Was actually glossing over a question and found myself struggling to perform something really simple.
If a string contains % I want to use a particular regex, else I want to use a different one.
I tried the following: https://regex101.com/r/UvFZpo/1/
Regex: (%)(?(1)[^$]+|[^%]+).
Test string: abc%
But I'm not getting the expected results.
I was expecting to see abc% matched as it contains %.
If the string was, abc$, I'd expect it to use the second expression.
Where am I going wrong?
Regex parses strings from left to right, position by position.
Once your pattern matches &, its index is at the end of string, hence, it fails since there are no more chars to be matched by the subsequent [^$]+ pattern.
You can use a mere alternation here:
^(?:([^$]*%[^$]*)|([^%]+))$
See the regex demo
If the string contains %, the Group 1 will be populated, else, Group 2 will.
Details
^ - start of string(?:([^$]*%[^$]*)|([^%]+)) - either of the two alternatives:
([^$]*%[^$]*) - Group 1: any 0+ chars other than $, as many as possible, % any 0+ chars other than $, as many as possible,| - or([^%]+) - any 1+ chars other than %, as many as possible$ - end of string.