Java - Split XML using XPath but with its Parent Tag

Question

I have following XML String :

<Aaaa>
    <Bbbb>
        <GroupC>
            <KeyId>10001</KeyId>
        </GroupC>
        <DetailC>
            <Dddd>
                <Eeee>Eeee 001</Eeee>
                <Ffff>Ffff 001</Ffff>
            </Dddd>
        </DetailC>
        <DetailC>
            <Dddd>
                <Eeee>Eeee 002</Eeee>
                <Ffff>Ffff 002</Ffff>
            </Dddd>
        </DetailC>
    </Bbbb>
</Aaaa>

I would like to split "DetailC" it into the smaller XML:

XML 01:

<Aaaa>
    <Bbbb>
        <GroupC>
            <KeyId>10001</KeyId>
        </GroupC>
        <DetailC>
            <Dddd>
                <Eeee>Eeee 001</Eeee>
                <Ffff>Ffff 001</Ffff>
            </Dddd>
        </DetailC>
    </Bbbb>
</Aaaa>

XML 02:

<Aaaa>
    <Bbbb>
        <GroupC>
            <KeyId>10001</KeyId>
        </GroupC>
        <DetailC>
            <Dddd>
                <Eeee>Eeee 002</Eeee>
                <Ffff>Ffff 002</Ffff>
            </Dddd>
        </DetailC>
    </Bbbb>
</Aaaa>

Can I know how can I do so using Java? Currently I only able to split into separate XML, but it is without <Aaaa>, <Bbbb>, <GroupC>

Java code:

package message;

import java.io.IOException;
import java.io.StringReader;
import java.io.StringWriter;
import java.nio.file.Files;
import java.nio.file.Paths;
import java.util.ArrayList;
import java.util.List;
import java.util.stream.Collectors;
import java.util.stream.Stream;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.transform.OutputKeys;
import javax.xml.transform.Transformer;
import javax.xml.transform.TransformerFactory;
import javax.xml.transform.dom.DOMSource;
import javax.xml.transform.stream.StreamResult;
import org.apache.xpath.CachedXPathAPI;
import org.w3c.dom.Document;
import org.w3c.dom.Node;
import org.w3c.dom.traversal.NodeIterator;
import org.xml.sax.InputSource;

public class mainClass {

    public static void main(String[] args) throws Exception{
        // TODO Auto-generated method stub

        String path = "D:\\abc.xml";
        String xml = readFile(path);


        List<String> xmlList2 = splitXML(xml, "/Aaaa/Bbbb/DetailC");

        for (String xmlC : xmlList2) {
            System.out.println("xmlC: " + xmlC);
        }
    }

    private static List<String> splitXML(String xmlMessage, String xPath) throws Exception {

        List<String> xmlList = new ArrayList<>();

        Transformer xform = TransformerFactory.newInstance().newTransformer();
        xform.setOutputProperty(OutputKeys.OMIT_XML_DECLARATION, "yes");

        DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
        DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
        InputSource parameterSource = new InputSource(new StringReader(xmlMessage));
        Document doc = dBuilder.parse(parameterSource);
        DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
        factory.setNamespaceAware(true); // never forget this!

        CachedXPathAPI cachedXPathAPI = new CachedXPathAPI();
        NodeIterator nl = cachedXPathAPI.selectNodeIterator(doc, xPath);

        Node node;
        while ((node = nl.nextNode()) != null) {
            StringWriter buf = new StringWriter();
            DOMSource dom = new DOMSource(node);
            xform.transform(dom, new StreamResult(buf));
            xmlList.add(buf.toString());
        }

        return xmlList;
    }

    private static String readFile(String path) {
        String content = "";
        try (Stream<String> lines = Files.lines(Paths.get(path))) {

            content = lines.collect(Collectors.joining(System.lineSeparator()));

        } catch (IOException e) {
            e.printStackTrace();
        }
        return content;
    }
}

Answer 1

If you use Saxon 9 HE (availabe on Sourceforge and Maven for Java) you can solve that with XSLT 3, see the approach from Split XML file into multiple files using XSLT where you can change the code to

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:xs="http://www.w3.org/2001/XMLSchema" version="3.0"
    exclude-result-prefixes="xs">

<xsl:template match="DetailC">
  <xsl:variable name="pos" as="xs:integer">
     <xsl:number/>
  </xsl:variable>
  <xsl:result-document href="XML{format-number($pos, '000')}.xml">
      <xsl:apply-templates select="/" mode="split">
         <xsl:with-param name="this-detail" select="." tunnel="yes"/>
      </xsl:apply-templates>
  </xsl:result-document>
</xsl:template>

<xsl:template match="@* | node()" mode="split">
  <xsl:copy>
    <xsl:apply-templates select="@* | node()" mode="#current"/>
  </xsl:copy>
</xsl:template>

<xsl:template match="DetailC" mode="split">
  <xsl:param name="this-detail" tunnel="yes"/>
  <xsl:if test=". is $this-detail">
    <xsl:next-match/>
  </xsl:if>
</xsl:template>

</xsl:stylesheet>

To run Saxon 9 in Java you can use either the JAXP transformation API http://saxonica.com/html/documentation/using-xsl/embedding/jaxp-transformation.html or the Saxon 9 specific s9api http://saxonica.com/html/documentation/using-xsl/embedding/s9api-transformation.html.

Keep in mind that Transformer can directly transform a file with StreamSource (e.g. https://docs.oracle.com/javase/8/docs/api/javax/xml/transform/stream/StreamSource.html#StreamSource-java.lang.String- or https://docs.oracle.com/javase/8/docs/api/javax/xml/transform/stream/StreamSource.html#StreamSource-java.io.File-) so there is no need to read in the file contents in a string or to build a DOM by hand, you can load any XML file directly as the input to XSLT.