JAVA XML:获取内容节点

我有一个这样的xml:

<root>
   <countries>
      <country id="98" nom="Espagne"/>
      <country id="76" nom="France"/>
   </countries>
</root>

我可以用以下内容读取root标记:

Document doc = DocumentBuilderFactory.newInstance()
                    .newDocumentBuilder().parse(XmlFile);

System.out.println("Root element :" + doc.getDocumentElement().getnodename());      

Node NodeCountries = doc.getElementsByTagName("countries").item(0);     

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


private static String nodeToString(Node node) throws Exception{
            StringWriter sw = new StringWriter();

              Transformer t = TransformerFactory.newInstance().newTransformer();
              t.setOutputProperty(OutputKeys.OMIT_XML_DeclARATION,"yes");
              t.setOutputProperty(OutputKeys.INDENT,"yes");
              t.transform(new DOMSource(node),new StreamResult(sw));

            return sw.toString();
          }

但是我不能像这样在国家标签内获取所有内容:

<country id="98" nom="Espagne"/>
<country id="76" nom="France"/>
rustybash 回答:JAVA XML:获取内容节点

以下示例将显示<country id="98" nom="Espagne"/><country id="76" nom="France"/>

import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.ParserConfigurationException;
import org.xml.sax.InputSource;
import java.io.StringReader;
import org.w3c.dom.Document;
import org.xml.sax.SAXException;

import org.w3c.dom.ls.DOMImplementationLS;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;
import org.w3c.dom.ls.LSSerializer;

...

String xml = "<root><countries><country id=\"98\" nom=\"Espagne\"/><country id=\"76\" nom=\"France\"/></countries></root>";
DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
DocumentBuilder builder = factory.newDocumentBuilder();
InputSource is = new InputSource(new StringReader(xml));
Document doc = builder.parse(is);
Node node = doc.getElementsByTagName("countries").item(0);
String innerXml = getInnerXml(node);
System.out.println(innerXml);

辅助方法getInnerXml(node)如下所示:

private String getInnerXml(Node node) {
    DOMImplementationLS lsImpl = (DOMImplementationLS) node.getOwnerDocument().getImplementation().getFeature("LS","3.0");
    LSSerializer lsSerializer = lsImpl.createLSSerializer();
    lsSerializer.getDomConfig().setParameter("xml-declaration",false);
    NodeList childNodes = node.getChildNodes();
    StringBuilder sb = new StringBuilder();
    for (int i = 0; i < childNodes.getLength(); i++) {
        sb.append(lsSerializer.writeToString(childNodes.item(i)));
    }
    return sb.toString();
}

让我知道我是否误解了要求(再次!)。

警告 在这里,这不是一个很好的解决方案。它涉及到“手工”构造XML(即字符串连接),并且如果输入出乎意料的不同或复杂,则会带来结果脆弱或什至损坏的风险。

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