regex - Non-greedy regular expression match for multicharacter delimiters in awk

Question

Consider the string "AB 1 BA 2 AB 3 BA". How can I match the content between "AB" and "BA" in a non-greedy fashion (in awk)?

I have tried the following:

awk '
BEGIN {
    str="AB 1 BA 2 AB 3 BA"
    regex="AB([^B][^A]|B[^A]|[^B]A)*BA"
    if (match(str,regex))
        print substr(str,RSTART,RLENGTH)
}'

with no output. I believe the reason for no match is that there is an odd number of characters between "AB" and "BA". If I replace str with "AB 11 BA 22 AB 33 BA" the regex seems to work..

Answer 1

Merge your two negated character classes and remove the [^A] from the second alternation:

regex = "AB([^AB]|B|[^B]A)*BA"

This regex fails on the string ABABA, though - not sure if that is a problem.

Explanation:

AB       # Match AB
(        # Group 1 (could also be non-capturing)
 [^AB]   # Match any character except A or B
|        # or
 B       # Match B
|        # or
 [^B]A   # Match any character except B, then A
)*       # Repeat as needed
BA       # Match BA

Since the only way to match an A in the alternation is by matching a character except B before it, we can safely use the simple B as one of the alternatives.