I've searched for questions like this, but all the cases I found were solved in a problem-specific manner, like using !g in vi to negate the regex matches, or matching other things, without a regex negation.
Thus, I'm interested in a “pure” solution to this:
Having a set of strings I need to filter them with a regular expression matcher so that it only leaves (matches) the strings lacking a given substring. For example, filtering out "Foo" in:
Boo
Foo
Bar
FooBar
BooFooBar
Baz
Would result in:
Boo
Bar
Baz
I tried constructing it with negative look aheads/behinds (?!regex)
/(?<!regex)
, but couldn't figure it out. Is that even possible?
Try this regular expression:
^(?:(?!Foo).)*$
This will consume one character at a time and test if there is no Foo ahead. The same can be done with a negative look-behind:
^(?:.(?<!Foo))*$
But you can also do the same without look-around assertions:
^(?:[^F]*|F(?:$|[^o].|o(?:$|[^o])))*$
This matches any character except F or an F that is either not followed by a o or if followed by an o not followed by another o.
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