Question

我正在尝试使用Apache Commons JXPath解析XML。但由于某种原因，它在解析xml之后无法识别子节点。这是示例代码：

private static void processUrl(String seed){
    String test = "<?xml version=\"1.0\" encoding=\"UTF-8\"?><feed xmlns=\"http://www.w3.org/2005/Atom\" xmlns:media=\"http://search.yahoo.com/mrss/\" xmlns:openSearch=\"http://a9.com/-/spec/opensearchrss/1.0/\" xmlns:gd=\"http://schemas.google.com/g/2005\" xmlns:yt=\"http://gdata.youtube.com/schemas/2007\"><id>http://gdata.youtube.com/feeds/api/videos</id><logo>http://www.youtube.com/img/pic_youtubelogo_123x63.gif</logo><link rel=\"alternate\" type=\"text/html\" href=\"http://www.youtube.com\"/><author><name>YouTube</name><uri>http://www.youtube.com/</uri></author><generator version=\"2.1\" uri=\"http://gdata.youtube.com\">YouTube data API</generator><openSearch:totalResults>144</openSearch:totalResults><entry><id>http://gdata.youtube.com/feeds/api/videos/P1lDDu9L5YQ</id><published>2010-09-20T17:41:38.000Z</published><updated>2011-09-18T22:15:38.000Z</updated><category scheme=\"http://schemas.google.com/g/2005#kind\" term=\"http://gdata.youtube.com/schemas/2007#video\"/><link rel=\"alternate\" type=\"text/html\" href=\"http://www.youtube.com/watch?v=P1lDDu9L5YQ&amp;feature=youtube_gdata\"/></entry></feed>";
    Document doc = null;
    try{
        DocumentBuilder builder = DocumentBuilderFactory.newInstance().newDocumentBuilder();
        ByteArrayInputStream bais = new ByteArrayInputStream(test.toString().getBytes("UTF8"));
        doc = builder.parse(bais);
        bais.close();

        JXPathContext ctx = JXPathContext.newContext(doc);
        List entryNodes = ctx.selectNodes("/feed/entry");
        System.out.println("number of threadNodes " + entryNodes.size());
        int totalThreads = 0;
        for (Object each : entryNodes) {
            totalThreads++;
            Node eachEntryNode = (Node) each;
            JXPathContext msgCtx = JXPathContext.newContext(eachEntryNode);
            String title = (String) msgCtx.getValue("title");
        }
    }catch (Exception ex) {
        ex.printStackTrace();
    }
}

我之前使用过JXPath，从未遇到过任何问题。我调试了文档对象，它似乎没有子节点（）。我能看到的只是根元素。我也试过没有运气的DOMParser。

DOMParser parser = new DOMParser();
        Document doc = (Document) parser.parseXML(new ByteArrayInputStream(sb0.toString().getBytes("UTF-8")));

如果有人可以提供有关此用途的指示，我将不胜感激。

Answer 1

此问题与JXPath如何处理默认命名空间有关，该命名空间严格遵循XPath 1.0规范。这也解释了删除默认命名空间http://www.w3.org/2005/Atom后它的工作原理。为了使其能够使用默认命名空间，您可以执行以下操作：

JXPathContext ctx = JXPathContext.newContext(doc.getDocumentElement());
// Register the default namespace, giving it a prefix of your choice
ctx.registerNamespace("myfeed", "http://www.w3.org/2005/Atom");

// Now query for entry elements using the registered prefix
List entryNodes = ctx.selectNodes("myfeed:entry");

有关此问题的详细信息，请参阅以下链接。

http://markmail.org/message/7iqw4bjrkwerbh46

Make jxpath namespace aware

使用Commons JXPath进行XML解析的问题

1 个答案: