Question

我有一个xml如下。

<fieldConstraint>
    <fieldBinding>
        <fieldName>instanceId</fieldName>
        <classType>Integer</classType>
        <bindingName>$instanceId</bindingName>
        <genericType>Number</genericType>
    </fieldBinding>
    </fieldConstraint>
        <fieldConstraint>
        <fieldBinding>
        <fieldName></fieldName>
        <classType>String</classType>
        <bindingName>$alertText</bindingName>
        <genericType>String</genericType>
    </fieldBinding>
    <fieldBinding>
        <fieldName>alertText22</fieldName>
        <classType>String22</classType>
        <bindingName>$alertText</bindingName>
        <genericType>String</genericType>
    </fieldBinding>
</fieldConstraint>

我需要通过某个特殊符号（＃）分隔<fieldName>值和<classType>值。

例如：instanceId #Integer（所需的输出应如下所示）

我正在使用java尝试使用xpath。我无法获得所需的输出。

List importedFields =new ArrayList();
 DocumentBuilderFactory domFactory = 
  DocumentBuilderFactory.newInstance();
  domFactory.setNamespaceAware(true); 
  DocumentBuilder builder = domFactory.newDocumentBuilder();
  Document doc = builder.parse("abc.xml");
  XPath xpath = XPathFactory.newInstance().newXPath();
  NamespaceContext context = new NamespaceContextMap(
            "foo", "http://www.cisco.com/BRL");
  xpath.setNamespaceContext(context);
  XPathExpression expr = xpath.compile("concat(//foo:fieldConstraint/foo:fieldBinding/foo:fieldName/text(),//foo:fieldConstraint/foo:fieldBinding/foo:classType/text())");
  Object result = expr.evaluate(doc, XPathConstants.NODESET);
  NodeList nodes = (NodeList) result;
  for (int i = 0; i < nodes.getLength(); i++) {
      Node currentItem = nodes.item(i);
      String value = currentItem.getTextContent();
      importedFields.add(value);

      System.out.println("FIELD ARRAY list IS:"+importedFields);

我收到以下异常。

你的帮助赞赏。

Exception in thread "main" 
com.sun.org.apache.xpath.internal.XPathException: Can not convert #STRING to a NodeList!
    at com.sun.org.apache.xpath.internal.objects.XObject.error(Unknown Source)
    at com.sun.org.apache.xpath.internal.objects.XObject.nodelist(Unknown Source)
    at com.sun.org.apache.xpath.internal.jaxp.XPathExpressionImpl.getResultAsType(Unknown Source)
    at com.sun.org.apache.xpath.internal.jaxp.XPathExpressionImpl.eval(Unknown Source)
    at com.sun.org.apache.xpath.internal.jaxp.XPathExpressionImpl.evaluate(Unknown Source)
    at com.cisco.sample.Xpath.main(Xpath.java:47)
--------------- linked to ------------------
javax.xml.xpath.XPathExpressionException
    at com.sun.org.apache.xpath.internal.jaxp.XPathExpressionImpl.evaluate(Unknown Source)
    at com.cisco.sample.Xpath.main(Xpath.java:47)
Caused by: com.sun.org.apache.xpath.internal.XPathException: Can not convert #STRING to a NodeList!
    at com.sun.org.apache.xpath.internal.objects.XObject.error(Unknown Source)
    at com.sun.org.apache.xpath.internal.objects.XObject.nodelist(Unknown Source)
    at com.sun.org.apache.xpath.internal.jaxp.XPathExpressionImpl.getResultAsType(Unknown Source)
    at com.sun.org.apache.xpath.internal.jaxp.XPathExpressionImpl.eval(Unknown Source)
    ... 2 more

Answer 1

concat函数返回一个String，但您使用的xpath求值会尝试返回NodeList。

尝试：

Object result = expr.evaluate(doc, XPathConstants.STRING);
String s = (String)result;

更新

要获取所有元素，您可以执行以下操作：

NodeList nodeList = doc.getElementsByTagName("fieldBinding");
for (int i = 0; i < nodeList.getLength(); i++) {
  Node node = nodeList.item(i);
  if (node.getNodeType() == Node.ELEMENT_NODE) {
    Element element = (Element) node;
    Node fieldNameNode = element.getElementsByTagName("fieldName").item(0);
    String fieldName = fieldNameNode.getNodeValue();
    Node classTypeNode = element.getElementsByTagName("classType").item(0);
    String classType = classTypeNode.getNodeValue();
  }
}

Answer 2

将以下行更改为XPathConstants.STRING，因为您返回的字符串不是节点列表

Object result = expr.evaluate(doc, XPathConstants.STRING);

和XPATH

XPathExpression expr = xpath.compile("concat(//foo:fieldConstraint/foo:fieldBinding/foo:fieldName/text(),'#', //foo:fieldConstraint/foo:fieldBinding/foo:classType/text())");

使用java的xpath concat（）函数

2 个答案: