java - 使用预编译 xpath 的条件 xpath

Question

我在项目中使用 XPATH，我需要有条件地遍历节点

public static String getNodeContentForMultipleTag1(XPathExpression expr,Document doc) {
    try {

        NodeList typeResult = (NodeList) expr.evaluate(doc, XPathConstants.NODESET);
        for (int i = 0; i < typeResult.getLength(); i++) {
            Node typeResultNode = typeResult.item(i);
            System.out.println(typeResultNode.getTextContent());
        }
    } catch (XPathExpressionException e) {
        throw new RuntimeException("Failed parsing expression",e);
    }
    return null;
}

public static void main(String[] args) throws ParserConfigurationException, SAXException, IOException, XPathExpressionException {
    String s="<ex><DtTm><TxDtTm><Cd>ABCD</Cd><dt>1234</dt></TxDtTm><TxDtTm><Cd>XYZ</Cd><dt>891</dt></TxDtTm></DtTm></ex>";
    InputStream inputStream = new ByteArrayInputStream(s.getBytes(StandardCharsets.UTF_8));
    DocumentBuilder db= XpathInstanceUtil.getDocumentBuilderFactory().newDocumentBuilder();
    Document doc = db.parse(inputStream);
    XPath xpath = XpathInstanceUtil.getXPathFactory().newXPath();
    XPathExpression expr = xpath.compile("/ex/DtTm/TxDtTm");
    inputStream.close();
    long st = System.currentTimeMillis();
    getNodeContentForMultipleTag1(expr, doc);
    long end = System.currentTimeMillis();
    System.out.println(end-st);

    long st1 = System.currentTimeMillis();
    getNodeContentForMultipleTag1(expr, doc);
    long end1 = System.currentTimeMillis();
    System.out.println(end1-st1);

}

如果 Cd 值为 ABCD，我应该得到 1234 结果。 我尝试过以下方法

public static String getNodeContentForMultipleTag(String expresssion,String expectedNode,String expectedExpressionTag,Document doc) {
    try {
        XPath xpath = XpathInstanceUtil.getXPathFactory().newXPath();
        NodeList typeResult = (NodeList) evaluateXPath(doc,expresssion,xpath,XPathConstants.NODESET);
        NodeList valueResult= (NodeList) evaluateXPath(doc,expectedExpressionTag,xpath,XPathConstants.NODESET);
        //NodeList typeResult = (NodeList) xpath.evaluate(expresssion,doc, XPathConstants.NODESET);
        //NodeList valueResult = (NodeList) xpath.evaluate(expectedExpressionTag,doc, XPathConstants.NODESET);
        for (int i = 0; i < typeResult.getLength(); i++) {
            Node typeResultNode = typeResult.item(i);
            typeResultNode.getParentNode().removeChild(typeResultNode);
            Node valueResultNode = valueResult.item(i);
            if(typeResultNode.getTextContent().equals(expectedNode) && valueResultNode!=null){
                valueResultNode.getParentNode().removeChild(valueResultNode);
                return  valueResultNode.getTextContent();
            }
        }
    } catch (XPathExpressionException e) {
        throw new RuntimeException("Failed parsing expression"+expresssion,e);
    }
    return null;
}

这就是表达式的样子

public static void main(String[] args) throws ParserConfigurationException, SAXException, IOException {
        String s="<ex><DtTm><TxDtTm><Cd>ABCD</Cd><dt>1234</dt></TxDtTm><TxDtTm><Cd>XYZ</Cd><dt>891</dt></TxDtTm></DtTm></ex>";
        InputStream inputStream = new ByteArrayInputStream(s.getBytes(StandardCharsets.UTF_8));
        DocumentBuilder db= XpathInstanceUtil.getDocumentBuilderFactory().newDocumentBuilder();
        Document doc = db.parse(inputStream);
        String ss = getNodeContentForMultipleTag("/ex/DtTm/TxDtTm/Cd", "XYZ", "/ex/DtTm/TxDtTm/dt", doc);
        System.out.println(ss);
    }

但是它的性能很低，应该如何修改才能高效解析

Answer 1

这段代码看起来很奇怪。为什么使用 Java 而不是 XPath 来完成所有这些工作？为什么在搜索时要修改 DOM 树？

您只需执行 XPath 表达式 /ex/DtTm/TxDtTm[Cd='ABCD']/dt 即可。