将字符串中的所有特殊字符高效编码为实体

Question

I have a string like this "abcd !@&$%^^&*()<>!/". I have list of all the entity codes for characters in a separate string ie only encode those characters which are in another string "!=&4....^=9...". I want to convert all of special characters into their entities except alphanumeric by regex as using loop on characters on by one is too slow.

eg it should show "abc &#4..;&#4.." in other convert words all the special characters on keyboard.

Is there an efficient regex I can write ? I have tried this with loops but it is too slow to look at each character one by one and maintain a list of all special characters entities in other string

There are libraries but they do not convert all of the characters.

The code I wrote

// String to be encoded

String sDecoded = "abcd !@#$%^&*();'m,";
// Special character entity list to put instead to special character. It is     tokenized on cross and divide symbol as it cannot be entered by user on keyboard

String specialCharacters = "&÷$amp;×–÷–"


// Check the input
if (sDecoded == null || sDecoded.trim ().length () == 0)
  return (sDecoded);

// Use StringTokenizer which is faster than split method
StringTokenizer st = new StringTokenizer(specialCharacters, "×");
String[] reg = null;
String[] charactersArray = sDecoded.split("");
String sEncoded = "";

// now loop on it and in each iteration, we will be getting a decodedCharacter:EncodedEntity pair 


for(int i = 0; i < charactersArray.length; i++)
{   
    st = new StringTokenizer(specialCharacters, "×");


    while(st.hasMoreElements())
    {
        reg = st.nextElement().toString().split("÷");

         // This is an error, the character should not be blank ever because it will be character that we will encode
         if(StringUtils.isBlank(reg[0]))
            return sDecoded;

        String c = charactersArray[i];


        if(c.equalsIgnoreCase(reg[0]))
        {
            sEncoded = sEncoded + c.replace(reg[0], reg[1]);
            break;
        }

        if(st.countTokens() == 0)
            sEncoded = sEncoded + c.toString();

 }

}

    return (sEncoded);

Answer 1

I don't know what definition of "efficient" you are using, but there's the "don't reinvent the wheel" efficiency of using a simple call to Apache commons StringEscapeUtils utility class:

String encoded = StringEscapeUtils.escapeXml11(str);

or

String encoded = StringEscapeUtils.escapeHtml4(str);

and a variety of other similar methods, depending on which exact encoding you want.

Answer 2

Your approach is quite slow and inefficient. Maybe it looks elegant nowadays to use regex like a silver bullet for everything, but it is definitely not for this task. I see you are also using tokenizer which is also slow.Also loop inside a loop will degrade performance.

I would recomment using an iterative way with string builder which will produce blazing fast results, you will try for yourself. For each special character make an 'if' statement. Even if it looks too much code it will be very fast. Test yourself. Try this :

class Scratch {

    public static void main(String[] args) {
        System.out.println(escapeSpecials("abc &"));
    }

    public static String escapeSpecials(String origin) {
        StringBuilder result = new StringBuilder();
        char[] chars = origin.toCharArray();
        for (char c : chars) {
            if (c == '&') {
                result.append("&");
            } else if (c == '\u2013') {
                result.append("–");
            } else {
                // not a special character
                result.append(c);
            }
        }
        return result.toString();
    }
}