使用Java中的XPath和命名空间解析XML

Question

你能帮助我调整这段代码,以便它能够解析XML吗？如果我删除XML命名空间,它可以工作:

String webXmlContent = "<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n" +
                       "<foo xmlns=\"http://foo.bar/boo\"><bar>baz</bar></foo>";
DocumentBuilderFactory domFactory = DocumentBuilderFactory.newInstance();
domFactory.setNamespaceAware(true);
DocumentBuilder builder = domFactory.newDocumentBuilder();
org.w3c.dom.Document doc = builder.parse(new StringInputStream(webXmlContent));

NamespaceContextImpl namespaceContext = new NamespaceContextImpl();
namespaceContext.startPrefixMapping("foo", "http://www.w3.org/2001/XMLSchema-instance");
XPath xpath = XPathFactory.newInstance().newXPath();
xpath.setNamespaceContext(namespaceContext);

XPathExpression expr = xpath.compile("/foo/bar");
Object result = expr.evaluate(doc, XPathConstants.NODESET);
NodeList nodes = (NodeList) result;
System.out.println("Got " + nodes.getLength() + " nodes");

Answer 1

1. 你必须在你的XPath中使用一个前缀,例如:"/ my:foo/my:bar"你可以选择你喜欢的任何前缀 - 它与你在XML中使用或不使用的前缀没有任何关系文件 - 但您必须选择一个.这是XPath 1.0的限制.

2. 您必须执行从"my"到" http://foo.bar/boo "的前缀映射(而不是" http://www.w3.org/2001/XMLSchema-instance ")