2 
My xml is as follows...In this how can I get NAME attribute from status tag using DOM xml   
parsing...

 <tickets type="array">
    <ticket>
           <ticket-id type="integer">6</ticket-id>
           <status>
               <id type="integer">1706729</id>
               <name>Completed</name>
               <colour>black</colour>

           </status>
    </ticket>

    <ticket>
           <ticket-id type="integer">5</ticket-id>
           <status>
               <id type="integer">1706929</id>
               <name>New</name>
               <colour>White</colour>

           </status>
    </ticket>
</tickets>
4

1 回答 1

2

这样你应该能够得到你正在寻找的东西

String xml = "<tickets type=\"array\"><ticket><ticket-id type=\"integer\">6</ticket-id><status><id type=\"integer\">1706729</id><name>Completed</name><colour>black</colour></status></ticket><ticket><ticket-id type=\"integer\">5</ticket-id><status><id type=\"integer\">1706929</id><name>New</name><colour>White</colour></status></ticket></tickets>";
            Log.e("xml parsing ", xml);
            try {
                DocumentBuilderFactory domFactory = DocumentBuilderFactory.newInstance();
                domFactory.setNamespaceAware(true);
                domFactory.setIgnoringComments(true);
                domFactory.setIgnoringElementContentWhitespace(true);
                domFactory.setCoalescing(true);
                DocumentBuilder domBuilder = null;
                Document document = null;

                try {
                    domBuilder = domFactory.newDocumentBuilder();
                }
                catch (ParserConfigurationException parserConfigurationException) {
                    Logger.log(Logger.LEVEL_ERROR, FrameworkHelper.class.getName(), "getDocumentRoot", "ParserConfigurationException caught while creating DocumentBuilder object, " + parserConfigurationException.getMessage());
                }

                document = domBuilder.parse(new ByteArrayInputStream(xml.getBytes()));//try catch
                NodeList ticketList = document.getElementsByTagName("ticket");
                for (int i = 0; i < ticketList.getLength(); i++) {
                  Element ticketNode = (Element) ticketList.item(i);
                  String tickets = ticketNode.getNodeValue();
                  NodeList statusList = ticketNode.getElementsByTagName("status");
                  for (int j = 0; j < statusList.getLength(); j++) {
                    Element statusNode = (Element) statusList.item(j);
                    String statuses = statusNode.getNodeValue();
                    NodeList nameList = statusNode.getElementsByTagName("name");
                    for (int k = 0; k < nameList.getLength(); k++) {
                      Element nameNode = (Element) nameList.item(k);
                      String names = nameNode.getChildNodes().item(0).getNodeValue();
                      Toast.makeText(this, "name: " + names,
                            Toast.LENGTH_SHORT).show();
                    }
                  }
                }
            } catch (SAXException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            } catch (IOException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }

但是，如果层次结构对维护并不重要并且您只想独立处理名称，您也可以直接调用 getElementsByTagName("name")

于 2012-12-24T07:56:55.330 回答