在Java中使用XPATH过滤XML文档

Question

我有以下XML文档，我需要使用xpath表达式进行过滤：

<List>
<Item>
<Name>abc</Name>
<Category><Name>123</Name><value>aaa</value></Category>
</Item>
<Item>
<Name>def</Name>
<Category><Name>456</Name><value>bbb</value></Category>
</Item>
<Item>
<Name>xyz</Name>
<Category><Name>123</Name><value>ccc</value></Category>
</Item>
</List>

我提供以下表达式：“ / List / Item / Category [Name ='123']”，它给了我：

<Category>
<Name>123</Name>
<value>aaa</value>
<Category>
<Name>123</Name>
<value>ccc</value>
</Category>
</Category>

但是我需要这个输出：

<List>
    <Item>
    <Name>abc</Name>
    <Category><Name>123</Name><value>aaa</value></Category>
    </Item>
    <Item>
    <Name>xyz</Name>
    <Category><Name>123</Name><value>ccc</value></Category>
    </Item>
    </List>

Answer 1

xPath将从当前Document返回与您的条件匹配的节点

如果要生成Document的“过滤”表示，则需要创建第二个Document并将匹配的Node附加到它

例如...

try {
    DocumentBuilderFactory f = DocumentBuilderFactory.newInstance();
    DocumentBuilder b = f.newDocumentBuilder();
    Document original = b.parse(...);
    original.getDocumentElement().normalize();

    Document filtered = b.newDocument();
    Node root = filtered.createElement("List");
    filtered.appendChild(root);

    String expression = "/List/Item/Category[Name='123']";
    XPath xPath = XPathFactory.newInstance().newXPath();
    Object result = xPath.compile(expression).evaluate(original, XPathConstants.NODESET);

    NodeList nodes = (NodeList) result;
    for (int i = 0; i < nodes.getLength(); i++) {

        Node node = nodes.item(i);
        filtered.adoptNode(node);
        root.appendChild(node);

    }

    try (ByteArrayOutputStream os = new ByteArrayOutputStream()) {

        Transformer tf = TransformerFactory.newInstance().newTransformer();
        tf.setOutputProperty(OutputKeys.INDENT, "yes");
        tf.setOutputProperty(OutputKeys.METHOD, "xml");
        tf.setOutputProperty("{http://xml.apache.org/xslt}indent-amount", "4");

        DOMSource domSource = new DOMSource(filtered);
        StreamResult sr = new StreamResult(os);
        tf.transform(domSource, sr);

        String text = new String(os.toByteArray());
        System.out.println(text);

    } catch (TransformerException ex) {
        ex.printStackTrace();
    }

} catch (ParserConfigurationException | SAXException | IOException | XPathExpressionException | DOMException exp) {
    exp.printStackTrace();
}

因此，基于您的原始XML，这将输出

<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<List>
    <Category>
      <Name>123</Name>
      <value>aaa</value>
    </Category>
    <Category>
      <Name>123</Name>
      <value>ccc</value>
    </Category>
</List>

请注意，当将原始文档中的节点附加到新文档中时，这将删除它们。

我只想过滤结果而不丢失xml结构

XPath与SQL非常相似，它是一种查询语言，它的工作不是生成新结构。 话虽如此，您可以控制它返回哪个节点，例如，如果您将查询更改为类似这样的内容...

String expression = "/List/Item[Category[Name='123']]";

然后输出变得更像...

<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<List>
    <Item>
    <Name>abc</Name>
    <Category>
      <Name>123</Name>
      <value>aaa</value>
    </Category>
  </Item>
    <Item>
    <Name>xyz</Name>
    <Category>
      <Name>123</Name>
      <value>ccc</value>
    </Category>
  </Item>
</List>