[Java] 通过XPath获取XML中某个节点的属性

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/**      * Get PA Url      * @author jzhang6      * @return url      */     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;     }