本文介绍了解析java中的xml字符串?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
如何解析存储在java字符串对象中的xml?
how do you parse xml stored in a java string object?
Java的XMLReader仅解析URI或输入流中的XML文档。是不是可以从包含xml数据的String解析?
Java's XMLReader only parses XML documents from a URI or inputstream. is it not possible to parse from a String containing an xml data?
现在我有以下内容:
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();
}
在我的处理程序上我有这个:
And on my handler i have this:
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);
}
}
提前致谢
推荐答案
可以 。
InputSource 可以在
An InputSource can take a Reader in its constructor
因此,您可以通过
new InputSource(new StringReader("... your xml here....")));
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