How to replace tokens in a string without StringTokenizer

Question

Given a string like so:

 Hello {FIRST_NAME}, this is a personalized message for you.

Where FIRST_NAME is an arbitrary token (a key in a map passed to the method), to write a routine which would turn that string into:

Hello Jim, this is a personalized message for you.

given a map with an entry FIRST_NAME -> Jim.

It would seem that StringTokenizer is the most straight forward approach, but the Javadocs really say you should prefer to use the regex aproach. How would you do that in a regex based solution?

Answer 1

The docs mean that you should prefer writing a regex-based tokenizer, IIRC. What might work better for you is a standard regex search-replace.

Answer 2

String.replaceAll("{FIRST_NAME}", actualName);

Check out the javadocs for it here.

Answer 3

Well, I would rather use String.format(), or better MessageFormat.

Answer 4

Try this:

Note: The author's final solution builds upon this sample and is much more concise.

public class TokenReplacer {

    private Pattern tokenPattern;

    public TokenReplacer() {
        tokenPattern = Pattern.compile("\\{([^}]+)\\}");
    }

    public String replaceTokens(String text, Map<String, String> valuesByKey) {
        StringBuilder output = new StringBuilder();
        Matcher tokenMatcher = tokenPattern.matcher(text);

        int cursor = 0;
        while (tokenMatcher.find()) {
            // A token is defined as a sequence of the format "{...}".
            // A key is defined as the content between the brackets.
            int tokenStart = tokenMatcher.start();
            int tokenEnd = tokenMatcher.end();
            int keyStart = tokenMatcher.start(1);
            int keyEnd = tokenMatcher.end(1);

            output.append(text.substring(cursor, tokenStart));

            String token = text.substring(tokenStart, tokenEnd);
            String key = text.substring(keyStart, keyEnd);

            if (valuesByKey.containsKey(key)) {
                String value = valuesByKey.get(key);
                output.append(value);
            } else {
                output.append(token);
            }

            cursor = tokenEnd;
        }
        output.append(text.substring(cursor));

        return output.toString();
    }

}

Answer 5

The most straight forward would seem to be something along the lines of this:

public static void main(String[] args) {
 String tokenString = "Hello {FIRST_NAME}, this is a personalized message for you.";
 Map<String, String> tokenMap = new HashMap<String, String>();
 tokenMap.put("{FIRST_NAME}", "Jim");
 String transformedString = tokenString;
 for (String token : tokenMap.keySet()) {
  transformedString = transformedString.replace(token, tokenMap.get(token));
 }
 System.out.println("New String: " + transformedString);
}

It loops through all your tokens and replaces every token with what you need, and uses the standard String method for replacement, thus skipping the whole RegEx frustrations.

Answer 6

With import java.util.regex.*:

Pattern p = Pattern.compile("{([^{}]*)}");
Matcher m = p.matcher(line);  // line being "Hello, {FIRST_NAME}..."
while (m.find) {
  String key = m.group(1);
  if (map.containsKey(key)) {
    String value= map.get(key);
    m.replaceFirst(value);
  }
}

So, the regex is recommended because it can easily identify the places that require substitution in the string, as well as extracting the name of the key for substitution. It's much more efficient than breaking the whole string.

You'll probably want to loop with the Matcher line inside and the Pattern line outside, so you can replace all lines. The pattern never needs to be recompiled, and it's more efficient to avoid doing so unnecessarily.

Answer 7

Depending on how ridiculously complex your string is, you could try using a more serious string templating language, like Velocity. In Velocity's case, you'd do something like this:

Velocity.init();
VelocityContext context = new VelocityContext();
context.put( "name", "Bob" );
StringWriter output = new StringWriter();
Velocity.evaluate( context, output, "", 
      "Hello, #name, this is a personalized message for you.");
System.out.println(output.toString());

But that is likely overkill if you only want to replace one or two values.

Answer 8

import java.util.HashMap;

public class ReplaceTest {

  public static void main(String[] args) {
    HashMap<String, String> map = new HashMap<String, String>();

    map.put("FIRST_NAME", "Jim");
    map.put("LAST_NAME",  "Johnson");
    map.put("PHONE",      "410-555-1212");

    String s = "Hello {FIRST_NAME} {LAST_NAME}, this is a personalized message for you.";

    for (String key : map.keySet()) {
      s = s.replaceAll("\\{" + key + "\\}", map.get(key));
    }

    System.out.println(s);
  }

}

Answer 9

Thanks everyone for the answers!

Gizmo's answer was definitely out of the box, and a great solution, but unfortunately not appropriate as the format can't be limited to what the Formatter class does in this case.

Adam Paynter really got to the heart of the matter, with the right pattern.

Peter Nix and Sean Bright had a great workaround to avoid all of the complexities of the regex, but I needed to raise some errors if there were bad tokens, which that didn't do.

But in terms of both doing a regex and a reasonable replace loop, this is the answer I came up with (with a little help from Google and the existing answer, including Sean Bright's comment about how to use group(1) vs group()):

private static Pattern tokenPattern = Pattern.compile("\\{([^}]*)\\}");

public static String process(String template, Map<String, Object> params) {
    StringBuffer sb = new StringBuffer();
    Matcher myMatcher = tokenPattern.matcher(template);
    while (myMatcher.find()) {
        String field = myMatcher.group(1);
        myMatcher.appendReplacement(sb, "");
        sb.append(doParameter(field, params));
   }
    myMatcher.appendTail(sb);
    return sb.toString();
}

Where doParameter gets the value out of the map and converts it to a string and throws an exception if it isn't there.

Note also I changed the pattern to find empty braces (i.e. {}), as that is an error condition explicitly checked for.

EDIT: Note that appendReplacement is not agnostic about the content of the string. Per the javadocs, it recognizes $ and backslash as a special character, so I added some escaping to handle that to the sample above. Not done in the most performance conscious way, but in my case it isn't a big enough deal to be worth attempting to micro-optimize the string creations.

Thanks to the comment from Alan M, this can be made even simpler to avoid the special character issues of appendReplacement.

Answer 10

Generally we'd use MessageFormat in a case like this, coupled with loading the actual message text from a ResourceBundle. This gives you the added benefit of being G10N friendly.