Question

以下是我的XML文件：

<?xml version="1.0" encoding="UTF-8"?>
   <query xmlns:yahoo="http://www.yahooapis.com/v1/base.rng" yahoo:lang="en-GB">
      <results>

        <sector sectorid="1" sectorname="Basic Materials">
          <industry id="112" name="Agricultural Chemicals"/>
          <industry id="132" name="Aluminum"/>
          <industry id="110" name="Chemicals - Major Diversified"/>
          <industry id="131" name="Copper"/>
          <industry id="134" name="Gold"/>
          <industry id="121" name="Independent Oil and Gas"/>
          <industry id="120" name="Major Integrated Oil and Gas"/>
        </sector>

        <sector sectorid="2" sectorname="Conglomerates">
          <industry id="210" name="Conglomerates"/>
        </sector>

        <sector sectorid="7" sectorname="Services">
          <industry id="720" name="Advertising Agencies"/>
          <industry id="773" name="Air Delivery and Freight Services"/>
          <industry id="772" name="Air Services and Others"/>
          <industry id="730" name="Apparel Stores"/>
          <industry id="744" name="Auto Dealerships"/>
        </sector>

     </results>
   </query>

从上面的XML文件中，我希望将属性值sectorid，id和name存储在适当的变量中（我正在使用Java）。我一直在查看不同的XPath表达式，并且我提出了以下代码，但是，在存储java.lang.NumberFormatException: For input string: ""属性的值时会抛出id异常。这是我的代码：

public class XMLToDatabase {

    private static int __SectorID;
    private static int __IndustryID;
    private static String __IndustryName;

    public static void main(String[] args) throws SQLException, UnsupportedEncodingException, ParserConfigurationException, SAXException, IOException, XPathExpressionException {

        try {               
            File _XMLFile = new File("SectorsAndIndustries.xml");

            DocumentBuilderFactory _DocumentBuilderFactory = DocumentBuilderFactory.newInstance();
            _DocumentBuilderFactory.setNamespaceAware(true);

            DocumentBuilder _DocumentBuilder = _DocumentBuilderFactory.newDocumentBuilder();
            Document _Document = _DocumentBuilder.parse(_XMLFile);  

            _Document.getDocumentElement().normalize();

            XPath _XPath = XPathFactory.newInstance().newXPath();

            XPathExpression _XPathExpression = _XPath.compile("//sector | //industry");

            NodeList _NodeList = (NodeList) _XPathExpression.evaluate(_Document, XPathConstants.NODESET);


            for (int i = 0; i < _NodeList.getLength(); i++) {
                Node _Node = _NodeList.item(i);

                if(_Node.getNodeType() == Node.ELEMENT_NODE) {
                    Element _Element = (Element) _Node;

                    __SectorID = Integer.parseInt(_Element.getAttribute("sectorid"));
                    __IndustryID = Integer.parseInt(_Element.getAttribute("id"));
                    __IndustryName = _Element.getAttribute("name");
                }

            System.out.println(__SectorID + ", " + __IndustryID + ", " + __IndustryName);
            }
        } catch (Exception e) {
             e.printStackTrace();
        }

    }

}

有人可以帮我确定是否是我犯了错误的XPath Expression，或者我存储第二个变量__IndustryID的方式是否正确？因为第一个变量__SectorID正确存储了值1，但是为__IndustryID抛出了上述异常。理想情况下，我希望每次执行for循环时都存储所有3个属性的值，以将它们保存到数据库表中。如果需要更多信息，请告诉我。

Answer 1

据我所知，您正在使用sector或industry个元素的节点编译节点列表。对于其中的每一个，您都希望检索sectorid和id属性 - 但显然，没有任何元素都具有这两个属性。

更好的方法是

找到所有sector元素，并为每个元素打印出扇区ID
为每个sector元素遍历所有名为industry的子元素（这需要对每个sector元素应用第二个XPath表达式，但这是一个微不足道的元素：{{1} }）
并输出每个"industry"

Answer 2

Mathias提出了正确的方法，我已经为它提出了一个解决方案，稍作修改：

public class XMLToDatabase {

    private static int __SectorID;
    private static int __IndustryID;
    private static String __IndustryName;

    public static void main(String[] args) throws SQLException,
            UnsupportedEncodingException, ParserConfigurationException,
            SAXException, IOException, XPathExpressionException {

        try {
            File _XMLFile = new File("C:/Users/Sachin/Desktop/SectorsAndIndustries.xml");
            DocumentBuilderFactory _DocumentBuilderFactory = DocumentBuilderFactory.newInstance();
            DocumentBuilder _DocumentBuilder = _DocumentBuilderFactory.newDocumentBuilder();
            Document _Document = _DocumentBuilder.parse(_XMLFile);
            _Document.getDocumentElement().normalize();

            XPath _XPath = XPathFactory.newInstance().newXPath();

            NodeList _NodeList1 = (NodeList) _XPath.evaluate("/results/sector", _Document, XPathConstants.NODESET);

            for (int i = 0; i < _NodeList1.getLength(); i++) {
                Element _Element1 = (Element) _NodeList1.item(i);

                __SectorID = Integer.parseInt(_Element1.getAttribute("sectorid"));

                NodeList _NodeList2 = (NodeList) _XPath.evaluate("industry", _Element1, XPathConstants.NODESET);

                for (int k=0; k < _NodeList2.getLength(); k++) {
                    __IndustryID = Integer.parseInt(_XPath.evaluate("industry[position()=" + (k + 1) + "]/@id", _Element1));
                    __IndustryName = _XPath.evaluate("industry[position()=" + (k + 1) + "]/@name", _Element1);

                    System.out.println(__SectorID + ", " + __IndustryID + ", " + __IndustryName);
                }
                System.out.println("\n-----------\n");
            }

        } catch (Exception e) {
            e.printStackTrace();
        }
    }
}

XPath从父节点和子节点中选择属性值

2 个答案: