Question

我需要用一个独特的TAG替换一堆xml标签，我找到了一个解决方案，但我不确定它是那里最好的......你会建议什么？

/** Group tags replacement.
 * @param input <mytag><tag>val1</tag><tag>val2</tag></mytag>
 * @param tag tag
 * @param replacement <tag>val</tag>
 * @return <mytag><tag2>value</tag2></mytag>
 */
public static String replaceGroupTags(String input, String tag, String replacement) {
    Pattern replace = Pattern.compile("<" + tag + ">" + ".*" + "</" + tag + ">");
    Matcher matcher = replace.matcher(input);

    int start = Integer.MIN_VALUE;
    int end = Integer.MAX_VALUE;
    while (matcher.find()) {
        if (start == Integer.MIN_VALUE) start = matcher.start();
        if (end <= Integer.MAX_VALUE) end = matcher.end();
    }
    StringBuffer stringBuffer = new StringBuffer(input);
    return stringBuffer.replace(start, end, replacement).toString();
}

Answer 1

尝试使用正则表达式解析XML是一个坏主意。以下是使用DOM的解决方案：

public static void replaceGroupTags(File input, File output, String tag, String replacement) {
             Document doc = DocumentBuilderFactory.newInstance().newDocumentBuilder().parse(new InputSource(inputFile));

            XPath xpath = XPathFactory.newInstance().newXPath();
            NodeList nodes = (NodeList)xpath.evaluate("//SomeNode/" + tag, doc, XPathConstants.NODESET);

            // replace all occurences
            for (int i = 0; i < nodes.getLength(); i++) {
              nodes.item(i).setTextContent(replacement);
            }

            // save result to the file 
            Transformer xformer = TransformerFactory.newInstance().newTransformer();
            xformer.transform(new DOMSource(doc), new StreamResult(output);
    }

遍历文件，在tag节点内查找SomeNode所有出现的内容，并将其内容替换为replacement

使用Java替换一组重复的xml标记

1 个答案: