java - 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>

我正在使用 xpath 解析 xml:

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

但是，当我尝试解析 Dom Element 时遇到异常。有人可以建议一种简单的方法来在 Java 中实现此目的吗？

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 解析的示例 http://www.mkyong.com/java/how-to-read-xml-file-in-java-sax-parser/