从java中的字符串中删除无效的XML字符

Question

嗨 我想从字符串中删除所有无效的 XML 字符。 我想在 string.replace 方法中使用正则表达式。

喜欢

line.replace(regExp,"");

什么是正确的正则表达式？

无效的 XML 字符是不是这个的一切：

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

谢谢。

Answer 1

String xmlData = xmlData.codePoints().filter(c -> isValidXMLChar(c)).collect(StringBuilder::new,
                StringBuilder::appendCodePoint, StringBuilder::append).toString();

private boolean isValidXMLChar(int c) {
    if((c == 0x9) ||
       (c == 0xA) ||
       (c == 0xD) ||
       ((c >= 0x20) && (c <= 0xD7FF)) ||
       ((c >= 0xE000) && (c <= 0xFFFD)) ||
       ((c >= 0x10000) && (c <= 0x10FFFF)))
    {
        return true;
    }
    return false;
}

Answer 2

到目前为止，所有这些答案都只替换了字符本身。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;
  }

Answer 3

如果您想以类似 XML 的形式存储带有禁止字符的文本元素，您可以使用 XPL 代替。开发工具包提供并发 XPL 到 XML 和 XML 处理 - 这意味着从 XPL 到 XML 的转换没有时间成本。或者，如果您不需要 XML（命名空间）的全部功能，您可以使用 XPL。

Web Page: HLL XPL

Answer 4

来自Best way to encode text data for XML in Java?

String xmlEscapeText(String t) {
   StringBuilder sb = new StringBuilder();
   for(int i = 0; i < t.length(); i++){
      char c = t.charAt(i);
      switch(c){
      case '<': sb.append("&lt;"); break;
      case '>': sb.append("&gt;"); break;
      case '\"': sb.append("&quot;"); break;
      case '&': sb.append("&amp;"); break;
      case '\'': sb.append("&apos;"); break;
      default:
         if(c>0x7e) {
            sb.append("&#"+((int)c)+";");
         }else
            sb.append(c);
      }
   }
   return sb.toString();
}

Answer 5

Jun 的解决方案，简化。使用StringBuffer#appendCodePoint(int)，我不需要char current 或String#charAt(int)。我可以通过检查 codePoint 是否大于 0xFFFF 来判断代理对。

（没有必要执行 i++，因为低代理不会通过过滤器。但是如果将代码重新用于不同的代码点，它会失败。我更喜欢编程而不是黑客攻击。）

StringBuilder sb = new StringBuilder();
for (int i = 0; i < text.length(); i++) {
    int codePoint = text.codePointAt(i);
    if (codePoint > 0xFFFF) {
        i++;
    }
    if ((codePoint == 0x9) || (codePoint == 0xA) || (codePoint == 0xD)
            || ((codePoint >= 0x20) && (codePoint <= 0xD7FF))
            || ((codePoint >= 0xE000) && (codePoint <= 0xFFFD))
            || ((codePoint >= 0x10000) && (codePoint <= 0x10FFFF))) {
        sb.appendCodePoint(codePoint);
    }
}

Answer 6

Java's regex supports supplementary characters，因此您可以使用两个 UTF-16 编码字符指定这些高范围。

这是删除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"
                    + "]";

大多数人会想要 XML 1.0 版本。

这是删除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"
                    + "]+";

您需要使用String.replaceAll(...) 而不是String.replace(...)。

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

Answer 7

我们应该考虑代理字符吗？否则 '(current >= 0x10000) && (current

还测试了正则表达式的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 8

来自Mark McLaren's Weblog

  /**
   * This method ensures that the output String has only
   * valid XML unicode characters as specified by the
   * XML 1.0 standard. For reference, please see
   * <a href="http://www.w3.org/TR/2000/REC-xml-20001006#NT-Char">the
   * standard</a>. This method will return an empty
   * String if the input is null or empty.
   *
   * @param in The String whose non-valid characters we want to remove.
   * @return The in String, stripped of non-valid characters.
   */
  public static String stripNonValidXMLCharacters(String in) {
      StringBuffer out = new StringBuffer(); // Used to hold the output.
      char current; // Used to reference the current character.

      if (in == null || ("".equals(in))) return ""; // vacancy test.
      for (int i = 0; i < in.length(); i++) {
          current = in.charAt(i); // NOTE: No IndexOutOfBoundsException caught here; it should not happen.
          if ((current == 0x9) ||
              (current == 0xA) ||
              (current == 0xD) ||
              ((current >= 0x20) && (current <= 0xD7FF)) ||
              ((current >= 0xE000) && (current <= 0xFFFD)) ||
              ((current >= 0x10000) && (current <= 0x10FFFF)))
              out.append(current);
      }
      return out.toString();
  }   

Answer 9

相信下面的文章可以帮到你。

http://commons.apache.org/lang/api-2.1/org/apache/commons/lang/StringEscapeUtils.html http://www.javapractices.com/topic/TopicAction.do?Id=96

很快，尝试使用 Jakarta 项目中的 StringEscapeUtils。