Question

我有xml如下，

<students>
<Student><age>23</age><id>2000</id><name>PP2000</name></Student>
<Student><age>23</age><id>1000</id><name>PP1000</name></Student>
</students>

我有2个xpaths模板XPATH = students/Student将是模板节点，但是我不能硬编码这个xpath，因为它会改变其他XML，而且XML非常动态，可以扩展（但是相同基础XPATHs）所以，如果我使用模板节点再评估一个XPATH，我使用以下代码，

XPath xpathResource = XPathFactory.newInstance().newXPath();
    Document xmlDocument = //creating document;
    NodeList nodeList = (NodeList)xpathResource.compile("//students/Student").evaluate(xmlDocument, XPathConstants.NODESET);
    for (int nodeIndex = 0; nodeIndex < nodeList.getLength(); nodeIndex++) {
        Node currentNode = nodeList.item(nodeIndex);
        String xpathID = "//students/Student/id";
        String xpathName = "//students/Student/name";
        NodeList childID = (NodeList)xpathResource.compile(xpathID).evaluate(currentNode, XPathConstants.NODESET);
        NodeList childName = (NodeList)xpathResource.compile(xpathName).evaluate(currentNode, XPathConstants.NODESET);
        System.out.println("node ID " +childID.item(0).getTextContent());
        System.out.println("node Name " +childName.item(0).getTextContent());
    }

现在的问题是，这个for循环将执行2次，但两次我都得到2000 , PP2000作为ID值。有没有办法使用针对节点的通用XPATH迭代到子节点。我不能对整个XMLDocument使用通用XPATH，我有一些验证要做。我想使用XML nodelist作为结果集行，以便我可以验证XML值并完成我的工作。

Answer 1

    XPath xpathResource = XPathFactory.newInstance().newXPath();
    Document xmlDocument = //creating document;
    NodeList nodeList = (NodeList)xpathResource.compile("//students/Student/id").evaluate(xmlDocument, XPathConstants.NODESET);

for (int nodeIndex = 0; nodeIndex < nodeList.getLength(); nodeIndex++) {
        Node currentNode = nodeList.item(nodeIndex);

        System.out.println("node " +currentNode.getTextContent());
    }

针对子节点的XPATH评估

1 个答案: