java - 陷入 java XPath 困境

Question

有人能找出这段代码有什么问题吗？无论我选择什么 XPath，它总是返回空字符串

DocumentBuilderFactory domFactory = DocumentBuilderFactory.newInstance();
domFactory.setNamespaceAware(true);
DocumentBuilder builder = domFactory.newDocumentBuilder();
Document doc = builder.parse("chart.xml");
XPath xpath = XPathFactory.newInstance().newXPath();
String str = (String) xpath.evaluate("/row[@id='1']", doc.getDocumentElement(),    XPathConstants.STRING);
System.out.println("xml string is"+str);

我的chart.xml是

<?xml version="1.0" encoding="iso-8859-1"?>
 <chart>
  <row id="1">
    <Select numofobjects="0" id="1000" index="1">
      <Table alias="ConvertDetails" name="ConvertDetails"/>
   </Select>
 </row>
 <row id="2">
   <Select numofobjects="0" id="2000" index="2">
      <Table alias="ConvertDetails" name="ConvertDetails"/>
  </Select>
 </row>
</chart>

我的预期输出是

<Select numofobjects="0" id="1000" index="1">
      <Table alias="ConvertDetails" name="ConvertDetails"/>
   </Select>

Answer 1

正如 Martin 指出的，您需要选择一个节点，而不是它的字符串值:

DocumentBuilderFactory domFactory = DocumentBuilderFactory.newInstance();
domFactory.setNamespaceAware(true);
DocumentBuilder builder = domFactory.newDocumentBuilder();
Document doc = builder.parse("chart.xml");
XPath xpath = XPathFactory.newInstance().newXPath();
Node n = (Node) xpath.evaluate("row[@id='1']", doc.getDocumentElement(), 
                               XPathConstants.NODE);

然后您可以使用以下序列化辅助方法(借自 Get a node's inner XML as String in Java DOM ):

public static String innerXml(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(); 
}

像这样:

String xmlStr = "";
if (n != null) {
    xmlStr = innerXml(node);
}