java - 如何处理org.xml.sax.SAXParseException？

Question

A 尝试从 https://www.boardgamegeek.com/xmlapi/boardgame/13/catan 解析 XML并获得语言依赖性最高票数的值。

这是代码:

public class DomParserDemo {

    public static void main(String[] args) {

        try {

            DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
            DocumentBuilder dbBuilder = dbFactory.newDocumentBuilder();
            InputSource is = new InputSource(new StringReader("please paste XML from link");
                    Document doc = dbBuilder.parse(is);
            doc.getDocumentElement().normalize();
            System.out.println("Root element :" + doc.getDocumentElement().getNodeName());
            NodeList nodeList = doc.getElementsByTagName("result") ;

            String targetValue = "";
            int maxNumVotes = 0;
            for (int i = 0; i < nodeList.getLength(); i++) {
                Element element = (Element) nodeList.item(i);
                int numVotes = Integer.parseInt(element.getAttribute("numvotes"));
                if (numVotes > maxNumVotes) {
                    maxNumVotes = numVotes;
                    targetValue = element.getAttribute("value");
                }
            }
            System.out.println("Value: " + targetValue + " NumVotes: " + maxNumVotes);

        }
        catch (Exception e) {
            e.printStackTrace();
        }
    }
} 

输出:

[Fatal Error] :1:10703: The entity name must immediately follow the '&' in the entity reference.
org.xml.sax.SAXParseException; lineNumber: 1; columnNumber: 10703; The entity name must immediately follow the '&' in the entity reference.
    at java.xml/com.sun.org.apache.xerces.internal.parsers.DOMParser.parse(DOMParser.java:261)
    at java.xml/com.sun.org.apache.xerces.internal.jaxp.DocumentBuilderImpl.parse(DocumentBuilderImpl.java:339)
    at DomParserDemo.main(DomParserDemo.java:17)

Answer 1

如果您在浏览器中打开 URL 并搜索 &，第一个命中将找到:

BGTG 115 - Spiel des Jahres，然后和；现在

该 & 是有效的实体引用。

如果继续搜索，第二次点击填充找到:

卡坦岛:城市与骑士

这是无效的 XML。 & 后面必须跟有名称和 ;。要在值中包含 &，必须将其转义为 &。

简而言之，该 URL 返回的 XML 是无效，Java XML 解析器会告诉您这一点。