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我需要在 XML 文档中找到文本元素的确切 XPath。我认为这样做的一种方法是将 Document 转换为字符串，在子字符串周围添加一个临时标记，将其转换回 Document，然后找到 XPath。

这是我到目前为止所拥有的：

public String findXPathInXMLString(int startIndex, int endIndex, String string) throws IOException, ParserConfigurationException, SAXException {
    Conversion conversion = new Conversion();
    String xpath;

    //Step 1. Replace start to end index with temporary tag in string document
    StringBuilder stringBuilder = new StringBuilder(string);
    stringBuilder.replace(startIndex, endIndex, "<findXPathInXMLStringTemporaryTag>" + string.substring(startIndex, endIndex) + "</findXPathInXMLStringTemporaryTag>");

    //Step 2. Convert string document to DOM document & Find XPath of temporary tag in DOM document
    xpath = "/" + getXPath(conversion.stringToDocument(stringBuilder.toString()), "findXPathInXMLStringTemporaryTag");

    //Step 3. Cut off last part of the XPath
    //xpath = xpath.substring(0, 2).replace("/documentXPathTemporaryTag", "");

    //Step 4. Return the XPath
    return xpath;
}

public String getXPath(Document root, String elementName) {
    try {
        XPathExpression expr = XPathFactory.newInstance().newXPath().compile("//" + elementName);
        Node node = (Node) expr.evaluate(root, XPathConstants.NODE);

        if (node != null) {
            return getXPath(node);
        }
    } catch (XPathExpressionException e) {
    }

    return null;
}

public String getXPath(Node node) {
    if (node == null || node.getNodeType() != Node.ELEMENT_NODE) {
        return "";
    }
    return getXPath(node.getParentNode()) + "/" + node.getNodeName();
}

到目前为止，我遇到的问题是该方法getXPath没有放置，[x]因此返回的 XPath 是错误的，因为子字符串可能位于[3]特定标记的 rd 实例中，在这种情况下，XPath 将应用于所有具有相同路径的节点。我想得到一个只能引用一个特定元素的确切路径。

4

1 回答 1

2

好的，这是怎么回事（以ideone 为例）：

我改变了startIndex，endIndex只是index。临时节点可以附加到文本中的单个点。

public static String findXPathInXMLString(int index, String string) throws XPathExpressionException, SAXException, ParserConfigurationException, IOException {
    String xpath;

    //Step 1. Insert temporary tag in insert location
    StringBuilder stringBuilder = new StringBuilder(string);
    stringBuilder.insert(index, "<findXPathInXMLStringTemporaryTag />");

    Document document = DocumentBuilderFactory.newInstance().newDocumentBuilder().parse(
        new ByteArrayInputStream(stringBuilder.toString().getBytes())
      );

    //Step 2. Convert string document to DOM document & Find XPath of temporary tag in DOM document
    xpath = getXPath(document, "findXPathInXMLStringTemporaryTag");

    //Step 3. Cut off last part of the XPath
    xpath = xpath.replace("/findXPathInXMLStringTemporaryTag", "");

    //Step 4. Return the XPath
    return xpath;
}

private static String getXPath(Document root, String elementName) throws XPathExpressionException 
{
  XPathExpression expr = XPathFactory.newInstance().newXPath().compile("//"+elementName);
  Node node = (Node)expr.evaluate(root, XPathConstants.NODE);


  if(node != null) {
      return getXPath(node);
  }

  return null;
}

private static String getXPath(Node node) throws XPathExpressionException {
    if(node == null || node.getNodeType() != Node.ELEMENT_NODE) {
        return "";
    }

    return getXPath(node.getParentNode()) + "/" + node.getNodeName() + getIndex(node);
}

private static String getIndex(Node node) throws XPathExpressionException {
    XPathExpression expr = XPathFactory.newInstance().newXPath().compile("count(preceding-sibling::*[local-name() = '" + node.getNodeName() + "'])");
    int result = (int)(double)(Double)expr.evaluate(node, XPathConstants.NUMBER);

    if(result == 0){
        return "";
    }
    else {
        return "[" + (result + 1) + "]";
    }
}
于 2013-01-22T11:35:31.117 回答