你好!我花了一些时间用 XPath 解析 XML 文档。这似乎是一项简单的任务,但从一开始我就遇到了麻烦。我的代码是:
public class QueryXML3 {
public static void main(String[] args) {
DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
factory.setNamespaceAware(true);
DocumentBuilder builder;
Document doc = null;
try {
builder = factory.newDocumentBuilder();
//doc = builder.parse("SampleExample.xml");
InputStream is = QueryXML3.class.getClassLoader().getResourceAsStream("SampleXml.xml");
doc = builder.parse(is);
XPathFactory xpathFactory = XPathFactory.newInstance();
// Create XPath object
XPath xpath = xpathFactory.newXPath();
Node parNode = getParameterNode(doc, xpath);
System.out.println("parameter node:" + parNode);
NodeList res = getParameterNodeList(doc, xpath );
System.out.println("List of nodes" + res);
} catch (ParserConfigurationException | SAXException | IOException e) {
e.printStackTrace();
}
}
public static Node getParameterNode(Document doc, XPath xpath) {
Node res = null;
try {
res = (Node) xpath.evaluate("/definitions/process", doc, XPathConstants.NODE);
} catch (XPathExpressionException e) {
e.printStackTrace();
}
return res;
}
public static NodeList getParameterNodeList(Document doc, XPath xpath) {
NodeList nodeList = null;
try {
nodeList = (NodeList) xpath.evaluate("/definitions/process", doc, XPathConstants.NODESET);
for (int i = 0; i > nodeList.getLength(); i++) {
System.out.print(nodeList.item(i).getNodeName() + " ");
}
} catch (XPathExpressionException e) {
e.printStackTrace();
}
return nodeList;
}
}
结果我得到了这个:
参数 node:[process: null] 节点列表 com.sun.org.apache.xml.internal.dtm.ref.DTMNodeList@2f17aadf
我只想输出我的 xml 文件的所有节点及其属性...