[Java] 通过XPath获取XML中某个节点的属性

Posted 2020-06-13

public String getPAUrl(){
		String PAUrl = "";

		try {
			String filePath = System.getProperty ("user.dir").toString()+"/src/test/resources/config/environment.xml";
			logger.info("The path of environment.xml is : "+filePath);
			File file = new File(filePath);
			SAXReader saxReader = new SAXReader();
			Document document = saxReader.read(file);
			String currentEnv = Util.getEnvStr();
			Element el =XmlUtil.getSingleElement(document, "/root/environment[@type=‘" + currentEnv + "‘]/PAUrl");
			PAUrl = XmlUtil.getElementValue(el, "No default PA URL");
			logger.info("PA Url : "+PAUrl);
		} catch (DocumentException e) {
			e.printStackTrace();
		}
		
		return PAUrl;
	}