
       <child2>blahblah</child2>
    <child>
</element1>

There are many approaches, would like to see some ideas. I've been trying to use Java XML APIs for it, but it's not clear that there is a good way to do this.

thanks

Upvotes: 8

Views: 31400

Answers (8)

Hemant Thorat
Hemant Thorat

Reputation: 2456

You can use following function to extract xml block as string by passing proper xpath expression,

    private static String nodeToString(Node node) throws TransformerException
{
    StringWriter buf = new StringWriter();
    Transformer xform = TransformerFactory.newInstance().newTransformer();
    xform.setOutputProperty(OutputKeys.OMIT_XML_DECLARATION, "yes");
    xform.transform(new DOMSource(node), new StreamResult(buf));
    return(buf.toString());
}

    public static void main(String[] args) throws Exception
{
        DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
        DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
        Document doc = dBuilder.parse(inputFile);

        XPath xPath = XPathFactory.newInstance().newXPath();
        Node result = (Node)xPath.evaluate("A/B/C", doc, XPathConstants.NODE); //"A/B[id = '1']" //"//*[@type='t1']"

        System.out.println(nodeToString(result));

}

Upvotes: 0

Preksha Harne
Preksha Harne

Reputation: 21

public String getXML(String xmlContent, String tagName){

    String startTag = "<"+ tagName + ">";
    String endTag = "</"+ tagName + ">";
    int startposition = xmlContent.indexOf(startTag);
    int endposition = xmlContent.indexOf(endTag, startposition);
    if (startposition == -1){
        return "ddd";
    }
    startposition += startTag.length();
    if(endposition == -1){ 
        return "eee";
    }
    return xmlContent.substring(startposition, endposition);
}

Pass your xml as string to this method,and in your case pass 'element' as parameter tagname.

Upvotes: 2

Yuriy N.
Yuriy N.

Reputation: 6137

As question is actually about first occurrence of string inside another string, I would use String class methods, instead of XML parsers: 

public static String getElementAsString(String xml, String tagName){
    int beginIndex = xml.indexOf("<" + tagName);
    int endIndex = xml.indexOf("</" + tagName, beginIndex) + tagName.length() + 3;
    return xml.substring(beginIndex, endIndex);
}

Upvotes: 0

Monica
Monica

Reputation: 61

Since this is the top google answer and For those of you who just want the basic:

    public static String serializeXml(Element element) throws Exception
{
    ByteArrayOutputStream buffer = new ByteArrayOutputStream();
    StreamResult result = new StreamResult(buffer);

    DOMSource source = new DOMSource(element);
    TransformerFactory.newInstance().newTransformer().transform(source, result);

    return new String(buffer.toByteArray());
}

I use this for debug, which most likely is what you need this for

Upvotes: 6

TofuBeer
TofuBeer

Reputation: 61536

XMLBeans is an easy to use (once you get the hang of it) tool to deal with XML without having to deal with the annoyances of parsing.

It requires that you have a schema for the XML file, but it also provides a tool to generate a schema from an exisint XML file (depending on your needs the generated on is probably fine).

Upvotes: 1

Matt McMinn
Matt McMinn

Reputation: 16311

You're right, with the standard XML API, there's not a good way - here's one example (may be bug ridden; it runs, but I wrote it a long time ago).

import javax.xml.*;
import javax.xml.parsers.*;
import javax.xml.transform.*;
import javax.xml.transform.dom.*;
import javax.xml.transform.stream.*;
import org.w3c.dom.*;
import java.io.*;

public class Proc
{
    public static void main(String[] args) throws Exception
    {
        //Parse the input document
        DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
        DocumentBuilder builder = factory.newDocumentBuilder();
        Document doc = builder.parse(new File("in.xml"));

        //Set up the transformer to write the output string
        TransformerFactory tFactory = TransformerFactory.newInstance();
        Transformer transformer = tFactory.newTransformer();
        transformer.setOutputProperty("indent", "yes");
        StringWriter sw = new StringWriter();
        StreamResult result = new StreamResult(sw);

        //Find the first child node - this could be done with xpath as well
        NodeList nl = doc.getDocumentElement().getChildNodes();
        DOMSource source = null;
        for(int x = 0;x < nl.getLength();x++)
        {
            Node e = nl.item(x);
            if(e instanceof Element)
            {
                source = new DOMSource(e);
                break;
            }
        }

        //Do the transformation and output
        transformer.transform(source, result);
        System.out.println(sw.toString());
    }
}

It would seem like you could get the first child just by using doc.getDocumentElement().getFirstChild(), but the problem with that is if there is any whitespace between the root and the child element, that will create a Text node in the tree, and you'll get that node instead of the actual element node. The output from this program is:

D:\home\tmp\xml>java Proc
<?xml version="1.0" encoding="UTF-8"?>
<element1>
        <child attr1="blah">
           <child2>blahblah</child2>
       </child>
   </element1>

I think you can suppress the xml version string if you don't need it, but I'm not sure on that. I would probably try to use a third party XML library if at all possible.

Upvotes: 7

Mark
Mark

Reputation: 4716

If your xml has schema backing it, you could use xmlbeans or JAXB to generate pojo objects that help you marshal/unmarshal xml.

http://xmlbeans.apache.org/ https://jaxb.dev.java.net/

Upvotes: 0

duffymo
duffymo

Reputation: 308998

I would recommend JDOM. It's a Java XML library that makes dealing with XML much easier than the standard W3C approach.

Upvotes: 3

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