解析java中的xml字符串?
你如何解析存储在java字符串对象中的xml?解析java中的xml字符串?
Java的XMLReader仅解析来自URI或输入流的XML文档。是不是可以从包含xml数据的字符串解析?
现在,我有以下几点:
try {
SAXParserFactory factory = SAXParserFactory.newInstance();
SAXParser sp = factory.newSAXParser();
XMLReader xr = sp.getXMLReader();
ContactListXmlHandler handler = new ContactListXmlHandler();
xr.setContentHandler(handler);
xr.p
} catch (ParserConfigurationException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (SAXException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
而且对我的处理程序,我有这样的:提前
public class ContactListXmlHandler extends DefaultHandler implements Resources {
private List<ContactName> contactNameList = new ArrayList<ContactName>();
private ContactName contactItem;
private StringBuffer sb;
public List<ContactName> getContactNameList() {
return contactNameList;
}
@Override
public void startDocument() throws SAXException {
// TODO Auto-generated method stub
super.startDocument();
sb = new StringBuffer();
}
@Override
public void startElement(String uri, String localName, String qName,
Attributes attributes) throws SAXException {
// TODO Auto-generated method stub
super.startElement(uri, localName, qName, attributes);
if(localName.equals(XML_CONTACT_NAME)){
contactItem = new ContactName();
}
sb.setLength(0);
}
@Override
public void characters(char[] ch, int start, int length){
// TODO Auto-generated method stub
try {
super.characters(ch, start, length);
} catch (SAXException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
sb.append(ch, start, length);
}
@Override
public void endDocument() throws SAXException {
// TODO Auto-generated method stub
super.endDocument();
}
/**
* where the real stuff happens
*/
@Override
public void endElement(String uri, String localName, String qName)
throws SAXException {
// TODO Auto-generated method stub
//super.endElement(arg0, arg1, arg2);
if(contactItem != null){
if (localName.equalsIgnoreCase("title")) {
contactItem.setUid(sb.toString());
Log.d("handler", "setTitle = " + sb.toString());
} else if (localName.equalsIgnoreCase("link")) {
contactItem.setFullName(sb.toString());
} else if (localName.equalsIgnoreCase("item")){
Log.d("handler", "adding rss item");
contactNameList.add(contactItem);
}
sb.setLength(0);
}
}
感谢
的的InputSource可以采取读者在其constructor
所以,你可以通过StringReader
new InputSource(new StringReader("... your xml here....")));
ch eers mate。似乎工作。 – jonney 2010-10-11 14:06:45
看看这个:http://www.rgagnon.com/javadetails/java-0573.html
import javax.xml.parsers.*;
import org.xml.sax.InputSource;
import org.w3c.dom.*;
import java.io.*;
public class ParseXMLString {
public static void main(String arg[]) {
String xmlRecords =
"<data>" +
" <employee>" +
" <name>John</name>" +
" <title>Manager</title>" +
" </employee>" +
" <employee>" +
" <name>Sara</name>" +
" <title>Clerk</title>" +
" </employee>" +
"</data>";
try {
DocumentBuilderFactory dbf =
DocumentBuilderFactory.newInstance();
DocumentBuilder db = dbf.newDocumentBuilder();
InputSource is = new InputSource();
is.setCharacterStream(new StringReader(xmlRecords));
Document doc = db.parse(is);
NodeList nodes = doc.getElementsByTagName("employee");
// iterate the employees
for (int i = 0; i < nodes.getLength(); i++) {
Element element = (Element) nodes.item(i);
NodeList name = element.getElementsByTagName("name");
Element line = (Element) name.item(0);
System.out.println("Name: " + getCharacterDataFromElement(line));
NodeList title = element.getElementsByTagName("title");
line = (Element) title.item(0);
System.out.println("Title: " + getCharacterDataFromElement(line));
}
}
catch (Exception e) {
e.printStackTrace();
}
/*
output :
Name: John
Title: Manager
Name: Sara
Title: Clerk
*/
}
public static String getCharacterDataFromElement(Element e) {
Node child = e.getFirstChild();
if (child instanceof CharacterData) {
CharacterData cd = (CharacterData) child;
return cd.getData();
}
return "?";
}
}
你的XML可能是很简单的手工解析使用把解析XML字符串DOM或SAX API,但我仍然建议使用XML序列化API,例如JAXB,XStream或Simple,而不是因为编写自己的XML序列化/请求者ialization代码是一个拖动。
注意,XStream的常见问题错误地声称,您必须使用生成的类与JAXB:
如何XStream的比较JAXB(用于XML绑定的Java API)?
JAXB是一个Java绑定工具。它从一个模式生成Java代码,您可以将这些类转换为匹配处理后的模式的XML并返回。请注意,你不能使用自己的对象,你必须使用生成的东西。
似乎这是事实,但JAXB 2.0不再要求您使用从模式生成的Java类。
如果你走这条路,一定要检查出系列化的并排侧比较/编组我所提到的API:
http://blog.bdoughan.com/2010/10/how-does-jaxb-compare-to-xstream.html http://blog.bdoughan.com/2010/10/how-does-jaxb-compare-to-simple.html
你可以看看这个:http://*.com/questions/247161/how-do-i-turn-a-string-into-a-stream-in -java – foret 2010-10-11 14:00:58