Question

我在湄公河上看到了一个例子 - http://www.mkyong.com/java/how-to-read-xml-file-in-java-sax-parser/ 我尝试通过对上面代码中的代码进行以下修改来使其适用于xml文件（下面） -

1 - Have only two if blocks in startElement() and characters() methods.
2 - Change the print statements in above methods, ie 
    FIRSTNAME and First Name = passenger id
    LASTNAME and Last Name = name

问题是 - 在输出中，我看到乘客这个词而不是乘客id的值。我该如何解决这个问题？

<?xml version="1.0" encoding="utf-8"?>
<root xmlns:android="www.google.com">

<passenger id="001">
<name>Tom Cruise</name>
</passenger>

<passenger id="002">
<name>Tom Hanks</name>
</passenger>

</root>

Java代码

import javax.xml.parsers.SAXParser;
import javax.xml.parsers.SAXParserFactory;
import org.xml.sax.Attributes;
import org.xml.sax.SAXException;
import org.xml.sax.helpers.DefaultHandler;

public class ReadXMLFileSAX{

public static void main(String argv[]) {

try {

SAXParserFactory factory = SAXParserFactory.newInstance();
SAXParser saxParser = factory.newSAXParser();

DefaultHandler handler = new DefaultHandler() {

boolean bfname = false;
boolean blname = false;

public void startElement(String uri, String localName,String qName, 
            Attributes attributes) throws SAXException {

    System.out.println("Start Element :" + qName);

    if (qName.equalsIgnoreCase("passenger id")) {
        bfname = true;
    }

    if (qName.equalsIgnoreCase("name")) {
        blname = true;
    }

}

public void endElement(String uri, String localName,
    String qName) throws SAXException {

    System.out.println("End Element :" + qName);

}

public void characters(char ch[], int start, int length) throws SAXException {

    if (bfname) {
        System.out.println("passenger id : " + new String(ch, start,   length));
        bfname = false;
    }

    if (blname) {
        System.out.println("name : " + new String(ch, start, length));
        blname = false;
    }

}

 };

   saxParser.parse("c:\\flight.xml", handler);

 } catch (Exception e) {
   e.printStackTrace();
   }

   }

}

Answer 1

在startElement中，当＆＃34; passenger＆＃34;时，你获得的属性参数将具有该值。

public void startElement(String uri, String localName,
    String qName, Attributes attributes)
        throws SAXException {
    if (qName.equalsIgnoreCase("passenger") && attributes != null){
        System.out.println(attributes.getValue("id"));
    }
}

输出简单的SAX解析器时出错

1 个答案: