这是我正在解析的JAVA XML ..
<objects>
<object>...<class>A /<class>...</object>
<object>...<class>B</class>....</object>
<object>...<class>A /<class>...</object>
</objects>
现在我使用以下代码将XML拆分为3个基于object标签的XML。
DocumentBuilder builder = dbf.newDocumentBuilder();
Document doc = builder.parse("xml");
doc.getDocumentElement().normalize();
TransformerFactory tranFactory = TransformerFactory.newInstance();
Transformer aTransformer = tranFactory.newTransformer();
NodeList list =(NodeList) doc.getElementsByTagName("object");
System.out.println("XML SPLITED");
for (int i=0; i<list.getLength(); i++){
Node element = list.item(i).cloneNode(true);
if(element.hasChildNodes()){
Source src = new DOMSource(element);
FileOutputStream fs=new FileOutputStream("XML" + i + ".xml");
Result dest = new StreamResult(fs);
aTransformer.transform(src, dest);
fs.close();
}
我的要求是只获取带有类标记A的文件。所以我的输出只有2个XML。请发布你的答案。
答案 0 :(得分:2)
使用您的示例,您可以这样做:
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.IOException;
import java.util.logging.Level;
import java.util.logging.Logger;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.ParserConfigurationException;
import javax.xml.transform.Result;
import javax.xml.transform.Source;
import javax.xml.transform.Transformer;
import javax.xml.transform.TransformerConfigurationException;
import javax.xml.transform.TransformerException;
import javax.xml.transform.TransformerFactory;
import javax.xml.transform.dom.DOMSource;
import javax.xml.transform.stream.StreamResult;
import org.w3c.dom.Document;
import org.w3c.dom.Element;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;
import org.xml.sax.SAXException;
public class XmlSplitting {
private static final Logger logger = Logger.getLogger(XmlSplitting.class.getName());
private static final String FILE_PATH = "./";
private DocumentBuilder builder;
private Transformer transformer;
public XmlSplitting() throws ParserConfigurationException, TransformerConfigurationException {
DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
this.builder = factory.newDocumentBuilder();
TransformerFactory transfromerFactory = TransformerFactory.newInstance();
this.transformer = transfromerFactory.newTransformer();
}
public static void main(String[] args) {
try {
XmlSplitting s = new XmlSplitting();
s.run();
} catch (ParserConfigurationException | SAXException | IOException | TransformerException x) {
logger.log(Level.SEVERE, "Error", x);
}
}
private void run() throws ParserConfigurationException, SAXException, IOException, TransformerException {
File file = new File(FILE_PATH + "xml.xml");
if (file.exists()) {
Document document = this.builder.parse(file);
document.normalize();
NodeList list = document.getElementsByTagName("object");
for (int i = 0; i < list.getLength(); i++) {
Node node = list.item(i);
if (Node.ELEMENT_NODE == node.getNodeType()) {
Element object = (Element)node;
NodeList classes = object.getElementsByTagName("class");
if (1 == classes.getLength()) {
Node clazz = classes.item(0);
if (Node.ELEMENT_NODE == clazz.getNodeType()) {
this.copyNodeToNewFile(clazz.getTextContent(), node, i);
}
} else {
logger.log(Level.SEVERE, "Number of class nodes in node object is different than expected. Number of class nodes found: {0}.", classes.getLength());
}
}
}
} else {
logger.log(Level.SEVERE, "You provided wrong path for xml file.");
}
}
private void copyNodeToNewFile(String content, Node node, int i) throws IOException, TransformerException {
boolean copy = this.nodeShouldBeCopied(content);
logger.log(Level.INFO, "Node with content {0} will {1}be moved to separete file.", new Object[]{content, true == copy ? "" : "not "});
if (copy) {
String fileName = String.format("%sxml%d.xml", FILE_PATH, i);
try (FileOutputStream fos = new FileOutputStream(fileName)) {
Source source = new DOMSource(node);
Result destination = new StreamResult(fos);
this.transformer.transform(source, destination);
}
}
}
// here you can change condition to copy given node to file
private boolean nodeShouldBeCopied(String content) {
return content.contains("A");
}
}
答案 1 :(得分:-1)
如何使用巴西框架解析XML?它会让你的代码更具异国情调,你可以告诉你的朋友: http://jcoltrane.sourceforge.net/userguide/extending_parsing_process.html