使用Java解析xml

Question

我正在尝试解析dom元素。

元素元素：

<?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>

我正在尝试获取以下条目：

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

我通过使用xpath解析xml：

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

但是，当我尝试解析Dom Element时出现异常。 有人可以建议一种简单的方法来用Java实现吗？

请查看我的完整代码：

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);
        }
    }

严重：>> java.lang.Error：xpath表达式异常：javax.xml.xpath.XPathExpressionException

Answer 1

您可以使用SAX解析器。 这是SAX解析http://www.mkyong.com/java/how-to-read-xml-file-in-java-sax-parser/的示例

Answer 2

您可以利用NamespaceContext并执行以下操作：

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

我发现从URL提取xml时没有设置名称空间意识。

所以，

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

这样做可以解决我的问题。 如果不这样做，那么如我的示例所示，在解析xml时为XPathFactory实例设置名称空间上下文将无法正常工作。