所以我需要用 XPath 解析 XML 文件。 一切正常,但如果我有一个标 checkout 现多次,我的代码会显示标签内所有相同标签的所有子标签。 例如:我有一个
<a>
<b></b>
</a>
<c></c>
<a>
<b></b>
</a>
输出将是:
a
b
b
c
a
b
b
我的代码是这样的:
public void parseFileXPath() throws ParserConfigurationException, IOException, SAXException, XPathExpressionException {
DocumentBuilderFactory documentBuilderFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder documentBuilder = documentBuilderFactory.newDocumentBuilder();
Document doc = documentBuilder.parse(file);
doc.getDocumentElement().normalize();
XPathFactory factory = XPathFactory.newInstance();
XPath xpath = factory.newXPath();
String expression;
Node node;
NodeList nodeList;
expression="/*";
node = (Node) xpath.evaluate(expression,doc, XPathConstants.NODE);
//System.out.println(node.getNodeName());
expression="/frogans-fsdl/*";
nodeList = (NodeList) xpath.evaluate(expression,doc, XPathConstants.NODESET);
NodeList nodeList1;
for (int i = 0; i < nodeList.getLength(); i++) {
String aTag = nodeList.item(i).getNodeName();
System.out.println(aTag);
expression="/frogans-fsdl/"+aTag+"/*";
nodeList1 = (NodeList) xpath.evaluate(expression,doc,XPathConstants.NODESET);
for (int j = 0; j < nodeList1.getLength(); j++) {
System.out.println("\t"+nodeList1.item(j).getNodeName());
//TODO
//Display what's inside each tag
}
expression="";
}
}
请问有什么帮助吗?
最佳答案
以下内部循环代码可能有助于获得您想要的输出:
NodeList nodeList1;
for (int i = 0; i < nodeList.getLength(); i++) {
System.out.println(nodeList.item(i).getNodeName());
expression="*";
nodeList1 = (NodeList) xpath.evaluate(expression,nodeList.item(i),XPathConstants.NODESET);
for (int j = 0; j < nodeList1.getLength(); j++) {
System.out.println("\t"+nodeList1.item(j).getNodeName());
}
}
基本上,您需要将 nodeList.item(i) 作为后续 xpath 评估的第二个参数传递。
关于java - XPath 解析器 java 冗余,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36114652/