java - 在java中使用XPATH过滤XML文档

Question

我有以下 XML 文档，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']”，这给了我:

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

但我需要这个输出:

<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 将返回当前文档中与您的条件相匹配的节点

如果您想生成 Document 的“过滤”表示，那么您需要创建第二个 Document 并附加匹配的 Node > 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，根据您的原始 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，当节点附加到新文档时，这将从原始文档中删除节点。

i want to just filter the result without losing the xml structure

XPath 很像 SQL，它是一种查询语言，它的工作不是生成新的结构。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更像这样的内容...

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>