使用Java解析xml

Question

I am trying to parse a dom element.

Element element:

<?xml version="1.0" encoding="UTF-8"?>
<feed xmlns="http://www.w3.org/2005/Atom">
  <id>http://X/feed2</id>
  <title>Sample Feed</title>
  <entry>
    <id>http://X/feed2/104</id>
    <title>New Title</title>
  </entry>
</feed>

I am trying to fetch the following entry:

<entry>
  <id>http://top.cs.vt.edu/libx2/vsony7@vt.edu/feed2/104</id>
  <title>New Title</title>
</entry>

I am parsing the xml by using the xpath:

"/atom:feed/atom:entry[atom:id=\\"http://X/feed2/104\\"]"

But, I get an exception when I try to parse Dom Element. Can someone suggest a simple approach to achieve this in Java?

Please see my full code:

public static parseXml() {
        String externalEntryIdUrl = "http://theta.cs.vt.edu/~rupen/thirtylibapps/137";
        String externalFeedUrl = StringUtils.substringBeforeLast(externalEntryIdUrl, "/");
        try {
            URL url = new URL(externalFeedUrl);
            InputStream externalXml = new BufferedInputStream(url.openStream());
            DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
            DocumentBuilder db = dbf.newDocumentBuilder();
            Document doc = db.parse(externalXml);
            Element externalFeed = doc.getDocumentElement();
            String atomNameSpace = "xmlns:atom=\"http://www.w3.org/2005/Atom\"";
            String entryIdPath = String.format("//%s:entry[%s:id=%s]", atomNameSpace, atomNameSpace, externalEntryIdUrl);
            Element externalEntry = (Element) XPathSupport.evalNode(entryIdPath, externalFeed);
        } catch (Exception ex) {
            // Throw exception
        }
    }

static synchronized Node evalNode(String xpathExpr, Node node) {
    NodeList result = evalNodeSet(xpathExpr, node);
    if (result.getLength() > 1)
        throw new Error ("More than one node for:" + xpathExpr);
    else if (result.getLength() == 1)
        return result.item(0);
    else
        return null;
}

static synchronized NodeList evalNodeSet(String xpathExpr, Node node) {
        try {
                static XPath xpath = factory.newXPath();
                xpath.setNamespaceContext(context);

                static NamespaceContext context = new NamespaceContext() {
                    private Map<String, String> prefix2URI = new HashMap<String, String>();
                    {
                        prefix2URI.put("libx", "http://libx.org/xml/libx2");
                        prefix2URI.put("atom", "http://www.w3.org/2005/Atom");
                    }
                };

            XPathExpression expr = xpath.compile(xpathExpr);
            Object result = expr.evaluate(node, XPathConstants.NODESET);
            return (NodeList)result;
        } catch (XPathExpressionException xpee) {
            throw new Error ("An xpath expression exception: " + xpee);
        }
    }

SEVERE: >>java.lang.Error: An xpath expression exception: javax.xml.xpath.XPathExpressionException

Answer 1

You can use SAX parser. Here is a example for SAX parsing http://www.mkyong.com/java/how-to-read-xml-file-in-java-sax-parser/

Answer 2

You could leverage a NamespaceContext and do something like the following:

package forum9059851;

import java.io.FileInputStream;
import java.util.Iterator;
import javax.xml.namespace.NamespaceContext;
import javax.xml.xpath.*;
import org.w3c.dom.Element;
import org.xml.sax.InputSource;

public class Demo {

    public static void main(String[] args) {
        try {
            XPathFactory xpf = XPathFactory.newInstance();
            XPath xp = xpf.newXPath();
            xp.setNamespaceContext(new MyNamespaceContext());
            XPathExpression xpe = xp.compile("ns:feed/ns:entry");
            FileInputStream xmlStream = new FileInputStream("src/forum9059851/input.xml");
            InputSource xmlInput = new InputSource(xmlStream);
            Element result = (Element) xpe.evaluate(xmlInput, XPathConstants.NODE);
            System.out.println(result);
        } catch (Exception ex) {
            // Throw exception
        }
    }

    private static class MyNamespaceContext implements NamespaceContext {

        public String getNamespaceURI(String prefix) {
            if("ns".equals(prefix)) {
                return "http://www.w3.org/2005/Atom";
            }
            return null;
        }

        public String getPrefix(String namespaceURI) {
            return null;
        }

        public Iterator getPrefixes(String namespaceURI) {
            return null;
        }

    }

}

Answer 3

如果您不想重新发明轮子并想解析提要数据，我建议您使用已经可用的Rome库。

Answer 4

I figured that I didn't set the namespace awareness while fetching the xml from a URL.

So,

DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
dbf.setNamespaceAware(true);

Doing so fixes my issue. Without doing this, setting the namespace context for XPathFactory instance while parsing the xml as shown in my example doesn't work by itself.