从java中的字符串中删除无效的XML字符

Question

您好我想从字符串中删除所有无效的XML字符.我想使用string.replace方法的正则表达式.

喜欢

line.replace(regExp,"");

什么是正确的regExp使用？

无效的XML字符是不是这样的一切:

[#x1-#xD7FF] | [#xE000-#xFFFD] | [#x10000-#x10FFFF]

谢谢.

Answer 1

Java的正则表达式支持增补字符,因此您可以使用两个UTF-16编码的字符指定那些高范围.

以下是删除XML 1.0中非法字符的模式:

// XML 1.0
// #x9 | #xA | #xD | [#x20-#xD7FF] | [#xE000-#xFFFD] | [#x10000-#x10FFFF]
String xml10pattern = "[^"
                    + "\u0009\r\n"
                    + "\u0020-\uD7FF"
                    + "\uE000-\uFFFD"
                    + "\ud800\udc00-\udbff\udfff"
                    + "]";

大多数人都想要XML 1.0版本.

以下是删除XML 1.1中非法字符的模式:

// XML 1.1
// [#x1-#xD7FF] | [#xE000-#xFFFD] | [#x10000-#x10FFFF]
String xml11pattern = "[^"
                    + "\u0001-\uD7FF"
                    + "\uE000-\uFFFD"
                    + "\ud800\udc00-\udbff\udfff"
                    + "]+";

你需要使用String.replaceAll(...)而不是String.replace(...).

String illegal = "Hello, World!\0";
String legal = illegal.replaceAll(pattern, "");

Answer 2

我们应该考虑代理人物吗？否则'(当前> = 0x10000)&&(当前<= 0x10FFFF)'永远不会成立.

还测试了正则表达式方式似乎比以下循环慢.

if (null == text || text.isEmpty()) {
    return text;
}
final int len = text.length();
char current = 0;
int codePoint = 0;
StringBuilder sb = new StringBuilder();
for (int i = 0; i < len; i++) {
    current = text.charAt(i);
    boolean surrogate = false;
    if (Character.isHighSurrogate(current)
            && i + 1 < len && Character.isLowSurrogate(text.charAt(i + 1))) {
        surrogate = true;
        codePoint = text.codePointAt(i++);
    } else {
        codePoint = current;
    }
    if ((codePoint == 0x9) || (codePoint == 0xA) || (codePoint == 0xD)
            || ((codePoint >= 0x20) && (codePoint <= 0xD7FF))
            || ((codePoint >= 0xE000) && (codePoint <= 0xFFFD))
            || ((codePoint >= 0x10000) && (codePoint <= 0x10FFFF))) {
        sb.append(current);
        if (surrogate) {
            sb.append(text.charAt(i));
        }
    }
}

Answer 3

到目前为止，所有这些答案都只是替换了角色本身。但有时 XML 文档会包含无效的 XML 实体序列，从而导致错误。例如，如果您的 xml 中有，则 java xml 解析器将抛出Illegal character entity: expansion character (code 0x2 at ....

这是一个简单的java程序，可以替换那些无效的实体序列。

  public final Pattern XML_ENTITY_PATTERN = Pattern.compile("\\&\\#(?:x([0-9a-fA-F]+)|([0-9]+))\\;");

  /**
   * Remove problematic xml entities from the xml string so that you can parse it with java DOM / SAX libraries.
   */
  String getCleanedXml(String xmlString) {
    Matcher m = XML_ENTITY_PATTERN.matcher(xmlString);
    Set<String> replaceSet = new HashSet<>();
    while (m.find()) {
      String group = m.group(1);
      int val;
      if (group != null) {
        val = Integer.parseInt(group, 16);
        if (isInvalidXmlChar(val)) {
          replaceSet.add("&#x" + group + ";");
        }
      } else if ((group = m.group(2)) != null) {
        val = Integer.parseInt(group);
        if (isInvalidXmlChar(val)) {
          replaceSet.add("&#" + group + ";");
        }
      }
    }
    String cleanedXmlString = xmlString;
    for (String replacer : replaceSet) {
      cleanedXmlString = cleanedXmlString.replaceAll(replacer, "");
    }
    return cleanedXmlString;
  }

  private boolean isInvalidXmlChar(int val) {
    if (val == 0x9 || val == 0xA || val == 0xD ||
            val >= 0x20 && val <= 0xD7FF ||
            val >= 0x10000 && val <= 0x10FFFF) {
      return false;
    }
    return true;
  }