本示例说明如何使用SAX解析器获取XML文件中元素的属性。
package org.nhooo.example.xml; import org.xml.sax.Attributes; import org.xml.sax.SAXException; import org.xml.sax.helpers.DefaultHandler; import javax.xml.parsers.SAXParser; import javax.xml.parsers.SAXParserFactory; import java.io.InputStream; public class SAXElementAttribute { public static void main(String[] args) { SAXElementAttribute demo = new SAXElementAttribute(); demo.run(); } private void run() { try { // 创建SAXParserFactory实例和一个SAXParser SAXParserFactory factory = SAXParserFactory.newInstance(); SAXParser parser = factory.newSAXParser(); // 获取InputStream到elements.xml文件并进行解析 // 其内容使用SAXHandler。 InputStream is = getClass().getResourceAsStream("/elements.xml"); DefaultHandler handler = new SAXHandler(); parser.parse(is, handler); } catch (Exception e) { e.printStackTrace(); } } class SAXHandler extends DefaultHandler { @Override public void startElement(String uri, String localName, String qName, Attributes attributes) throws SAXException { int attributeLength = attributes.getLength(); if ("person".equals(qName)) { for (int i = 0; i < attributeLength; i++) { // 获取属性名称和值 String attrName = attributes.getQName(i); String attrVal = attributes.getValue(i); System.out.print(attrName + " = " + attrVal + "; "); } System.out.println(""); } } } }
该elements.xml文件如下:
<?xml version="1.0" encoding="UTF-8"?> <root> <persons> <person name="Foo" age="25"/> <person name="Bar" age="22"/> </persons> </root>