我用Java读取了这个代码的XML文件:
File file = new File("file.xml");
DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
DocumentBuilder db = dbf.newDocumentBuilder();
Document doc = db.parse(file);
NodeList nodeLst = doc.getElementsByTagName("record");
for (int i = 0; i < nodeLst.getLength(); i++) {
Node node = nodeLst.item(i);
...
}
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那么,我如何从节点实例中获取完整的xml内容?(包括所有标签,属性等)
谢谢.
Pau*_*ime 14
从stackoverflow中查看其他答案.
您将使用DOMSource(而不是StreamSource),并在构造函数中传递您的节点.
然后,您可以将节点转换为String.
快速样品:
public class NodeToString {
public static void main(String[] args) throws TransformerException, ParserConfigurationException, SAXException, IOException {
// just to get access to a Node
String fakeXml = "<!-- Document comment -->\n <aaa>\n\n<bbb/> \n<ccc/></aaa>";
DocumentBuilder docBuilder = DocumentBuilderFactory.newInstance().newDocumentBuilder();
Document doc = docBuilder.parse(new InputSource(new StringReader(fakeXml)));
Node node = doc.getDocumentElement();
// test the method
System.out.println(node2String(node));
}
static String node2String(Node node) throws TransformerFactoryConfigurationError, TransformerException {
// you may prefer to use single instances of Transformer, and
// StringWriter rather than create each time. That would be up to your
// judgement and whether your app is single threaded etc
StreamResult xmlOutput = new StreamResult(new StringWriter());
Transformer transformer = TransformerFactory.newInstance().newTransformer();
transformer.setOutputProperty(OutputKeys.OMIT_XML_DECLARATION, "yes");
transformer.transform(new DOMSource(node), xmlOutput);
return xmlOutput.getWriter().toString();
}
}
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