在 Java 中清理 HTML Id 属性的函数

Question

我已经编写了下一个函数。但肯定有人有更优雅的方式来执行这项任务。

/**
 * 
 * HTML 4 Specification
 * ID and NAME tokens must begin with a letter ([A-Za-z]) and may be followed by any number 
 * of letters, digits ([0-9]), hyphens ("-"), underscores ("_"), colons (":"), and periods (".").
 * @param s
 * @return
 */
public static String sanitizeHTMLIdAttribute(String s) {
  String sanitize = "";
  if(s!=null) {
    for(int i = 0; i < s.length(); i++) {
      if(s.charAt(i) == '-' || s.charAt(i) == '_' || s.charAt(i) == ':' || 
         s.charAt(i) == '.' || s.charAt(i) == '0' || s.charAt(i) == '1' || 
         s.charAt(i) == '2' || s.charAt(i) == '3' || s.charAt(i) == '4' || 
         s.charAt(i) == '5' || s.charAt(i) == '6' || s.charAt(i) == '7' || 
         s.charAt(i) == '8' || s.charAt(i) == '9' || 
         s.charAt(i) == 'a' || s.charAt(i) == 'b' || s.charAt(i) == 'c' || 
         s.charAt(i) == 'd' || s.charAt(i) == 'e' || s.charAt(i) == 'f' || 
         s.charAt(i) == 'g' || s.charAt(i) == 'h' || s.charAt(i) == 'i' || 
         s.charAt(i) == 'j' || s.charAt(i) == 'k' || s.charAt(i) == 'l' ||  
         s.charAt(i) == 'm' || s.charAt(i) == 'n' || s.charAt(i) == 'o' ||
         s.charAt(i) == 'p' || s.charAt(i) == 'q' || s.charAt(i) == 'r' || 
         s.charAt(i) == 's' || s.charAt(i) == 't' || s.charAt(i) == 'u' || 
         s.charAt(i) == 'w' || s.charAt(i) == 'x' || s.charAt(i) == 'y' ||
     s.charAt(i) == 'z' || 
         s.charAt(i) == 'A' || s.charAt(i) == 'B' || s.charAt(i) == 'C' || 
         s.charAt(i) == 'D' || s.charAt(i) == 'E' || s.charAt(i) == 'F' || 
         s.charAt(i) == 'G' || s.charAt(i) == 'H' || s.charAt(i) == 'I' ||
         s.charAt(i) == 'J' || s.charAt(i) == 'K' || s.charAt(i) == 'L' ||
         s.charAt(i) == 'M' || s.charAt(i) == 'N' || s.charAt(i) == 'O' ||
         s.charAt(i) == 'P' || s.charAt(i) == 'Q' || s.charAt(i) == 'R' || 
         s.charAt(i) == 'S' || s.charAt(i) == 'T' || s.charAt(i) == 'U' || 
         s.charAt(i) == 'W' ||  s.charAt(i) == 'X' || s.charAt(i) == 'Y' || 
         s.charAt(i) == 'Z') {
              sanitize += s.charAt(i);
      }
    }
    if(sanitize.length()>0) {
      while(sanitize.charAt(0) == '0' || sanitize.charAt(0) == '1' || 
            sanitize.charAt(0) == '2' || sanitize.charAt(0) == '3' || 
            sanitize.charAt(0) == '4' || sanitize.charAt(0) == '5' || 
            sanitize.charAt(0) == '6' || sanitize.charAt(0) == '7' || 
            sanitize.charAt(0) == '8' || sanitize.charAt(0) == '9') {
                sanitize = sanitize.substring(1, sanitize.length()); 
      }
    }
    return sanitize;
  } 
  return null;      
}

Answer 1

我会这样做：

/**
 * 
 * HTML 4 Specification ID and NAME tokens must begin with a letter
 * ([A-Za-z]) and may be followed by any number of letters, digits ([0-9]),
 * hyphens ("-"), underscores ("_"), colons (":"), and periods (".").
 * 
 * @param s
 * @return
 */
public static String sanitizeHTMLIdAttribute(String s) {
    if (s == null) return null;
    StringBuilder sb = new StringBuilder();
    int firstLegal = 0;
    while (firstLegal < s.length() && !isAZ(s.charAt(firstLegal)))
        ++firstLegal;
    for (int i = firstLegal; i < s.length(); ++i){
        final char ch = s.charAt(i);
        if (isOkIdInnerChar(ch)) sb.append(ch);
    }
    return sb.length() == s.length()? s : sb.toString();
}

private static boolean isOkIdInnerChar(char ch) {
    return isAZ(ch) || isNum(ch) || isSpecial(ch);
}

private static boolean isSpecial(char ch) {
    switch (ch) {
    case '-': case '_':
    case ':': case '.':
        return true;
    default:
        return false;
    }
}

private static boolean isAZ(char ch) {
    return ('A' <= ch && ch <= 'Z') || ('a' <= ch && ch <= 'z'); 
}

private static boolean isNum(char ch) {
    return '0' <= ch && ch <= '9';
}

...除了如果s == null 和IllegalArgumentException 如果s 不包含合法字符，我可能更愿意抛出NullPointerException，但这是一个偏好问题，当然。一些额外的功能：

• 如果 s 是一个有效的 ID，它会按原样返回以节省空间（更少的字符串实例浮动）和时间（String 构造很昂贵——是的，我知道分配很便宜，但还有更多就在那里比分配）。
• 我不使用Character.isDigit，因为它对所有 Unicode 数字返回true，包括像“٣”这样的东西
• 我不使用Character.isLetter，因为它对所有 Unicode 字母返回true，包括像“å”这样的东西

Answer 2

使用Character.isLetterOrDigit(char) 可以显着缩短代码；例如

for(int i = 0; i < s.length(); i++) {
  char c = s.charAt(i);

  if (Character.isLetterOrDigit(c) || c == '.' || etc ...) {

  }
}

您可以结合将允许的标点字符存储在Set 中来执行此操作；例如

private static final Set<Character> ALLOWED =
  new HashSet<Character>(Arrays.asList('.', '-', '_', ':'));

for(int i = 0; i < s.length(); i++) {
  char c = s.charAt(i);

  if (ALLOWED.contains(c)) {

  }
}

Answer 3

您可能想要检查表达式（正则表达式不是我的强项），但这应该会删除无效字符：

  private static Pattern INVALID_LEADING = Pattern.compile("^[^a-zA-Z]+");
  private static Pattern INVALID = Pattern
      .compile("[^\\w\\u002e\\u003a\\u002d\\u005f]+");

  private static String sanitize(String id) {
    Matcher matcher = INVALID_LEADING.matcher(id);
    if (matcher.find()) {
      id = matcher.replaceFirst("");
    }
    Matcher invalid = INVALID.matcher(id);
    if (invalid.find()) {
      id = invalid.replaceAll("");
    }
    return id;
  }

如果您对正则表达式不满意，请注意您的许多字符都在连续范围内，因此可以使用以下方法检测到：

  private static boolean isLatinDigit(char ch) {
    return ch >= '0' && ch <= '9';
  }