Regex: match pattern as long as it's not in the beginning
You can use a look behind to make sure it is not at the beginning. (?<!^)aaa
Since I came here via Google search, and was interested in a solution that is not using a lookbehind, here are my 2 cents.
The [^^]aaa
pattern matches a character other than ^
and then 3 a
s anywhere inside a string. The [^...]
is a negated character class where ^
is not considered a special character. Note the first ^
that is right after [
is special as it denotes a negation, and the second one is just a literal caret symbol.
Thus, a ^
cannot be inside [...]
to denote the start of string.
A solution is to use any negative lookaround, these two will work equally well:
(?<!^)aaa
and a lookahead:
(?!^)aaa
Why lookahead works, too? Lookarounds are zero-width assertions, and anchors are zero-width, too - they consume no text. Literally speaking, (?<!^)
checks if there is no start of string position immediately to the left of the current location, and (?!^)
checks if there is no start of string position immediately to the right of the current location. The same locations are being checked, that is why both work well.
If you don't want to use lookbehind then use this regex:
/.(aaa)/
And use matched group # 1
.