Java 如何提取完整的 XML 块

Question

使用这个 XML 示例：

<A>
  <B>
    <id>0</id>
  </B>
  <B>
    <id>1</id>
  </B>
</A>

我想要一个简单的方法来提取节点B的XML块，返回XML字符串：

<B>
  <id>1</id>
</B>

要检索这个节点，我应该使用一些 Java XPath 库，如 XOM 或 Java XPath，但我找不到如何获取完整的 XML 字符串。

我使用 C# 找到了两个等效的已回答问题： C# How to extract complete xml node set 和 how can I extract an XML block from an XML document?

Answer 1

引用第二个B 元素所需的表达式应如下所示：

/*/B[id='1']

或者，如果目标节点在文档中的未知位置，使用：

//B[id='1']

完整的 Java 示例（假设 XML 位于名为 workbook.xml 的文件中）：

DocumentBuilderFactory domFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder builder = domFactory.newDocumentBuilder();
Document doc = builder.parse("workbook.xml");
XPath xpath = XPathFactory.newInstance().newXPath();
XPathExpression expr = xpath.compile("//B[id='1']");        

NodeList nodes = (NodeList) expr.evaluate(doc, XPathConstants.NODESET);
for (int i = 0; i < nodes.getLength(); i++) {
    System.out.println("[" + nodes.item(i) + "]");
}

Answer 2

添加到lwburk的解决方案，将DOM节点转换为字符串形式，可以使用Transformer：

private static String nodeToString(Node node)
throws TransformerException
{
    StringWriter buf = new StringWriter();
    Transformer xform = TransformerFactory.newInstance().newTransformer();
    xform.setOutputProperty(OutputKeys.OMIT_XML_DECLARATION, "yes");
    xform.transform(new DOMSource(node), new StreamResult(buf));
    return(buf.toString());
}

完整示例：

public static void main(String... args)
throws Exception
{
    String xml = "<A><B><id>0</id></B><B><id>1</id></B></A>";
    DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
    Document doc = dbf.newDocumentBuilder().parse(new InputSource(new StringReader(xml)));

    XPath xPath = XPathFactory.newInstance().newXPath();
    Node result = (Node)xPath.evaluate("A/B[id = '1']", doc, XPathConstants.NODE);

    System.out.println(nodeToString(result));
}

private static String nodeToString(Node node)
throws TransformerException
{
    StringWriter buf = new StringWriter();
    Transformer xform = TransformerFactory.newInstance().newTransformer();
    xform.setOutputProperty(OutputKeys.OMIT_XML_DECLARATION, "yes");
    xform.transform(new DOMSource(node), new StreamResult(buf));
    return(buf.toString());
}