在Java中使用XPath进行XML解析
[英]XML parsing using XPath in Java
问题描述
Hi! 
嗨！ I've spent some time to parse an XML document with XPath. 我花了一些时间用XPath解析XML文档。 It seeams to be a simple task but I got in troubles since the begining. 这似乎是一个简单的任务，但是自从一开始我就遇到了麻烦。 My code is : 我的代码是：
public class QueryXML3 {
public static void main(String[] args) {
DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
factory.setNamespaceAware(true);
DocumentBuilder builder;
Document doc = null;
try {
builder = factory.newDocumentBuilder();
//doc = builder.parse("SampleExample.xml");
InputStream is = QueryXML3.class.getClassLoader().getResourceAsStream("SampleXml.xml");
doc = builder.parse(is);
XPathFactory xpathFactory = XPathFactory.newInstance();
// Create XPath object
XPath xpath = xpathFactory.newXPath();
Node parNode = getParameterNode(doc, xpath);
System.out.println("parameter node:" + parNode);
NodeList res = getParameterNodeList(doc, xpath );
System.out.println("List of nodes" + res);
} catch (ParserConfigurationException | SAXException | IOException e) {
e.printStackTrace();
}
}
public static Node getParameterNode(Document doc, XPath xpath) {
Node res = null;
try {
res = (Node) xpath.evaluate("/definitions/process", doc, XPathConstants.NODE);
} catch (XPathExpressionException e) {
e.printStackTrace();
}
return res;
}
public static NodeList getParameterNodeList(Document doc, XPath xpath) {
NodeList nodeList = null;
try {
nodeList = (NodeList) xpath.evaluate("/definitions/process", doc, XPathConstants.NODESET);
for (int i = 0; i > nodeList.getLength(); i++) {
System.out.print(nodeList.item(i).getNodeName() + " ");
}
} catch (XPathExpressionException e) {
e.printStackTrace();
}
return nodeList;
}
}
As a result i get this: 结果我得到这个：
parameter node:[process: null] List of nodes com.sun.org.apache.xml.internal.dtm.ref.DTMNodeList@2f17aadf 参数node：[process：null]节点列表com.sun.org.apache.xml.internal.dtm.ref.DTMNodeList@2f17aadf
I just want to output all the nodes of my xml file and theire attributes... 我只想输出我的xml文件和theire属性的所有节点...
2 个解决方案
解决方案1
0 2013-11-13 11:13:38
You are really asking how to serialize an Element to a string - use either a Transformer or DOMImplementationLS . 您实际上是在问如何将Element序列化为字符串-使用Transformer或DOMImplementationLS 。
The NodeList type has no toString() contract and the implementation does not override the default Object.toString() . NodeList类型没有toString()协定，并且实现不覆盖默认的Object.toString() 。 You need to iterate over the nodes and serialize each Element as above. 您需要遍历节点并如上所述序列化每个Element 。
解决方案2
0 2013-11-13 11:19:39
You could easily parse an XML file in java using a 3rd party package such as JSoup or JDom . 您可以使用第3方软件包（例如JSoup或JDom）轻松地在Java中解析XML文件。
As an example, here is some simple output of an XML files elements using JSoup: 例如，这是使用JSoup的XML文件元素的一些简单输出：
<note>
<to>Tove</to>
<from>Jani</from>
<heading>Reminder</heading>
<body>Don't forget me this weekend!</body>
</note>
Java code printing all elements and the selected <from> -element: Java代码打印所有元素和选定的<from> -element：
String xml = "<note>\n"
+ "<to>Tove</to>\n"
+ "<from>Jani</from>\n"
+ "<heading>Reminder</heading>\n"
+ "<body>Don't forget me this weekend!</body>\n"
+ "</note>";
Document doc = Jsoup.parse(xml, "", Parser.xmlParser());
for (Element e : doc.children()) {
System.out.println(e);
}
Element fromElement = doc.select("from").first();
System.out.println("\nThis is the <from>-element content:\n" + fromElement);
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