xml Java获取节点的标记名

Question

我需要读取一个小xml文件，并根据一个硬编码的HashMap验证其内容，其中key=tag和value=text位于标记内。 我无法获取节点的标记名。 如果将节点转换为元素，则会出现强制转换异常

我正在使用末日课程阅读：

DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
        DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
        Document doc = dBuilder.parse(xmlFile);

        NodeList list = doc.getElementsByTagName("MergeOptions");
        if (list.getLength() == 0)
        {
        //throw
        }
        NodeList config = list.item(0).getChildNodes();
        for (int i = 0; i <= config.getLength() - 1; i++)
        {

            Node setting = config.item(i);
            String nodeName = setting.getNodeValue();
            String value = setting.getTextContent();
            if (defaultMergeOptions.containsKey(nodeName) == false)
            {
                    //throw
            }

            if (defaultMergeOptions.get(nodeName).equals(value))
            {
                    //throw
            }

Xml文件：

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
  <MergeOptions>
   <sometagName>false</sometagName>
 </MergeOptions>

Answer 1

我做了一些不同的事情

似乎有效：

IntegrationTest.getInstance().getLogger().log(Level.INFO, "Reading merge-que file: " + xmlFile.getAbsolutePath());
    try
    {
        DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
        DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
        Document doc = dBuilder.parse(xmlFile);

        for (Entry<String, String> entry : defaultMergeOptions.entrySet())
        {
            String tagName = entry.getKey();

            NodeList list = doc.getElementsByTagName(tagName);
            if (list.getLength() != 1)
            {
                IntegrationTest.getInstance().getLogger().log(Level.SEVERE, TestResult.FAIL, "Merge option [{0}] has invalid content. Tag [{1}] missing or to many",
                        new Object[] { xmlFile.getName(), tagName });
                result = TestResult.FAIL;
                continue;
            }

            if (!defaultMergeOptions.get(tagName).equals(list.item(0).getTextContent()))
            {
                IntegrationTest.getInstance().getLogger().log(Level.WARNING, TestResult.FAIL, "Merge option [{0}] has diffrent content for tag [{1}].",
                        new Object[] { xmlFile.getCanonicalPath(), tagName });
                result = TestResult.FAIL;
            }

        }



    }
    catch (Exception e)
    {
        IntegrationTest.getInstance().getLogger().log(Level.SEVERE, SBUtil.stackTraceToString(e.getStackTrace()));
        throw new IntegrationTestException(e);

    }
}

Answer 2

我已经试过运行你的代码，它运行得很好，没有类强制转换异常。 请注意我是如何在for循环中使用元素来获取可能的子元素的名称、值或存在性的

final String xml = "<?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"yes\"?>\n" +
            "<MergeOptions>\n<sometagName>false</sometagName>\n</MergeOptions>";

final InputStream xsmlStream = new ByteArrayInputStream(xml.getBytes());

final DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
final DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
final Document doc = dBuilder.parse(xsmlStream);

final NodeList nodes = doc.getElementsByTagName("MergeOptions");

for (int i = 0; i < nodes.getLength(); i++) {
    final Element element = (Element) nodes.item(i);
    System.out.println(element.hasChildNodes());
    System.out.println(element.getNodeValue());
    System.out.println(element.getTagName());
}

使用哈希映射将节点名称作为键有点棘手，因为如果XML文件有多个具有相同名称和不同值的节点名称，哈希映射将只存储一个唯一键，从而只验证一个同名节点。其他名称相同但值不同的节点将无效

Answer 3

我正在帮助您完成以下代码结构。一旦看到标记名和值，就可以应用逻辑从HashMap键或值进行比较

import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;

import org.w3c.dom.Document;
import org.w3c.dom.Element;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;

public class Test1 {
    public static void main(String[] args) throws Exception {
        String xmlFile = "test.xml";
        DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
        DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
        Document doc = dBuilder.parse(xmlFile);
        Element root = doc.getDocumentElement();
        System.out.println(root.getNodeName());

        NodeList list = root.getChildNodes();

        for (int i = 0; i < list.getLength(); i++) {
            Node node = list.item(i);
            if (node.getNodeType() == Node.ELEMENT_NODE)
             {
                System.out.println(node.getNodeName() + " : " + node.getTextContent());
             }
        }

    }
}