Matching a whole word with leading or trailing spe

Question

I can replace dollar signs by using Matcher.quoteReplacement. I can replace words by adding boundary characters:

from = "\\b" + from + "\\b"; 
outString = line.replaceAll(from, to);

But I can't seem to combine them to replace words with dollar signs.

Here's an example. I am trying to replace "$temp4" (NOT $temp40) with "register1".

        String line = "add, $temp4, $temp40, 42";
        String to = "register1";
        String from = "$temp4";
        String outString;


        from = Matcher.quoteReplacement(from);
        from = "\\b" + from + "\\b";  //do whole word replacement

        outString = line.replaceAll(from, to);
        System.out.println(outString);

Outputs

"add, $temp4, $temp40, 42"

How do I get it to replace $temp4 and only $temp4?

Answer 1

Use unambiguous word boundaries, (?<!\w) and (?!\w), instead of \b that are context dependent:

from = "(?<!\\w)" + Pattern.quote(from) + "(?!\\w)";

See the regex demo.

The (?<!\w) is a negative lookbehind that fails the match if there is a non-word char immediately to the left of the current location and (?!\w) is a negative lookahead that fails the match if there is a non-word char immediately to the right of the current location. The Pattern.quote(from) is necessary to escape any special chars in the from variable.

See the Java demo:

String line = "add, $temp4, $temp40, 42";
String to = "register1";
String from = "$temp4";
String outString;

from = "(?<!\\w)" + Pattern.quote(from) + "(?!\\w)";

outString = line.replaceAll(from, to);
System.out.println(outString);
// => add, register1, $temp40, 42

Answer 2

$ has special meaning in regex (it means “end of input”). To remove any special meaning from characters in your target, wrap it in regex quote/unquote expressions \Q...\E. Also, because $ is not ”word” character, the word boundary won’t wiork, so use look arounds instead:

line = line.replaceAll("(?<!\\S)\\Q" + from + "\\E(?![^ ,])", to);

Answer 3

Matcher.quoteReplacement() is for the replacement string (to), not the regex (from). To include a string literal in the regex, use Pattern.quote():

from = Pattern.quote(from);

Answer 4

Normally, Pattern.quote is the way to go to escape characters that may be specially interpreted by the regex engine.

However, the regular expression is still incorrect, because there is no word boundary before the $ in line; space and $ are both non-word characters. You need to place the word boundary after the $ character. There is no need for Pattern.quote here, because you're escaping things yourself.

String from = "\\$\\btemp4\\b";

Or more simply, because you know there is a word boundary between $ and temp4 already:

String from = "\\$temp4\\b";

The from variable can be constructed from the expression to replace. If from has "$temp4", then you can escape the dollar sign and add a word boundary.

from = "\\" + from + "\\b";

Output:

add, register1, $temp40, 42