Java escaping chars that already exist in the string

I need to escape chars that exist within a string but the escape chars are part of the escape code ie.

    Map<String, String> escapeChars = new HashMap<String, String>();
        escapeChars.put("^", "^94;");
        escapeChars.put("%", "^37;");
        escapeChars.put("'", "^39;");

    public Stri开发者_JAVA技巧ng findEscapeChars(String x) {
    for(Map.Entry<String, String> entry : escapeChars.entrySet()) {
        if(x.contains(entry.getKey()) ) {   
            x = x.replaceAll(entry.getKey(), entry.getValue());
        }
    }
    return x;
    }

    //result should match
    assertEquals("^37;", findEscapeChars("%")); //example of the caret symbol escaping
    assertEquals("^37; 1 2 ^39; ^94; 4 ^37;", findEscapeChars("% 1 2 ' ^ 4 %"));

It really needs to done in one loop and only one if condition. The problem is that the hash map doesn't traverse in a certain order therfore the caret symbol could try escape itself in the string. It seems to be tricky without numerous loops and if conditions, a simple solution would be ideal.

Thanks D

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I'm going to assume you only need to escape a single character at a time. Then you can do:

public String findEscapeChars(String original) {
    StringBuilder builder = new StringBuilder();
    for (int i = 0; i < original.length(); i++)
    {
        char c = original.charAt(i);
        String escaped = escapeMap.get(c);
        if (escaped != null)
        {
            builder.append(escaped);
        }
        else
        {
            builder.append(c);
        }
    }
    return builder.toString();
}

Note that unless you really need to have the dynamic nature of a map, a simple switch would probably be simpler and more efficient.

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First, if you need order in map use LinkedHashMap instead. Second, you even can solve everything without maps at all but using regular experession with lookahead. See http://download.oracle.com/javase/6/docs/api/java/util/regex/Pattern.html