未找到属性时返回null的JAVA Xpath表达式

在下面的xml示例中

      <employee>
        <payment contractType="1">
          <income type="0">
            <gr amount="2063.00" kae="211" code="1" />
            <gr amount="400.00" kae="215" code="6" />
            <et amount="47.55" kae="292" code="4012501" />
            <et amount="105.21" kae="293" code="4052000" />
            <de amount="88.15" code="4003101" />
          </income>
         </payment>
      </employee>
      <employee>
        <payment contractType="1">
          <income type="0">
            <gr amount="70.00" kae="213" code="4" />
            <gr amount="1560.00" kae="211" code="1" />
          </income>
       </payment>
    </employee>

我需要为“code”=“4”获取“amount”的值。如果income节点不包含此类数据（gr with code=“4”），我需要返回null或boolean false之类的值。目的是查看xml文件中的所有员工，如果他们没有任何金额，则将其加载到带有0的Arraylist中，代码为4，否则加载金额值

我用于此部件的代码：

    public class ReadXMLfile {
    public static void main(String[] args) {

    try {
        FileInputStream file = new FileInputStream(new File("E2015_1_1.xml"));

        DocumentBuilderFactory builderFactory = DocumentBuilderFactory.newInstance();

        DocumentBuilder builder =  builderFactory.newDocumentBuilder();

        Document xmlDocument = builder.parse(file);

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


    expression = "/psp/body/organizations/organization/employees/employee/payment/income[@type='0']/gr";
    nodeList = (NodeList) xPath.compile(expression).evaluate(xmlDocument, XPathConstants.NODESET);
    ArrayList<String> gr4incomeList = new ArrayList<String>();

    for (int i = 0; i < nodeList.getLength(); i++) {

          String acode = (String)nodeList.item(i).getAttributes().getNamedItem("code").getNodeValue();
          System.out.println("acode = '" + acode + "'");

               if (acode.equals("4")){
                    System.out.println(nodeList.item(i).getAttributes().getNamedItem("amount").getNodeValue());
                    gr4incomeList.add(nodeList.item(i).getAttributes().getNamedItem("amount").getNodeValue());
                    System.out.println("array = " + gr4incomeList.get(i));
        }else
                    gr4incomeList.add("0000");

    }

     } catch (FileNotFoundException e) {
        e.printStackTrace();
    } catch (SAXException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    } catch (ParserConfigurationException e) {
        e.printStackTrace();
    } catch (XPathExpressionException e) {
        e.printStackTrace();
    }
}

公共类ReadXMLfile{
公共静态void main（字符串[]args）{
试一试{
FileInputStream文件=新的FileInputStream（新文件（“E2015_1_1.xml”）；
DocumentBuilderFactory builderFactory=DocumentBuilderFactory.newInstance（）；
DocumentBuilder=builderFactory.newDocumentBuilder（）；
Document xmlDocument=builder.parse（文件）；
XPath=XPathFactory.newInstance（）.newXPath（）；
expression=“/psp/body/organization/organization/employees/employees/payment/income[@type='0']/gr”；
nodeList=（nodeList）xPath.compile（expression）.evaluate（xmlDocument，XPathConstants.NODESET）；
ArrayList gr4incomeList=新的ArrayList（）；
for（int i=0；i

}

问题是它会在arraylist中为找到的任何gr写入“0000”，但代码为“4”的gr除外

我真的被卡住了

---

有什么想法吗？谢谢你们

GR4收入表添加（“0000”）；如果你想写“0”，你不应该使用gr4incomeList.add（“0”）；是否可以有多个代码为“4”的项目，还是只有一个？@jlrise每位员工只能有一个代码为“4”的项目。感谢you@nafas我需要将其存储为4位字符串，thanks@Bonzay似乎你的代码只适用于第一个标记为code=“4”的员工。我说得对吗？

XPathExpression exp = xpath.compile("/employees/employee");
NodeList nodeList = (NodeList)exp.evaluate(xmlDocument, XPathConstants.NODESET);
XPathExpression grexp = xpath.compile("payment/income/gr[@code='4']");
XPathExpression amexp = xpath.compile("payment/income/gr[@code='4']/@amount");
for( int i = 0; i < nodeList.getLength(); ++i ){
    Node item = nodeList.item( i );
    Object resexp1 = grexp.evaluate( item, XPathConstants.NODE );
    if( resexp1 != null ){
        String resexp2 = amexp.evaluate( item );
        System.out.println( resexp2 );
    } else {
        System.out.println( "0000" );
    }
}

0000
70.00