Question

I am having problems with the nested ‘+’/’-‘ lookahead/lookbehind in regex.

Let’s say that I want to change the '*' in a string with '%' and let’s say that '\' escapes the next character. (Turning a regex to sql like command ^^).

So the string

• '*test*' should be changed to '%test%',
• '\\*test\\*' -> '\\%test\\%', but
• '\*test\*' and '\\\*test\\\*' should stay the same.

I tried:

(?<!\\)(?=\\\\)*\*      but this doesn't work
(?<!\\)((?=\\\\)*\*)    ...
(?<!\\(?=\\\\)*)\*      ...
(?=(?<!\\)(?=\\\\)*)\*  ...

What is the correct regex that will match the ‘*’s in examples given above?

What is the difference between (?<!\\(?=\\\\)*)\* and (?=(?<!\\)(?=\\\\)*)\* or if these are essentially wrong the difference between regex that have such a visual construction?

Answer 1

To find an unescaped character, you would look for a character that is preceded by an even number of (or zero) escape characters. This is relatively straight-forward.

(?<=(?<!\\)(?:\\\\)*)\*        # this is explained in Tim Pietzcker' answer

Unfortunately, many regex engines do not support variable-length look-behind, so we have to substitute with look-ahead:

(?=(?<!\\)(?:\\\\)*\*)(\\*)\*  # also look at ridgerunner's improved version

Replace this with the contents of group 1 and a % sign.

Explanation

(?=           # start look-ahead
  (?<!\\)     #   a position not preceded by a backslash (via look-behind)
  (?:\\\\)*   #   an even number of backslashes (don't capture them)
  \*          #   a star
)             # end look-ahead. If found,
(             # start group 1
  \\*         #   match any number of backslashes in front of the star
)             # end group 1
\*            # match the star itself

The look-ahead makes sure only even numbers of backslashes are taken into account. Anyway, there is no way around matching them into a group, since the look-ahead does not advance the position in the string.