Match a string that doesn't contain another string in Bash? - regex

Question

I want to match a string that contains some text in the beginning and end but doesn't contain a different text in the middle. For example: starts with a word (\w+) and ends with another one but doesn't contain NOT in between:
some_YES_text // ok
other_COOL_string // also ok
some_NOT_string // don't want to match this
Normally, I could do that with negative lookahead:
\w+_(?!NOT)\w+_\w+
But I'm writing a script in Bash which doesn't support it. What is the easiest way to achieve the same effect?
Edit: I wasn't precise before - I still need to use regex, not just plain text matching.

Answer 1

You may match abc_NOT_def or abc_anywordhere_def and capture one of them, or part of them, and upon a match, check if that capture is not empty. Then, just implement the logic you need:
s="other_NOT_string"
rx='^([[:alnum:]_]+_(NOT)_[[:alnum:]_]+|[[:alnum:]_]+_[[:alnum:]_]+_[[:alnum:]_]+)$'
if [[ "$s" =~ $rx ]]; then
if [ -z ${BASH_REMATCH[2]} ]; then
echo "MATCH: ${BASH_REMATCH[0]}"
else
echo "No match"
fi;
else
echo "No match"
fi;
Details
^ - start of string
( - Start of Group 1:
[[:alnum:]_]+_ - 1+ word chars (POSIX ERE \w equivalent) and a _
(NOT) - Group 2: NOT
_[[:alnum:]_]+ - _ and 1+ word chars
| - or
[[:alnum:]_]+_[[:alnum:]_]+_[[:alnum:]_]+ - 1+ word chars, _, 1+ word chars, _ and again 1+ word chars
) - end of Group 1.
$ - end of string
With [ -z ${BASH_REMATCH[2]} ] condition, we check if NOT was matched. If it was, there is no valid match, else, there is one.