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。 我想用string.replace方法使用正则表达式。

像

line.replace(regExp,"");

什么是使用权的正则表达式？

无效的XML字符是一切，是不是这样的：

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

感谢。

Answer 1

Java's regex supports supplementary characters，所以你可以用两个UTF-16编码字符来指定那些高范围。

这里是用于去除在XML 1.0都是非法字符模式：

// 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" 
        + "]"; 

大多数人都希望XML 1.0版。

这里是用于去除在XML 1.1都是非法字符模式：

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

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

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

Answer 2

我相信以下文章可能对您有所帮助。

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

不久，尝试使用StringEscapeUtils从Jakarta项目。

Answer 3

从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 4

我们应该考虑替代字符？否则'（当前> = 0x10000）& &（当前< = 0x10FFFF）'永远不会成立。

还测试了正则表达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 5

君的解决方案，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来告诉代理对。 （这是没有必要做的i ++，因为低代理不会通过过滤器，但然后一个代码将重新使用不同的代码点，它会失败。我更喜欢编程黑客。）

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

从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 7

如果你想存储在类似XML的形式禁字符的文本元素，你可以使用XPL代替。 dev-kit为XML和XML处理提供了并发XPL，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的全部功能（名称空间），则可以使用XPL。

Web Page: HLL XPL

Answer 8

所有这些问题的答案至今只替换字符本身。但是有时候XML文档会有无效的XML实体序列导致错误。例如，如果在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 9

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; 
}