如何获取具有特定属性值的特定XML元素？

Question

我试图通过取所有" <Type>"元素来解析URL中的XML文件,其中参数type_id ="4218"??

XML文档:

<BSQCUBS Version="0.04" Date="Fri Dec 9 11:43:29 GMT 2011" MachineDate="Fri, 09 Dec 2011 11:43:29 +0000">
  <Class class_id="385">
    <Title>Football Matches</Title>
    <Type type_id="4264" type_minbet="0.1" type_maxbet="2000.0">
      ...
    </Type>
    <Type type_id="5873" type_minbet="0" type_maxbet="0">
      ...
    </Type>
    <Type type_id="4725" type_minbet="0.1" type_maxbet="2000.0">
      ...
    </Type>
    <Type type_id="4218" type_minbet="0.1" type_maxbet="2000.0">
      ...
    </Type>
    <Type type_id="4221" type_minbet="0.1" type_maxbet="2000.0">
      ...
    </Type>
    <Type type_id="4218" type_minbet="0.1" type_maxbet="2000.0">
      ...
    </Type>
    <Type type_id="4299" type_minbet="0.1" type_maxbet="2000.0">
      ...
    </Type>
  </Class>
</BSQCUBS>

这是我的Java代码:

 DocumentBuilder db = dbf.newDocumentBuilder();
 Document doc = db.parse(new URL("http://cubs.bluesq.com/cubs/cubs.php?action=getpage&thepage=385.xml").openStream());

 doc.getDocumentElement().normalize();

 NodeList nodeList = doc.getElementsByTagName("Type");
 System.out.println("ukupno:"+nodeList.getLength());
 if (nodeList != null && nodeList.getLength() > 0) {
   for (int j = 0; j < nodeList.getLength(); j++) {
     Element el = (org.w3c.dom.Element) nodeList.item(j);
     type_id = Integer.parseInt(el.getAttribute("type_id"));
     System.out.println("type id:"+type_id);
   }
 }

这段代码给了我所有的元素,我不想要那个,我想要属性type_id ="4218"的所有元素!

Answer 1

XPath是您的正确选择:

DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
DocumentBuilder builder = factory.newDocumentBuilder();
Document doc = builder.parse("<Your xml doc uri>");
XPathFactory xPathfactory = XPathFactory.newInstance();
XPath xpath = xPathfactory.newXPath();
XPathExpression expr = xpath.compile("//Type[@type_id=\"4218\"]");
NodeList nl = (NodeList) expr.evaluate(doc, XPathConstants.NODESET);

并迭代 nl

Answer 2

你在循环中错过了一个条件:

 if(nodeList != null && nodeList.getLength() > 0){
     for (int j = 0; j < nodeList.getLength(); j++) {
         Element el = (org.w3c.dom.Element) nodeList.item(j);
         if (el.hasAttribute("type_id") && el.getAttribute("type_id").equals("4218")) {
              type_id = Integer.parseInt(el.getAttribute("type_id"));

              System.out.println("type id:"+type_id);
         }
     }
}

此外,您不需要测试getElementsByTagName返回的NodeList是否为null,因此您可以在循环之前删除if.

一般情况下,使用XPath可能会更好.

Answer 3

您可以使用XPath.XPath用于浏览XML文档中的元素和属性.在Java中有一些很好的Xpath实现.

对你而言

XPath xpath = XPathFactory.newInstance().newXPath();
XPathExpression expr = xpath.compile("//Type[@type_id=\"4218\"]");
Object exprResult = expr.evaluate(doc, XPathConstants.NODESET);
NodeList nodeList = (NodeList) exprResult;